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22373
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\section{Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3}
Tags: Summation to n of Summation to Index, Summation of Products of n Numbers taken m at a time with Repetitions, Summations
\begin{theorem}
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Then:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$
\end{theorem}
\begin{proof}
Let:
{{begin-eqn}}
{{eqn | n = a
| l = A
| o = :=
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \sum_{k \mathop = 0}^j a_i a_j a_k
| c =
}}
{{eqn | r = \sum_{0 \mathop \le i \mathop \le j \mathop \le k \mathop \le n} a_i a_j a_k
| c =
}}
{{eqn | n = b
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n \sum_{k \mathop = j}^n a_i a_j a_k
| c =
}}
{{eqn | n = c
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n \sum_{k \mathop = i}^j a_i a_j a_k
| c =
}}
{{eqn | n = d
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \sum_{k \mathop = i}^j a_i a_j a_k
| c =
}}
{{eqn | n = e
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \sum_{k \mathop = i}^n a_i a_j a_k
| c =
}}
{{end-eqn}}
Also, let:
{{begin-eqn}}
{{eqn | l = S_1
| o = :=
| r = \sum_{i \mathop = 0}^n a_i
| c =
}}
{{eqn | l = S_2
| o = :=
| r = \sum_{i \mathop = 0}^n {a_i}^2
| c =
}}
{{eqn | l = S_3
| o = :=
| r = \sum_{i \mathop = 0}^n {a_i}^3
| c =
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = 2 A
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n \sum_{k \mathop = j}^n a_i a_j a_k + \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n \sum_{k \mathop = i}^j a_i a_j a_k
| c = $(b) + (c)$
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n \left({\sum_{k \mathop = j}^n a_i a_j a_k + \sum_{k \mathop = i}^j a_i a_j a_k}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n \left({\sum_{k \mathop = i}^n a_i a_j a_k + a_i a_j a_j}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n a_i \left({\sum_{j \mathop = i}^n a_j}\right)^2 + \sum_{i \mathop = 0}^n a_i \left({\sum_{j \mathop = i}^n {a_j}^2}\right)
| c =
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = A_1
| o = :=
| r = \sum_{i \mathop = 0}^n a_i \left({\sum_{j \mathop = i}^n a_j}\right)^2
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n a_i \sum_{j \mathop = i}^n a_j \sum_{k \mathop = i}^n a_k
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n \sum_{k \mathop = i}^n a_i a_j a_k
| c =
}}
{{eqn | l = A_3
| o = :=
| r = \sum_{i \mathop = 0}^n a_i \left({\sum_{j \mathop = i}^n {a_j}^2}\right)
| c =
}}
{{end-eqn}}
as calculated above.
Thus:
:$(1): \quad 2 A = A_1 + A_3$
Similarly:
{{begin-eqn}}
{{eqn | l = 2 A
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \sum_{k \mathop = 0}^j a_i a_j a_k + \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \sum_{k \mathop = j}^i a_i a_j a_k
| c = $(a) + (d)$
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \left({\sum_{k \mathop = 0}^j a_i a_j a_k + \sum_{k \mathop = j}^i a_i a_j a_k}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \left({\sum_{k \mathop = 0}^i a_i a_j a_k + a_i a_j a_j}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n a_i \left({\sum_{j \mathop = 0}^i a_j}\right)^2 + \sum_{i \mathop = 0}^n a_i \left({\sum_{j \mathop = 0}^i {a_j}^2}\right)
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = A_1 + A
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n \sum_{k \mathop = i}^n a_i a_j a_k + \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \sum_{k \mathop = i}^n a_i a_j a_k
| c = using $(e)$
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{k \mathop = i}^n \left({\sum_{j \mathop = i}^n a_i a_j a_k + \sum_{j \mathop = 0}^i a_i a_j a_k}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{k \mathop = i}^n \left({\sum_{j \mathop = 0}^n a_i a_j a_k + {a_i}^2 a_k}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n a_i \sum_{j \mathop = 0}^n \sum_{k \mathop = j}^n a_j a_k + \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n {a_i}^2 a_j
| c =
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = A_2
| o = :=
| r = \sum_{i \mathop = 0}^n a_i \sum_{j \mathop = 0}^n \sum_{k \mathop = j}^n a_j a_k
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^n \sum_{k \mathop = j}^n a_i a_j a_k
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^n \sum_{k \mathop = 0}^j a_i a_j a_k
| c =
}}
{{eqn | l = A_4
| o = :=
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n {a_i}^2 a_j
| c =
}}
{{end-eqn}}
as calculated above.
Thus:
:$(2): \quad A_1 + A = A_2 + A_4$
Then:
{{begin-eqn}}
{{eqn | l = 2 A_2
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^n \sum_{k \mathop = j}^n a_i a_j a_k + \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^n \sum_{k \mathop = 0}^j a_i a_j a_k
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^n \left({\sum_{k \mathop = j}^n a_i a_j a_k + \sum_{k \mathop = 0}^j a_i a_j a_k}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^n \left({\sum_{k \mathop = 0}^n a_i a_j a_k + a_i {a_j}^2}\right)
| c =
}}
{{eqn | r = \left({\sum_{i \mathop = 0}^n a_i}\right)^3 + \sum_{i \mathop = 0}^n a_i \sum_{i \mathop = 0}^n {a_i}^2
| c =
}}
{{eqn | n = 3
| r = {S_1}^3 + S_1 S_2
| c =
}}
{{end-eqn}}
Now we have that:
{{begin-eqn}}
{{eqn | l = A_3
| r = \sum_{i \mathop = 0}^n a_i \left({\sum_{j \mathop = i}^n {a_j}^2}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n {a_i}^2 \left({\sum_{j \mathop = 0}^i a_j}\right)
| c =
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = A_3 + A_4
| r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i {a_i}^2 a_j + \sum_{i \mathop = 0}^n \sum_{j \mathop = i}^n {a_i}^2 a_j
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \left({\sum_{j \mathop = 0}^n {a_i}^2 a_j + {a_i}^3}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n {a_i}^2 \sum_{i \mathop = 0}^n a_i + \sum_{i \mathop = 0}^n {a_i}^3
| c =
}}
{{eqn | n = 4
| r = S_2 S_1 + S_3
| c =
}}
{{end-eqn}}
Finally:
{{begin-eqn}}
{{eqn | l = 2 A
| r = A_1 + A_3
| c = from $(1)$
}}
{{eqn | l = A + A_1
| r = A_2 + A_4
| c = from $(2)$
}}
{{eqn | l = 2 A_2
| r = {S_1}^3 + S_1 S_2
| c = from $(3)$
}}
{{eqn | l = A_3 + A_4
| r = S_1 S_2 + S_3
| c = from $(4)$
}}
{{eqn | ll= \leadsto
| l = 3 A + A_1
| r = A_1 + A_2 + A_3 + A_4
| c =
}}
{{eqn | ll= \leadsto
| l = 3 A
| r = A_2 + A_3 + A_4
| c =
}}
{{eqn | ll= \leadsto
| l = 6 A
| r = 2 A_2 + 2 \left({A_3 + A_4}\right)
| c =
}}
{{eqn | r = {S_1}^3 + S_1 S_2 + 2 S_1 S_2 + 2 S_3
| c =
}}
{{eqn | r = {S_1}^3 + 3 S_1 S_2 + 2 S_3
| c =
}}
{{eqn | ll= \leadsto
| l = A
| r = \frac { {S_1}^3} 6 + \frac {S_1 S_2} 2 + \frac {S_3} 3
| c =
}}
{{end-eqn}}
{{qed}}
{{Proofread}}
\end{proof}
|
22374
|
\section{Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 4}
Tags: Summation to n of Summation to Index, Summation of Products of n Numbers taken m at a time with Repetitions
\begin{theorem}
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Then:
:$\ds \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4} = \dfrac { {S_1}^4} {24} + \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 + \dfrac {S_4} 4$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$.
\end{theorem}
\begin{proof}
From Summation of Products of n Numbers taken m at a time with Repetitions:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$
where:
:$S_j = \ds \sum_{k \mathop = a}^b {x_k}^j$ for $j \in \Z_{\ge 0}$.
Setting $m = 4$:
{{begin-eqn}}
{{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4}
| r = \sum_{\substack {k_1, k_2, k_3, k_4 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + 3 k_3 \mathop + 4 k_4 \mathop = 4} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \dfrac { {S_4}^{k_4} } {4^{k_4} k_4 !}
| c =
}}
{{end-eqn}}
We need to find all sets of $k_1, k_2, k_3, k_4 \in \Z_{\ge 0}$ such that:
:$k_1 + 2 k_2 + 3 k_3 + 4 k_4 = 4$
Thus $\tuple {k_1, k_2, k_3, k_4}$ can be:
:$\tuple {4, 0, 0, 0}$
:$\tuple {2, 1, 0, 0}$
:$\tuple {0, 2, 0, 0}$
:$\tuple {1, 0, 1, 0}$
:$\tuple {0, 0, 0, 1}$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}
| r = \dfrac { {S_1}^4 {S_2}^0 {S_3}^0 {S_4}^0} {\paren {1^4 \times 4!} \paren {2^0 \times 0!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }
| c =
}}
{{eqn | o =
| ro= +
| r = \dfrac { {S_1}^2 {S_2}^1 {S_3}^0 {S_4}^0} {\paren {1^2 \times 2!} \paren {2^1 \times 1!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }
| c =
}}
{{eqn | o =
| ro= +
| r = \dfrac { {S_1}^0 {S_2}^2 {S_3}^0 {S_4}^0} {\paren {1^0 \times 0!} \paren {2^2 \times 2!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }
| c =
}}
{{eqn | o =
| ro= +
| r = \dfrac { {S_1}^1 {S_2}^0 {S_3}^1 {S_4}^0} {\paren {1^1 \times 1!} \paren {2^0 \times 0!} \paren {3^1 \times 1!} \paren {4^0 \times 0!} }
| c =
}}
{{eqn | o =
| ro= +
| r = \dfrac { {S_1}^0 {S_2}^0 {S_3}^0 {S_4}^1} {\paren {1^0 \times 0!} \paren {2^0 \times 0!} \paren {3^0 \times 0!} \paren {4^1 \times 1!} }
| c =
}}
{{eqn | r = \dfrac { {S_1}^4} {24} + \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 + \dfrac {S_4} 4
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22375
|
\section{Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula}
Tags: Summation of Products of n Numbers taken m at a time with Repetitions
\begin{theorem}
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
{{begin-eqn}}
{{eqn | l = h_m
| r = \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} \paren {\prod_{k \mathop = 1}^m x_{j_k} }
| c =
}}
{{eqn | r = \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m}
| c =
}}
{{end-eqn}}
That is, $h_m$ is the product of all $m$-tuples of elements of $U$ taken $m$ at a time.
For $r \in \Z_{> 0}$, let:
:$S_r = \ds \sum_{j \mathop = a}^b {x_j}^r$
Let $S_m$ be expressed in the form:
:$S_m = \ds \sum_{k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} A_m {h_1}^{k_1} {h_2}^{k_2} \cdots {h_m}^{k_m}$
for $k_1, k_2, \ldots, k_m \ge 0$.
Then :
:$A_m = \paren {-1}^{k_1 + k_2 + \cdots + k_m - 1} \dfrac {m \paren {k_1 + k_2 + \cdots + k_m - 1}! } {k_1! \, k_2! \, \cdots k_m!}$
\end{theorem}
\begin{proof}
Let $\map G z$ be the generating function for the sequence $\sequence {h_m}$.
{{begin-eqn}}
{{eqn | l = \sum_{m \mathop \ge 1} \dfrac {S_m z^m} m
| r = \map \ln {\map G z}
| c = Summation of Products of n Numbers taken m at a time with Repetitions: Lemma 2
}}
{{eqn | r = \map \ln {1 + h_1 z + h_2 z^2 + \cdots}
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 1} \dfrac {\paren {-1}^{k - 1} } k \paren {h_1 z + h_2 z^2 + \cdots}^k
| c = Power Series Expansion for Logarithm of 1 + x
}}
{{eqn | ll= \leadsto
| l = S_m
| r = m \sum_{k \mathop \ge 1} \dfrac {\paren {-1}^{k - 1} } k \paren {\sum_{j \mathop = 1} h_j z^j}^k
| c =
}}
{{end-eqn}}
{{finish|Not sure where this is going}}
\end{proof}
|
22376
|
\section{Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1}
Tags: Summation of Products of n Numbers taken m at a time with Repetitions
\begin{theorem}
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
{{begin-eqn}}
{{eqn | l = h_m
| r = \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} \paren {\prod_{k \mathop = 1}^m x_{j_k} }
| c =
}}
{{eqn | r = \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m}
| c =
}}
{{end-eqn}}
That is, $h_m$ is the product of all $m$-tuples of elements of $U$ taken $m$ at a time.
Let $\map G z$ be the generating function for the sequence $\sequence {h_m}$.
Then:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \prod_{k \mathop = a}^b \dfrac 1 {1 - x_k z}
| c =
}}
{{eqn | r = \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \cdots \paren {1 - x_b z} }
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
For each $k \in \set {a, a + 1, \ldots, b}$, the product of $x_k$ taken $m$ at a time is simply ${x_k}^m$.
Thus for $n = 1$ we have:
:$h_m = {x_k}^m$
Let the generating function for such a $\sequence {h_m}$ be $\map {G_k} z$.
From Generating Function for Sequence of Powers of Constant:
:$\map {G_k} z = \dfrac 1 {1 - x_k z}$
By Product of Summations, we have:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \prod_{k \mathop = a}^b \sum_{j \mathop = 1}^m x_j$
Hence:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{k \mathop \ge 0} h_k z^k
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \prod_{k \mathop = a}^b \dfrac 1 {1 - x_k z}
| c = Product of Generating Functions: General Rule
}}
{{eqn | r = \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \dotsm \paren {1 - x_b z} }
| c =
}}
{{end-eqn}}
{{qed}}
{{Proofread}}
\end{proof}
|
22377
|
\section{Summation of Sum of Mappings on Finite Set}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f, g: S \to \mathbb A$ be mappings.
Let $h = f + g$ be their sum.
Then we have the equality of summations on finite sets:
:$\ds \sum_{s \mathop \in S} \map h s = \sum_{s \mathop \in S} \map f s + \sum_{s \mathop \in S} \map g s$
\end{theorem}
\begin{proof}
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following equality of indexed summations:
:$\ds \sum_{i \mathop = 0}^{n - 1} \map h {\map \sigma i} = \sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i} + \sum_{i \mathop = 0}^{n - 1} \map g {\map \sigma i}$
By Sum of Mappings Composed with Mapping, $h \circ \sigma = f \circ \sigma + g \circ \sigma$.
The above equality now follows from Indexed Summation of Sum of Mappings.
{{qed}}
\end{proof}
|
22378
|
\section{Summation of Summation over Divisors of Function of Two Variables}
Tags: Divisors, Divisibility, Summations
\begin{theorem}
Let $c, d, n \in \Z$.
Then:
:$\ds \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{d \mathop \divides \paren {n / c} } \map f {c, c d}$
where $c \divides d$ denotes that $c$ is a divisor of $d$.
\end{theorem}
\begin{proof}
From Exchange of Order of Summation with Dependency on Both Indices:
{{:Exchange of Order of Summation with Dependency on Both Indices}}
We have that:
:$\map R d$ is the propositional function:
::$d \divides n$
:$\map S {d, c}$ is the propositional function:
::$c \divides d$
Thus $\map {R'} {d, c}$ is the propositional function:
::Both $d \divides n$ and $c \divides d$
This is the same as:
::$c \divides n$ and $\dfrac d c \divides \dfrac n c$
Similarly, $\map {S'} c$ is the propositional function:
::$\exists d$ such that both $d \divides n$ and $c \divides d$
This is the same as:
::$c \divides n$
This gives:
:$\ds \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{\paren {d / c} \mathrel \divides \paren {n / c} } \map f {c, d}$
Replacing $d / c$ with $d$:
:$\ds \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{d \mathop \divides \paren {n / c} } \map f {c, c d}$
{{qed}}
\end{proof}
|
22379
|
\section{Summation of Zero/Finite Set}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $0 : S \to \mathbb A$ be the zero mapping.
{{explain|Presumably the above is a constant mapping on $0$ -- needs to be made explicit.}}
Then the summation of $0$ over $S$ equals zero:
:$\ds \sum_{s \mathop \in S} 0 \left({s}\right) = 0$
\end{theorem}
\begin{proof}
At least three proofs are possible:
:using the definition of summation and Indexed Summation of Zero
:using Indexed Summation of Sum of Mappings
:using Summation of Multiple of Mapping on Finite Set.
{{ProofWanted}}
Category:Summations
\end{proof}
|
22380
|
\section{Summation of Zero/Indexed Summation}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N,\Z,\Q,\R,\C$.
Let $a, b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f_0 : \closedint a b \to \mathbb A$ be the zero mapping.
Then the indexed summation of $0$ from $a$ to $b$ equals zero:
:$\ds \sum_{i \mathop = a}^b \map {f_0} i = 0$
\end{theorem}
\begin{proof}
At least three proofs are possible:
* by induction, using Identity Element of Addition on Numbers
* using Indexed Summation of Multiple of Mapping
* using Indexed Summation of Sum of Mappings
{{ProofWanted}}
Category:Summations
\end{proof}
|
22381
|
\section{Summation of Zero/Set}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a set.
Let $0: S \to \mathbb A$ be the zero mapping.
Then the summation with finite support of $0$ over $S$ equals zero:
:$\ds \sum_{s \mathop \in S} \map 0 s = 0$
\end{theorem}
\begin{proof}
By Support of Zero Mapping, the support of $0$ is empty.
By Empty Set is Finite, the support of $0$ is indeed finite.
By Summation over Empty Set:
:$\ds \sum_{s \mathop \in S} \map 0 s = \sum_{s \mathop \in \O} \map 0 s = 0$
{{qed}}
Category:Summations
\end{proof}
|
22382
|
\section{Summation of i from 1 to n of Summation of j from 1 to i}
Tags: Summation of i from 1 to n of Summation of j from 1 to i, Summations
\begin{theorem}
:$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$
\end{theorem}
\begin{proof}
:$\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i$
can be expressed as:
:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} a_{i j}$
where:
:$R \left({i}\right)$ is the propositional function $1 \le i \le n$
:$S \left({i, j}\right)$ is the propositional function $1 \le j \le i$
We wish to find a propositional function $S' \left({j}\right)$ which is to be:
:there exists an $i$ such that both $1 \le i \le n$ and $1 \le j \le i$
This is satisfied by the propositional function:
:$S' \left({j}\right) := 1 \le j \le n$
Next we wish to find a propositional function $R' \left({i, j}\right)$ which is to be:
:both $1 \le i \le n$ and $1 \le j \le i$
This is satisfied by the propositional function:
:$R' \left({i, j}\right) := j \le i \le n$
Hence the result, from Exchange of Order of Summation with Dependency on Both Indices.
{{qed}}
\end{proof}
|
22383
|
\section{Summation over Finite Set Equals Summation over Support}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $\map \supp f$ be its support.
Then we have an equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S} \map f s = \sum_{s \mathop \in \map \supp f} \map f s$
\end{theorem}
\begin{proof}
Note that by Subset of Finite Set is Finite, $\map \supp f$ is indeed finite.
The result now follows from:
* Sum over Complement of Finite Set
* Sum of Zero over Finite Set
* Identity Element of Addition on Numbers
{{qed}}
\end{proof}
|
22384
|
\section{Summation over Finite Set is Well-Defined}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $n$ be the cardinality of $S$.
let $\N_{<n}$ be an initial segment of the natural numbers.
Let $g, h: \N_{<n} \to S$ be bijections.
Then we have an equality of indexed summations of the compositions $f \circ g$ and $f \circ h$:
:$\ds \sum_{i \mathop = 0}^{n - 1} \map f {\map g i} = \sum_{i \mathop = 0}^{n - 1} \map f {\map h i}$
That is, the definition of summation over a finite set does not depend on the choice of the bijection $g: S \to \N_{< n}$.
\end{theorem}
\begin{proof}
By Inverse of Bijection is Bijection, $h^{-1} : \N_{<n} \to S$ is a bijection.
By Composite of Bijections is Bijection, the composition $h^{-1}\circ g$ is a permutation of $\N_{<n}$.
By Indexed Summation does not Change under Permutation, we have an equality of indexed summations:
:$\ds \sum_{i \mathop = 0}^{n - 1} \map {\paren {f \circ h} } i = \sum_{i \mathop = 0}^{n - 1} \map {\paren {f \circ h} \circ \paren {h^{-1} \circ g} } i$
By Composition of Mappings is Associative and Composite of Bijection with Inverse is Identity Mapping, the {{RHS}} equals $\ds \sum_{i \mathop = 0}^{n - 1} \map f {\map g i}$.
{{qed}}
\end{proof}
|
22385
|
\section{Summation over Interval equals Indexed Summation}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a, b \in \Z$ be integers.
Let $\closedint a b$ be the integer interval between $a$ and $b$.
Let $f: \closedint a b \to \mathbb A$ be a mapping.
Then the summation over the finite set $\closedint a b$ equals the indexed summation from $a$ to $b$:
:$\ds \sum_{k \mathop \in \closedint a b} \map f k = \sum_{k \mathop = a}^b \map f k$
\end{theorem}
\begin{proof}
By Cardinality of Integer Interval, $\closedint a b$ has cardinality $b - a + 1$.
By Translation of Integer Interval is Bijection, the mapping $T : \closedint 0 {b - a} \to \closedint a b$ defined as:
:$\map T k = k + a$
is a bijection.
By definition of summation:
:$\ds \sum_{k \mathop \in \closedint a b} \map f k = \sum_{k \mathop = 0}^{b - a} \map f {k + a}$
By Indexed Summation over Translated Interval:
:$\ds \sum_{k \mathop = 0}^{b - a} \map f {k + a} = \sum_{k \mathop = a}^b \map f k$
{{qed}}
Category:Summations
\end{proof}
|
22386
|
\section{Summation over Lower Index of Unsigned Stirling Numbers of the First Kind}
Tags: Stirling Numbers, Factorials
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\ds \sum_k {n \brack k} = n!$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$n!$ denotes the factorial of $n$.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P N$ be the proposition:
:$\ds \sum_k {n \brack k} = n!$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k {0 \brack k}
| r = \sum_k \delta_{0 k}
| c = Unsigned Stirling Number of the First Kind of 0
}}
{{eqn | r = 1
| c = all terms vanish but for $k = 0$
}}
{{eqn | r = 0!
| c = {{Defof|Factorial}}
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
\end{proof}
|
22387
|
\section{Summation over Lower Index of Unsigned Stirling Numbers of the First Kind with Alternating Signs}
Tags: Stirling Numbers, Factorials
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\ds \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$\delta_{n 0}$ denotes the Kronecker delta.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k \paren {-1}^k {0 \brack k}
| r = \sum_k \delta_{0 k}
| c = Unsigned Stirling Number of the First Kind of 0
}}
{{eqn | r = 1
| c = all terms vanish but for $k = 0$
}}
{{eqn | r = \delta_{0 0} - \delta_{0 1}
| c = {{Defof|Kronecker Delta}}
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k \paren {-1}^k {1 \brack k}
| r = \sum_k \paren {-1}^k \delta_{1 k}
| c = Unsigned Stirling Number of the First Kind of 1
}}
{{eqn | r = -1
| c = all terms vanish but for $k = 1$
}}
{{eqn | r = \delta_{1 0} - \delta_{1 1}
| c = {{Defof|Kronecker Delta}}
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
\end{proof}
|
22388
|
\section{Summation over k of Ceiling of k over 2}
Tags: Ceiling Function, Summations
\begin{theorem}
:$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \ceiling {\dfrac {n \paren {n + 2} } 4}$
\end{theorem}
\begin{proof}
By Permutation of Indices of Summation:
:$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \sum_{k \mathop = 1}^n \ceiling {\dfrac {n + 1 - k} 2}$
and so:
:$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2} }$
First take the case where $n$ is even.
For $k$ odd:
:$\ceiling {\dfrac k 2} = \dfrac k 2 + \dfrac 1 2$
and:
:$\ceiling {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2}
| r = \dfrac k 2 + \dfrac 1 2 + \dfrac {n + 1 - k} 2
| c =
}}
{{eqn | r = \dfrac {k + 1 + n + 1 - k} 2
| c =
}}
{{eqn | r = \dfrac {n + 2} 2
| c =
}}
{{end-eqn}}
For $k$ even:
:$\ceiling {\dfrac k 2} = \dfrac k 2$
and:
:$\ceiling {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2 + \dfrac 1 2 = \dfrac {n - k + 2} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2}
| r = \dfrac k 2 + \dfrac {n - k + 2} 2
| c =
}}
{{eqn | r = \dfrac {k + n - k + 2} 2
| c =
}}
{{eqn | r = \dfrac {n + 2} 2
| c =
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2}
| r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2} }
| c =
}}
{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\dfrac {n + 2} 2}
| c =
}}
{{eqn | r = \dfrac 1 2 n \paren {\dfrac {n + 2} 2}
| c =
}}
{{eqn | r = \dfrac {n \paren {n + 2} } 4
| c =
}}
{{eqn | r = \ceiling {\dfrac {n \paren {n + 2} } 4}
| c = as $\dfrac {n \paren {n + 2} } 4$ is an integer
}}
{{end-eqn}}
{{qed|lemma}}
Next take the case where $n$ is odd.
For $k$ odd:
:$\ceiling {\dfrac k 2} = \dfrac k 2 + \dfrac 1 2$
and:
:$\ceiling {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2 + \dfrac 1 2$
Hence:
{{begin-eqn}}
{{eqn | l = \ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2}
| r = \dfrac k 2 + \dfrac 1 2 + \dfrac {n + 1 - k} 2 + \dfrac 1 2
| c =
}}
{{eqn | r = \dfrac {k + 1 + n + 1 - k + 1} 2
| c =
}}
{{eqn | r = \dfrac {n + 3} 2
| c =
}}
{{end-eqn}}
For $k$ even:
:$\ceiling {\dfrac k 2} = \dfrac k 2$
and:
:$\ceiling {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2}
| r = \dfrac k 2 + \dfrac {n - k + 1} 2
| c =
}}
{{eqn | r = \dfrac {k + n - k + 1} 2
| c =
}}
{{eqn | r = \dfrac {n + 1} 2
| c =
}}
{{end-eqn}}
Let $n = 2 t + 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2}
| r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2} }
| c =
}}
{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^{2 t + 1} \paren {\ceiling {\dfrac k 2} + \ceiling {\dfrac {2 t + 2 - k} 2} }
| c =
}}
{{eqn | r = \dfrac t 2 \dfrac {\paren {2 t + 1} + 1} 2 + \dfrac {t + 1} 2 \dfrac {\paren {2 t + 1} + 3} 2
| c = there are $t$ even terms and $t + 1$ odd terms
}}
{{eqn | r = \dfrac {2 t^2 + 2 t} 4 + \dfrac {2 t^2 + 6 t + 4} 4
| c = multiplying out
}}
{{eqn | r = \dfrac {4 t^2 + 8 t + 3} 4 + \dfrac 1 4
| c =
}}
{{eqn | r = \dfrac {\paren {2 t + 1} \paren {2 t + 3} } + \dfrac 1 4
| c =
}}
{{eqn | r = \dfrac {n \paren {n + 2} } 4 + \dfrac 1 4
| c =
}}
{{eqn | r = \ceiling {\dfrac {n \paren {n + 2} } 4}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22389
|
\section{Summation over k of Floor of k over 2}
Tags: Summations, Floor Function
\begin{theorem}
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \floor {\dfrac {n^2} 4}$
\end{theorem}
\begin{proof}
By Permutation of Indices of Summation:
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \sum_{k \mathop = 1}^n \floor {\dfrac {n + 1 - k} 2}$
and so:
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }$
First take the case where $n$ is even.
For $k$ odd:
:$\floor {\dfrac k 2} = \dfrac k 2 - \dfrac 1 2$
and:
:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
| r = \dfrac k 2 - \dfrac 1 2 + \dfrac {n + 1 - k} 2
| c =
}}
{{eqn | r = \dfrac {k - 1 + n + 1 - k} 2
| c =
}}
{{eqn | r = \dfrac n 2
| c =
}}
{{end-eqn}}
For $k$ even:
:$\floor {\dfrac k 2} = \dfrac k 2$
and:
:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2 - \dfrac 1 2 = \dfrac {n - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
| r = \dfrac k 2 + \dfrac {n - k} 2
| c =
}}
{{eqn | r = \dfrac {k + n - k} 2
| c =
}}
{{eqn | r = \dfrac n 2
| c =
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \floor {\dfrac k 2}
| r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }
| c =
}}
{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\dfrac n 2}
| c =
}}
{{eqn | r = \dfrac 1 2 n \dfrac n 2
| c =
}}
{{eqn | r = \dfrac {n^2} 4
| c =
}}
{{eqn | r = \floor {\dfrac {n^2} 4}
| c = as $\dfrac {n^2} 4$ is an integer
}}
{{end-eqn}}
{{qed|lemma}}
Next take the case where $n$ is odd.
For $k$ odd:
:$\floor {\dfrac k 2} = \dfrac k 2 - \dfrac 1 2$
and:
:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2 - \dfrac 1 2$
Hence:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
| r = \dfrac k 2 - \dfrac 1 2 + \dfrac {n + 1 - k} 2 - \dfrac 1 2
| c =
}}
{{eqn | r = \dfrac {k - 1 + n + 1 - k - 1} 2
| c =
}}
{{eqn | r = \dfrac {n - 1} 2
| c =
}}
{{end-eqn}}
For $k$ even:
:$\floor {\dfrac k 2} = \dfrac k 2$
and:
:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
| r = \dfrac k 2 + \dfrac {n - k + 1} 2
| c =
}}
{{eqn | r = \dfrac {k + n - k + 1} 2
| c =
}}
{{eqn | r = \dfrac {n + 1} 2
| c =
}}
{{end-eqn}}
Let $n = 2 t + 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \floor {\dfrac k 2}
| r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }
| c =
}}
{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^{2 t + 1} \paren {\floor {\dfrac k 2} + \floor {\dfrac {2 t + 2 - k} 2} }
| c =
}}
{{eqn | r = \dfrac t 2 \dfrac {\paren {2 t + 1} + 1} 2 + \dfrac {t + 1} 2 \dfrac {\paren {2 t + 1} - 1} 2
| c = there are $t$ even terms and $t + 1$ odd terms
}}
{{eqn | r = \dfrac {2 t^2 + 2 t} 4 + \dfrac {2 t^2 + 2 t} 4
| c = multiplying out
}}
{{eqn | r = \dfrac {4 t^2 + 4 t} 4 + \dfrac 1 4 - \dfrac 1 4
| c =
}}
{{eqn | r = \dfrac {\paren {2 t + 1}^2} 4 - \dfrac 1 4
| c =
}}
{{eqn | r = \dfrac {n^2} 4 - \dfrac 1 4
| c =
}}
{{eqn | r = \floor {\dfrac {n^2} 4}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22390
|
\section{Summation over k of Floor of mk+x over n}
Tags: Summations, Floor Function
\begin{theorem}
Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {m k + x} n} = \dfrac {\paren {m - 1} \paren {n - 1} } 2 + \dfrac {d - 1} 2 + d \floor {\dfrac x d}$
where:
:$\floor x$ denotes the floor of $x$
:$d$ is the greatest common divisor of $m$ and $n$.
\end{theorem}
\begin{proof}
By definition of modulo 1:
:$\ds \sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {m k + x} n} = \sum_{0 \mathop \le k \mathop < n} \dfrac {m k + x} n - \sum_{0 \mathop \le k \mathop < n} \fractpart {\dfrac {m k + x} n}$
where $\fractpart y$ in this context denotes the fractional part of $y$.
First we have:
{{begin-eqn}}
{{eqn | l = \sum_{0 \mathop \le k \mathop < n} \dfrac {m k + x} n
| r = \frac m n \sum_{0 \mathop \le k \mathop < n} k + \sum_{0 \mathop \le k \mathop < n} \dfrac x n
| c =
}}
{{eqn | r = \frac m n \frac {n \paren {n - 1} } 2 + n \dfrac x n
| c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {m \paren {n - 1} } 2 + x
| c =
}}
{{end-eqn}}
Let $S$ be defined as:
:$\ds S := \sum_{0 \mathop \le k \mathop < n} \fractpart {\dfrac {m k + x} n}$
Thus:
:$(1): \quad \ds \sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {m k + x} n} = \dfrac {m \paren {n - 1} } 2 + x - S$
Let $d = \gcd \set {m, n}$.
Let:
{{begin-eqn}}
{{eqn | l = t
| r = \frac n d
}}
{{eqn | l = u
| r = \frac m d
}}
{{eqn | ll= \leadsto
| l = \frac m n
| r = \frac u t
| c =
}}
{{eqn | ll= \leadsto
| l = m t
| r = u n
| c =
}}
{{eqn | ll= \leadsto
| l = u
| r = \frac {m t} n
| c =
}}
{{end-eqn}}
We have that:
{{begin-eqn}}
{{eqn | l = \fractpart {\dfrac {m k + x} n}
| r = \fractpart {\dfrac {m k + x} n + u}
| c = {{Defof|Fractional Part}}: $u$ is an integer
}}
{{eqn | r = \fractpart {\dfrac {m k + x} n + \frac {m t} n}
| c =
}}
{{eqn | r = \fractpart {\dfrac {m \paren {k + t} + x} n}
| c =
}}
{{end-eqn}}
Thus $S$ consists of $d$ copies of the same summation:
{{begin-eqn}}
{{eqn | l = S
| r = \sum_{0 \mathop \le k \mathop < n} \fractpart {\dfrac {m k + x} n}
| c =
}}
{{eqn | r = d \sum_{0 \mathop \le k \mathop < t} \fractpart {\dfrac {m k + x} n}
| c =
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \sum_{0 \mathop \le k \mathop < t} \fractpart {\dfrac {m k + x} n}
| r = \sum_{0 \mathop \le k \mathop < t} \fractpart {\dfrac x n + \dfrac {u k} t}
| c = substituting $\dfrac u t$ for $\dfrac m n$
}}
{{eqn | r = \sum_{0 \mathop \le k \mathop < t} \fractpart {\dfrac {x \bmod d} n + \dfrac k t}
| c = as $t \perp u$
}}
{{eqn | r = \sum_{0 \mathop \le k \mathop < t} \dfrac {x \bmod d} n + \dfrac k t
| c = as $\dfrac {x \bmod d} n < \dfrac 1 t$
}}
{{eqn | r = t \dfrac {x \bmod d} n + \frac 1 t \sum_{0 \mathop \le k \mathop < t} k
| c =
}}
{{eqn | r = \dfrac {t \paren {x \bmod d} } n + \frac 1 t \frac {t \paren {t - 1} } 2
| c = Closed Form for Triangular Numbers
}}
{{eqn | r = \dfrac {t \paren {x \bmod d} } n + \frac {t - 1} 2
| c =
}}
{{eqn | ll= \leadsto
| l = S
| r = d \paren {\dfrac {t \paren {x \bmod d} } n + \frac {t - 1} 2}
| c =
}}
{{eqn | r = \dfrac {n \paren {x \bmod d} } n + \frac {n - d} 2
| c = as $n = d t$
}}
{{eqn | r = x \bmod d + \frac {n - d} 2
| c =
}}
{{end-eqn}}
{{explain|Greater detail needed as to why $\ds \sum_{0 \mathop \le k \mathop < t} \fractpart {\dfrac x n + \dfrac {u k} t} {{=}} \sum_{0 \mathop \le k \mathop < t} \fractpart {\dfrac {x \bmod d} n + \dfrac k t}$}}
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {m k + x} n}
| r = \frac {m \paren {n - 1} } 2 + x - S
| c = from $(1)$
}}
{{eqn | r = \frac {m \paren {n - 1} } 2 + x - d \paren {\dfrac {t \paren {x \bmod d} } n + \frac {t - 1} 2}
| c =
}}
{{eqn | r = \frac {m \paren {n - 1} } 2 + x - x \bmod d - \frac {n - d} 2
| c =
}}
{{eqn | r = \frac {m \paren {n - 1} } 2 + x - x + d \floor {\frac x d} - \frac {n - 1} 2 + \frac {d - 1} 2
| c = {{Defof|Modulo Operation}} and algebra
}}
{{eqn | r = \frac {\paren {m - 1} \paren {n - 1} } 2 + \frac {d - 1} 2 + d \floor {\frac x d}
| c = simplification
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22391
|
\section{Summation over k of Floor of x plus k over y}
Tags: Summations, Floor Function
\begin{theorem}
Let $x, y \in \R$ such that $y > 0$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < y} \floor {x + \dfrac k y} = \floor {x y + \floor {x + 1} \paren {\ceiling y - y} }$
\end{theorem}
\begin{proof}
When $x$ increases by $1$, both sides increase by $\ceiling y$.
So we can assume $0 \le x < 1$.
When $x = 0$, both sides are equal to $0$.
When $x$ increases past $1 - \dfrac k y$ for $0 \le k < y$, both sides increase by $1$.
Hence the result.
{{qed}}
\end{proof}
|
22392
|
\section{Summation over k to n of Product of kth with n-kth Fibonacci Numbers}
Tags: Fibonacci Numbers
\begin{theorem}
:$\ds \sum_{k \mathop = 0}^n F_k F_{n - k} = \dfrac {\paren {n - 1} F_n + 2n F_{n - 1} } 5$
where $F_n$ denotes the $n$th Fibonacci number.
\end{theorem}
\begin{proof}
From Generating Function for Fibonacci Numbers, a generating function for the Fibonacci numbers is:
:$\map G z = \dfrac z {1 - z - z^2}$
Then:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \dfrac z {1 - z - z^2}
| c =
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }
| c = Partial Fraction Expansion
}}
{{end-eqn}}
where:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \paren {\map G z}^2
| r = \paren {\dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} } }^2
| c =
}}
{{eqn | r = \dfrac 1 5 \paren {\paren {\dfrac 1 {1 - \phi z} }^2 + \paren {\dfrac 1 {1 - \hat \phi z} }^2 - 2 \dfrac 1 {1 - \phi z} \dfrac 1 {1 - \hat \phi z} }
| c =
}}
{{eqn | r = \dfrac 1 5 \paren {\dfrac 1 {\paren {1 - \phi z}^2} + \dfrac 1 {\paren {1 - \hat \phi z}^2} - 2 \dfrac {1 - \hat \phi z + 1 - \phi z} {\paren {1 - \phi z} \paren {1 - \hat \phi z} } }
| c =
}}
{{eqn | r = \dfrac 1 5 \paren {\dfrac 1 {\paren {1 - \phi z}^2} + \dfrac 1 {\paren {1 - \hat \phi z}^2} - \dfrac 2 {\paren {1 - \phi z} \paren {1 - \paren {1 - \phi} z} } }
| c = Definition of $\hat \phi$
}}
{{eqn | r = \dfrac 1 5 \paren {\dfrac 1 {\paren {1 - \phi z}^2} + \dfrac 1 {\paren {1 - \hat \phi z}^2} - \dfrac 2 {1 - z - z^2} }
| c = after algebra
}}
{{eqn | r = \dfrac 1 5 \paren {\sum_{n \mathop = 0}^\infty \paren {n + 1} \paren {\phi z}^n + \sum_{n \mathop = 0}^\infty \paren {n + 1} \paren {\hat \phi z}^n - \dfrac 2 z \map G z}
| c = Power Series Expansion for Square of Reciprocal of 1-z
}}
{{eqn | r = \dfrac 1 5 \paren {\sum_{n \mathop = 0}^\infty \paren {n + 1} \paren {\phi^n + \hat \phi^n} z^n - \dfrac 2 z \sum_{n \mathop = 0}^\infty F_n z^n}
| c = Generating Function for Fibonacci Numbers
}}
{{eqn | r = \dfrac 1 5 \paren {\sum_{n \mathop = 0}^\infty \paren {n + 1} \paren {\phi^n + \hat \phi^n} z^n - 2 \sum_{n \mathop = 0}^\infty F_n z^{n - 1} }
| c =
}}
{{eqn | r = \dfrac 1 5 \paren {\sum_{n \mathop = 0}^\infty \paren {n + 1} \paren {\phi^n + \hat \phi^n} z^n - 2 \sum_{n \mathop = 1}^\infty F_{n + 1} z^n}
| c = Translation of Index Variable of Summation
}}
{{eqn | r = \dfrac 1 5 \sum_{n \mathop = 0}^\infty \paren {\paren {n + 1} \paren {\phi^n + \hat \phi^n} - 2 F_{n + 1} } z^n
| c =
}}
{{end-eqn}}
Since the coefficient of $z^n$ in $\paren {\map G z}^2$ is $\ds \sum_{k \mathop = 0}^n F_k F_{n - k}$, by comparing coefficients:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n F_k F_{n - k}
| r = \dfrac 1 5 \paren {\paren {n + 1} \paren {\phi^n + \hat \phi^n} - 2 F_{n + 1} }
| c =
}}
{{eqn | r = \dfrac 1 5 \paren {\paren {n + 1} \paren {\dfrac {\phi^{2 n} - \hat \phi^{2 n} } {\sqrt 5} \times \dfrac {\sqrt 5} {\phi^n - \hat \phi^n} } - 2 F_{n + 1} }
| c = Difference of Two Squares
}}
{{eqn | r = \dfrac 1 5 \paren {\paren {n + 1} \dfrac {F_{2 n} } {F_n} - 2 F_{n + 1} }
| c = Euler-Binet Formula
}}
{{eqn | r = \dfrac 1 5 \paren {\paren {n + 1} \dfrac {F_{n - 1} F_n + F_n F_{n + 1} } {F_n} - 2 F_{n + 1} }
| c = Fibonacci Number in terms of Smaller Fibonacci Numbers
}}
{{eqn | r = \dfrac 1 5 \paren {\paren {n + 1} \paren {F_{n - 1} + F_{n + 1} } - 2 F_{n + 1} }
| c =
}}
{{eqn | r = \dfrac 1 5 \paren {\paren {n + 1} \paren {F_n + 2 F_{n - 1} } - 2 \paren {F_n + F_{n - 1} } }
| c = {{Defof|Fibonacci Numbers}}
}}
{{eqn | r = \dfrac {\paren {n - 1} F_n + 2n F_{n - 1} } 5
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22393
|
\section{Summation to n of Power of k over k}
Tags: Summation to n of Power of k over k, Harmonic Numbers, Binomial Coefficients
\begin{theorem}
:$\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$
where:
:$H_n$ denotes the $n$th harmonic number
:$\dbinom n k$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
The proof proceeds by induction over $n$.
For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:
:$\displaystyle \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \displaystyle \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\left({x - 1}\right)^k} k$
\end{proof}
|
22394
|
\section{Summation to n of Square of kth Harmonic Number}
Tags: Harmonic Numbers
\begin{theorem}
:$\ds \sum_{k \mathop = 1}^n {H_k}^2 = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n$
where $H_k$ denotes the $k$th harmonic number.
\end{theorem}
\begin{proof}
Consider:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}
| r = \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} - 1} {k + 1}
| c = factorizing
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}
| c = Sum of Summations equals Summation of Sum
}}
{{eqn | r = n - 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}
| c = Summation of Unity over Elements
}}
{{eqn | r = n - 1 - \paren {H_n - 1}
| c = Harmonic Number and compensate for $\dfrac 1 1$ term
}}
{{eqn | n = 1
| r = n - H_n
| c = simplification
}}
{{end-eqn}}
{{refactor|The above result is worth publishing on its own page as a result in its own right. I'm surprised we don't already have it.}}
Using Summation by Parts:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n {H_k}^2
| r = H_n \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^{n - 1} \paren {\paren {\sum_{i \mathop = 1}^k {H_i} } \paren {H_{k + 1} - H_k} }
| c =
}}
{{eqn | r = H_n \paren {\paren {n + 1} H_n - n} - \sum_{k \mathop = 1}^{n - 1} \paren {\paren {k + 1} H_k - k} \paren {H_{k + 1} - H_k}
| c = Sum of Sequence of Harmonic Numbers twice
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} H_k - k} {k + 1}
| c = $H_{k + 1} - H_k = \dfrac 1 {k+1}$
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \sum_{k \mathop = 1}^{n - 1} H_k + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}
| c = Sum of Summations equals Summation of Sum
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren { \sum_{k \mathop = 1}^n H_k - H_n} + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}
| c = increase range of summation and compensate
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren {\paren {n + 1} H_n - n - H_n } + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}
| c = Sum of Sequence of Harmonic Numbers
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren {n + 1} H_n + n + H_n + n - H_n
| c = $(1)$
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n
| c = simplification
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22395
|
\section{Summation to n of kth Harmonic Number over k}
Tags: Harmonic Numbers, General Harmonic Numbers
\begin{theorem}
:$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} k = \dfrac { {H_n}^2 + H_n^{\paren 2} } 2$
where:
:$H_n$ denotes the $n$th harmonic number
:$H_n^{\paren 2}$ denotes a general harmonic number.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} k
| r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 {j k}
| c = {{Defof|Harmonic Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 j \dfrac 1 k
| c =
}}
{{eqn | r = \dfrac 1 2 \paren {\paren {\sum_{k \mathop = 1}^n \dfrac 1 k}^2 + \paren {\sum_{k \mathop = 1}^n \dfrac 1 {k^2} } }
| c = Summation of Products of n Numbers taken m at a time with Repetitions
}}
{{eqn | r = \dfrac { {H_n}^2 + H_n^{\paren 2} } 2
| c = {{Defof|Harmonic Number}} and {{Defof|General Harmonic Numbers}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22396
|
\section{Summation to n of kth Harmonic Number over k+1}
Tags: Harmonic Numbers, General Harmonic Numbers
\begin{theorem}
:$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1} = \dfrac { {H_{n + 1} }^2 - H_{n + 1}^{\paren 2} } 2$
where:
:$H_n$ denotes the $n$th harmonic number
:$H_n^{\paren 2}$ denotes a general harmonic number.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1}
| r = \sum_{k \mathop = 1}^n \paren {\dfrac {H_{k + 1} } {k + 1} - \dfrac 1 {\paren {k + 1} \paren {k + 1} } }
| c = {{Defof|Harmonic Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \dfrac {H_{k + 1} } {k + 1} - \sum_{k \mathop = 1}^n \dfrac 1 {\paren {k + 1}^2}
| c =
}}
{{eqn | r = \sum_{k \mathop = 2}^{n + 1} \dfrac {H_k} k - \sum_{k \mathop = 2}^{n + 1} \dfrac 1 {k^2}
| c = Translation of Index Variable of Summation
}}
{{eqn | r = \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - \dfrac {H_1} 1 - \paren{\sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2} - \dfrac 1 {1^2} }
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - 1 - \paren{\sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2} - 1}
| c = Harmonic Number $H_1$
}}
{{eqn | r = \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - \sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2}
| c = simplifying
}}
{{eqn | r = \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - H_{n + 1}^{\paren 2}
| c = {{Defof|General Harmonic Numbers}}
}}
{{eqn | r = \dfrac { {H_{n + 1} }^2 + H_{n + 1}^{\paren 2} } 2 - H_{n + 1}^{\paren 2}
| c = Summation to n of kth Harmonic Number over k
}}
{{eqn | r = \dfrac { {H_{n + 1} }^2 + H_{n + 1}^{\paren 2} - 2 H_{n + 1}^{\paren 2} } 2
| c =
}}
{{eqn | r = \dfrac { {H_{n + 1} }^2 - H_{n + 1}^{\paren 2} } 2
| c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22397
|
\section{Sums of Consecutive Sequences of Squares that equal Squares}
Tags: Sums of Sequences, Sum of Sequence of Squares, Square Numbers
\begin{theorem}
The $24$th square pyramidal number is the only one which is square:
:$1^2 + 2^2 + 3^2 + \cdots + 24^2 = 70^2$
while there are several Sum of Sequence of Squares which are square, for example:
:$18^2 + 19^2 + \cdots + 28^2 = 77^2$
and:
:$25^2 + 26^2 + \cdots + 624^2 = 9010^2$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 1^2 + 2^2 + 3^2 + \cdots + 24^2
| r = \dfrac {24 \times \paren {24 + 1} \times \paren {2 \times 24 + 1} } 6
| c = Sum of Sequence of Squares
}}
{{eqn | r = \dfrac {24 \times 25 \times 49} 6
| c =
}}
{{eqn | r = \dfrac {2^3 \times 3 \times 5^2 \times 7^2} {2 \times 3}
| c =
}}
{{eqn | r = 2^2 \times 5^2 \times 7^2
| c =
}}
{{eqn | r = \paren {2 \times 5 \times 7}^2
| c =
}}
{{eqn | r = 70^2
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = 18^2 + 19^2 + \cdots + 28^2
| r = \dfrac {28 \times \paren {28 + 1} \times \paren {2 \times 28 + 1} } 6 - \dfrac {17 \times \paren {17 + 1} \times \paren {2 \times 17 + 1} } 6
| c = Sum of Sequence of Squares
}}
{{eqn | r = \dfrac {28 \times 29 \times 57 - 17 \times 18 \times 35} 6
| c =
}}
{{eqn | r = \dfrac {\paren {2^2 \times 7} \times 29 \times \paren {3 \times 19} - 17 \times \paren {2 \times 3^2} \times \paren {5 \times 7} } {2 \times 3}
| c =
}}
{{eqn | r = 2 \times 7 \times 29 \times 19 - 17 \times 3 \times 5 \times 7
| c =
}}
{{eqn | r = 7 \times \paren {1102 - 255}
| c =
}}
{{eqn | r = 7 \times 847
| c =
}}
{{eqn | r = 7 \times 7 \times 11^2
| c =
}}
{{eqn | r = 77^2
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = 25^2 + 26^2 + \cdots + 624^2
| r = \dfrac {624 \times \paren {624 + 1} \times \paren {2 \times 624 + 1} } 6 - \dfrac {24 \times \paren {24 + 1} \times \paren {2 \times 24 + 1} } 6
| c = Sum of Sequence of Squares
}}
{{eqn | r = \dfrac {624 \times 625 \times 1249 - 24 \times 25 \times 49} 6
| c =
}}
{{eqn | r = \dfrac {\paren {2^4 \times 3 \times 13} \times 5^4 \times 1249 - \paren {2^3 \times 3} \times 5^2 \times 7^2} {2 \times 3}
| c =
}}
{{eqn | r = 2^3 \times 5^4 \times 13 \times 1249 - 2^2 \times 5^2 \times 7^2
| c =
}}
{{eqn | r = 2^2 \times 5^2 \times \paren {2 \times 5^2 \times 13 \times 1249 - 7^2}
| c =
}}
{{eqn | r = 2^2 \times 5^2 \times \paren {811 \, 850 - 49}
| c =
}}
{{eqn | r = 2^2 \times 5^2 \times \paren {811 \, 801}
| c =
}}
{{eqn | r = 2^2 \times 5^2 \times \paren {17^2 \times 53^2}
| c =
}}
{{eqn | r = 9010^2
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22398
|
\section{Sums of Partial Sequences of Squares}
Tags: Sums of Sequences, Sums of Partial Sequences of Squares, Square Numbers
\begin{theorem}
Let $n \in \Z_{>0}$.
Consider the odd number $2 n + 1$ and its square $\paren {2 n + 1}^2 = 2 m + 1$.
Then:
:$\ds \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$
That is:
:the sum of the squares of the $n + 1$ integers up to $m$
equals:
:the sum of the squares of the $n$ integers from $m + 1$ upwards.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$
where $\paren {2 n + 1})^2 = 2 m + 1$.
First it is worth rewriting this so as to eliminate $m$.
{{begin-eqn}}
{{eqn | l = \paren {2 n + 1}^2
| r = 4 n^2 + 4 n + 1
| c =
}}
{{eqn | r = 2 \paren {2 n^2 + 2 n} + 1
| c =
}}
{{end-eqn}}
Thus the statement to be proved can be expressed:
:$\ds \sum_{j \mathop = 0}^n \paren {2 n^2 + 2 n - j}^2 = \sum_{j \mathop = 1}^n \paren {2 n^2 + 2 n + j}^2$
\end{proof}
|
22399
|
\section{Sums of Sequences of Consecutive Squares which are Square}
Tags: Sums of Sequences, Square Numbers
\begin{theorem}
The sums of the following sequences of successive squares are themselves square:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 7}^{29} k^2
| r = 7^2 + 8^2 + \cdots + 29^2
| c =
}}
{{eqn | l = \sum_{i \mathop = 7}^{39} k^2
| r = 7^2 + 8^2 + \cdots + 39^2
| c =
}}
{{eqn | l = \sum_{i \mathop = 7}^{56} k^2
| r = 7^2 + 8^2 + \cdots + 56^2
| c =
}}
{{eqn | l = \sum_{i \mathop = 7}^{190} k^2
| r = 7^2 + 8^2 + \cdots + 190^2
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
From Sum of Sequence of Squares:
{{:Sum of Sequence of Squares}}
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 7}^{29} i^2
| r = \sum_{i \mathop = 1}^{29} i^2 - \sum_{i \mathop = 1}^6 i^2
| c =
}}
{{eqn | r = \frac {29 \left({29 + 1}\right) \left({2 \times 29 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6
| c =
}}
{{eqn | r = \frac {29 \times 30 \times 59} 6 - \frac {6 \times 7 \times 13} 6
| c =
}}
{{eqn | r = 8 \, 555 - 91
| c =
}}
{{eqn | r = 8 \, 464
| c =
}}
{{eqn | r = 92^2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 7}^{39} i^2
| r = \sum_{i \mathop = 1}^{39} i^2 - \sum_{i \mathop = 1}^6 i^2
| c =
}}
{{eqn | r = \frac {39 \left({39 + 1}\right) \left({2 \times 39 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6
| c =
}}
{{eqn | r = \frac {39 \times 40 \times 79} 6 - \frac {6 \times 7 \times 13} 6
| c =
}}
{{eqn | r = 20 \, 540 - 91
| c =
}}
{{eqn | r = 20 \, 449
| c =
}}
{{eqn | r = 143^2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 7}^{56} i^2
| r = \sum_{i \mathop = 1}^{56} i^2 - \sum_{i \mathop = 1}^6 i^2
| c =
}}
{{eqn | r = \frac {56 \left({56 + 1}\right) \left({2 \times 56 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6
| c =
}}
{{eqn | r = \frac {56 \times 57 \times 113} 6 - \frac {6 \times 7 \times 13} 6
| c =
}}
{{eqn | r = 60 \, 116 - 91
| c =
}}
{{eqn | r = 60 \, 025
| c =
}}
{{eqn | r = 245^2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 7}^{190} i^2
| r = \sum_{i \mathop = 1}^{190} i^2 - \sum_{i \mathop = 1}^6 i^2
| c =
}}
{{eqn | r = \frac {190 \left({190 + 1}\right) \left({2 \times 190 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6
| c =
}}
{{eqn | r = \frac {190 \times 191 \times 381} 6 - \frac {6 \times 7 \times 13} 6
| c =
}}
{{eqn | r = 2 \, 304 \, 415 - 91
| c =
}}
{{eqn | r = 2 \, 304 \, 324
| c =
}}
{{eqn | r = 1 \, 518^2
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22400
|
\section{Sums of Squares in Lines of Order 3 Magic Square}
Tags: Magic Squares
\begin{theorem}
Consider the order 3 magic square:
{{:Magic Square/Examples/Order 3}}
: The sums of the squares of the top and bottom rows are equal, and differ by $18$ from the sums of the squares of the middle row
: The sums of the squares of the left and right columns are equal , and differ by $18$ from the sums of the squares of the middle column.
\end{theorem}
\begin{proof}
For the rows:
{{begin-eqn}}
{{eqn | l = 2^2 + 7^2 + 6^2
| r = 4 + 49 + 36
| c =
}}
{{eqn | r = 89
| c =
}}
{{eqn | l = 4^2 + 3^2 + 8^2
| r = 16 + 9 + 64
| c =
}}
{{eqn | r = 89
| c =
}}
{{eqn | l = 9^2 + 5^2 + 1^2
| r = 81 + 25 + 1
| c =
}}
{{eqn | r = 107
| c =
}}
{{eqn | r = 89 + 18
| c =
}}
{{end-eqn}}
For the colums:
{{begin-eqn}}
{{eqn | l = 2^2 + 9^2 + 4^2
| r = 4 + 81 + 16
| c =
}}
{{eqn | r = 101
| c =
}}
{{eqn | l = 6^2 + 1^2 + 8^2
| r = 36 + 1 + 64
| c =
}}
{{eqn | r = 101
| c =
}}
{{eqn | l = 7^2 + 5^2 + 3^2
| r = 49 + 25 + 9
| c =
}}
{{eqn | r = 83
| c =
}}
{{eqn | r = 101 - 18
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22401
|
\section{Sums of Squares of Diagonals of Order 3 Magic Square}
Tags: Magic Squares
\begin{theorem}
Consider the order 3 magic square:
{{:Magic Square/Examples/Order 3}}
The sums of the squares of the diagonals, when expressed as $3$-digit decimal numbers, are equal to the sums of the squares of those same diagonals of that same order 3 magic square when reversed.
{{improve|Find a way to describe the "diagonals" accurately, as what is being demonstrated here does not match the description.}}
\end{theorem}
\begin{proof}
For the top-left to bottom-right diagonals:
{{begin-eqn}}
{{eqn | l = 258^2 + 714^2 + 693^2
| r = 66564 + 509796 + 480249
| c =
}}
{{eqn | r = 1056609
| c =
}}
{{eqn | l = 852^2 + 417^2 + 396^2
| r = 725904 + 173889 + 156816
| c =
}}
{{eqn | r = 1056609
| c =
}}
{{end-eqn}}
For the bottom-left to top-right diagonals:
{{begin-eqn}}
{{eqn | l = 456^2 + 312^2 + 897^2
| r = 207936 + 97344 + 804609
| c =
}}
{{eqn | r = 1109889
| c =
}}
{{eqn | l = 654^2 + 213^2 + 798^2
| r = 427716 + 45369 + 636804
| c =
}}
{{eqn | r = 1109889
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22402
|
\section{Sums of Squares of Lines of Order 3 Magic Square}
Tags: Magic Squares
\begin{theorem}
Consider the order 3 magic square:
{{:Magic Square/Examples/Order 3}}
: The sums of the squares of the rows, when expressed as $3$-digit decimal numbers, are equal to the sums of the squares of those same rows of that same order 3 magic square when reflected in a vertical axis:
:$\begin{array}{|c|c|c|}
\hline 6 & 7 & 2 \\
\hline 1 & 5 & 9 \\
\hline 8 & 3 & 4 \\
\hline
\end{array}$
Similarly:
: The sums of the squares of the columns, when expressed as $3$-digit decimal numbers, are equal to the sums of the squares of those same columns of that same order 3 magic square when reflected in a horizontal axis:
:$\begin{array}{|c|c|c|}
\hline 4 & 3 & 8 \\
\hline 9 & 5 & 1 \\
\hline 2 & 7 & 6 \\
\hline
\end{array}$
\end{theorem}
\begin{proof}
For the rows:
{{begin-eqn}}
{{eqn | l = 276^2 + 951^2 + 438^2
| r = 76176 + 904401 + 191844
| c =
}}
{{eqn | r = 1172421
| c =
}}
{{eqn | l = 672^2 + 159^2 + 834^2
| r = 451584 + 25281 + 695556
| c =
}}
{{eqn | r = 1172421
| c =
}}
{{end-eqn}}
For the columns:
{{begin-eqn}}
{{eqn | l = 492^2 + 357^2 + 816^2
| r = 242064 + 127449 + 665856
| c =
}}
{{eqn | r = 1035369
| c =
}}
{{eqn | l = 294^2 + 753^2 + 618^2
| r = 86436 + 567009 + 381924
| c =
}}
{{eqn | r = 1035369
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22403
|
\section{Superabundant Numbers are Infinite in Number}
Tags: Superabundant Numbers
\begin{theorem}
There are infinitely many superabundant numbers.
\end{theorem}
\begin{proof}
{{AimForCont}} the set $S$ of superabundant numbers is finite.
Let $m$ be the greatest element of $S$.
By definition of superabundant, $m$ has the largest abundancy index of all the elements of $S$.
Consider the integer $2 m$.
From Abundancy Index of Product is greater than Abundancy Index of Proper Factors, $2 m$ has a higher abundancy index than $m$.
There are two possibilities:
:$(1): \quad 2 m$ is the smallest integer greater that $n$ which has a higher abundancy index than $m$.
By definition, that would make $m$ superabundant.
:$(2) \quad$ There exists a finite set $T := \set {n \in \Z: m < n < 2 m: \map A n > \map A m}$, where $\map A n$ denotes the abundancy index of $n$.
The smallest element $t$ of $T$ therefore has an abundancy index greater than all smaller positive integers.
Thus by definition $t$ is superabundant.
In either case, there exists a superabundant number not in $S$.
Thus $S$ cannot contain all superabundant numbers.
But this contradicts our initial assumption that the set $S$, containing all superabundant numbers is finite.
It follows by Proof by Contradiction that $S$ is infinite.
Hence the result.
{{qed}}
\end{proof}
|
22404
|
\section{Superset Relation is Compatible with Subset Product}
Tags: Compatible Relations
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $\circ_\PP$ be the subset product on $\powerset S$, the power set of $S$.
Then the superset relation $\supseteq$ is compatible with $\circ_\PP$.
\end{theorem}
\begin{proof}
By Subset Relation is Compatible with Subset Product, the subset relation $\subseteq$ is compatible with $\circ_\PP$.
From Inverse of Subset Relation is Superset, the inverse of $\subseteq$ is $\supseteq$.
The result follows from Inverse of Relation Compatible with Operation is Compatible.
{{qed}}
Category:Compatible Relations
\end{proof}
|
22405
|
\section{Superset of Co-Countable Set}
Tags: Set Theory
\begin{theorem}
Every superset of a co-countable set is co-countable.
\end{theorem}
\begin{proof}
Let $S$ be a set.
Let $A$ be co-countable in $S$, and let $B$ be such that $A \subseteq B \subseteq S$.
From Relative Complement inverts Subsets, it follows that:
:$\complement_S \left({B}\right) \subseteq \complement_S \left({A}\right)$
As $A$ is co-countable, $\complement_S \left({A}\right)$ is countable.
By Subset of Countably Infinite Set is Countable, it follows that $\complement_S \left({B}\right)$ is also countable.
Therefore, $B$ is also co-countable, and the result follows.
{{qed}}
\end{proof}
|
22406
|
\section{Superset of Dependent Set is Dependent}
Tags: Matroid Theory
\begin{theorem}
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $A, B \subseteq S$ such that $A \subseteq B$
If $A$ is a dependent subset then $B$ is a dependent subset.
\end{theorem}
\begin{proof}
From the contrapositive statement of matroid axiom $(\text I 2)$:
:$A \notin \mathscr I \implies B \notin \mathscr I$
By the definition of a dependent subset:
:If $A$ is not an dependent subset then $B$ is not an dependent subset.
{{qed}}
Category:Matroid Theory
\end{proof}
|
22407
|
\section{Superset of Dependent Set is Dependent/Corollary}
Tags: Matroid Theory
\begin{theorem}
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x \in A$.
If $x$ is a loop then $A$ is dependent.
\end{theorem}
\begin{proof}
Let $x$ be a loop.
By definition of a loop:
:$\set x \notin \mathscr I$
By definition of a dependent subset:
:$\set x$ is a dependent subset
From Singleton of Element is Subset:
:$\set x \subseteq A$
From Superset of Dependent Set is Dependent:
:$A$ is a dependent subset
{{qed}}
\end{proof}
|
22408
|
\section{Superset of Infinite Set is Infinite}
Tags: Set Theory
\begin{theorem}
Let $S$ be an infinite set.
Let $T \supseteq S$ be a superset of $S$.
Then $T$ is also infinite.
\end{theorem}
\begin{proof}
Suppose $T$ were finite.
Then by Set Finite iff Injection to Initial Segment of Natural Numbers, there is an injection $f: T \to \N_{<n}$ for some $n \in \N$.
But then by Restriction of Injection is Injection, also the restriction of $f$ to $S$:
:$f {\restriction_S}: S \to \N_{<n}$
is an injection.
Again by Set Finite iff Injection to Initial Segment of Natural Numbers, this contradicts the assumption that $S$ is infinite.
Hence $T$ is infinite.
{{qed}}
\end{proof}
|
22409
|
\section{Superset of Linearly Dependent Set is Linearly Dependent}
Tags: Unitary Modules, Linear Algebra, Proofs by Contraposition, Modules
\begin{theorem}
Let $S$ be a set of elements of a unitary module.
Let $S$ contain a subset $T$ which is linearly dependent.
Then $S$ is also linearly dependent.
\end{theorem}
\begin{proof}
Let $S$ be a linearly independent set.
Let $T$ be a subset of $S$.
By Subset of Linearly Independent Set is Linearly Independent, $T$ is also linearly independent.
Thus by Proof by Contraposition, if $T$ is linearly dependent, then so must $S$ be.
{{qed}}
\end{proof}
|
22410
|
\section{Superset of Neighborhood in Metric Space is Neighborhood}
Tags: Neighborhoods
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$ be a point in $M$.
Let $N$ be a neighborhood of $a$ in $M$.
Let $N \subseteq N' \subseteq A$.
Then $N'$ is a neighborhood of $a$ in $M$.
\end{theorem}
\begin{proof}
By definition of neighborhood:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq N$
where $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$ in $M$.
By Subset Relation is Transitive:
:$\map {B_\epsilon} a \subseteq N'$
The result follows by definition of neighborhood of $a$.
{{qed}}
\end{proof}
|
22411
|
\section{Superset of Neighborhood in Topological Space is Neighborhood}
Tags: Neighborhoods
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $N$ be a neighborhood of $x$ in $T$.
Let $N \subseteq N' \subseteq S$.
Then $N'$ is a neighborhood of $x$ in $T$.
That is:
:$\forall x \in S: \forall N \in \NN_x: N' \supseteq N \implies N' \in \NN_x$
where $\NN_x$ is the neighborhood filter of $x$.
\end{theorem}
\begin{proof}
By definition of neighborhood:
:$\exists U \in \tau: x \in U \subseteq N \subseteq S$
where $U$ is an open set of $T$.
By Subset Relation is Transitive:
:$U \subseteq N'$
The result follows by definition of neighborhood of $x$.
{{qed}}
\end{proof}
|
22412
|
\section{Superset of Order Generating is Order Generating}
Tags: Order Generating, Complete Lattices
\begin{theorem}
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X, Y$ be subsets of $S$ such that
:$X$ is order generating
and
:$X \subseteq Y$
Then $Y$ is order generating.
\end{theorem}
\begin{proof}
Let $x \in S$.
Thus by definition of complete lattice:
:$x^\succeq \cap Y$ admits an infimum.
By definition of complete lattice:
:$x^\succeq$ admits an infimum
and
:$x^\succeq \cap X$ admits an infimum.
By Intersection is Subset:
:$x^\succeq \cap Y \subseteq x^\succeq$
By Infimum of Subset:
:$\map \inf {x^\succeq} \preceq \map \inf {x^\succeq \cap Y}$
By Infimum of Upper Closure of Element:
:$x \preceq \map \inf {x^\succeq \cap Y}$
By Set Intersection Preserves Subsets/Corollary:
:$x^\succeq \cap X \subseteq x^\succeq \cap Y$
By Infimum of Subset:
:$\map \inf {x^\succeq \cap Y} \preceq \map \inf {x^\succeq \cap X}$
By definition of order generating:
:$\map \inf {x^\succeq \cap Y} \preceq x$
Thus by definition of antisymmetry:
:$x = \map \inf {x^\succeq \cap Y}$
{{qed}}
\end{proof}
|
22413
|
\section{Superset of Unsatisfiable Set is Unsatisfiable}
Tags: Formal Semantics
\begin{theorem}
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be an $\mathscr M$-unsatisfiable set of formulas from $\LL$.
Let $\FF'$ be a superset of $\FF$.
Then $\FF'$ is also $\mathscr M$-unsatisfiable.
\end{theorem}
\begin{proof}
By assumption, $\FF$ is unsatisfiable.
Suppose now $\FF'$ were satisfiable.
Then it would follow from Subset of Satisfiable Set is Satisfiable that $\FF$ were also satisfiable.
We conclude that $\FF'$ must be unsatisfiable.
{{qed}}
Category:Formal Semantics
\end{proof}
|
22414
|
\section{Superspace of Homeomorphic Subspaces may not have Homeomorphism to Itself containing Subspace Homeomorphism}
Tags: Homeomorphisms
\begin{theorem}
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.
Let $H_1$ and $H_2$ be a homeomorphic.
Then it may be the case that there does not exist a homeomorphism $g: T_1 \to T_2$ such that:
:$g \restriction_{H_1} = f$
where:
:$g \restriction_{H_1}$ is the restriction of $g$ to $H_1$
:$f: H_1 \to H_2$ is a homeomorphism.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $H_1 := \set 0 \cup \closedint 1 2 \cup \set 3$, where $\closedint 1 2$ denotes the closed interval from $1$ to $2$.
Let $H_2 := \closedint 0 1 \cup \set 2 \cup \set 3$.
$H_1$ and $H_2$ are homeomorphic, as can be demonstrated by the mapping $\phi: H_1 \to \H_2$ defined as:
:$\forall x \in H_1: \map \phi x = \begin {cases} 2 & : x = 0 \\ 3 & : x = 3 \\ x - 1 & : x \in \closedint 1 2 \end {cases}$
which is trivially a homeomorphism.
So, let $g$ be a homeomorphism from $H_1$ and $H_2$.
Each of the singletons in $H_1$ has to map to one of the singletons in $H_2$.
As a result, $g$ is not monotone.
Let $f: \R \to \R$ be a homeomorphism such that $g$ is a restriction of $f$.
From Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone, $f$ is monotone.
{{explain|We need another result for the above over the whole range $\R$ -- for some reason this does not exist. We have Surjective Monotone Function is Continuous but not in the other direction.}}
Hence $f$ cannot be a bijection.
Hence $f$ cannot be a homeomorphism.
{{qed}}
\end{proof}
|
22415
|
\section{Supplementary Interior Angles implies Parallel Lines}
Tags: Parallel Lines, Angles, Lines, Transversals (Geometry)
\begin{theorem}
Given two infinite straight lines which are cut by a transversal, if the interior angles on the same side of the transversal are supplementary, then the lines are parallel.
{{:Euclid:Proposition/I/28}}
\end{theorem}
\begin{proof}
:200px
Let $AB$ and $CD$ be infinite straight lines.
Let $EF$ be a transversal that cuts them.
Let at least one pair of interior angles on the same side of the transversal be supplementary.
{{WLOG}}, let those interior angles be $\angle BGH$ and $\angle DHG$.
So, by definition, $\angle DHG + \angle BGH$ equals two right angles.
Also, from Two Angles on Straight Line make Two Right Angles, $\angle AGH + \angle BGH$ equals two right angles.
Then from Euclid's first and third common notion and Euclid's fourth postulate:
:$\angle AGH = \angle DHG$
Finally, by Equal Alternate Angles implies Parallel Lines:
:$AB \parallel CD$
{{qed}}
{{Euclid Note|28|I|{{EuclidNoteConverse|prop = 29|title = Parallelism implies Supplementary Interior Angles|part = third}}|part = second}}
\end{proof}
|
22416
|
\section{Suprema Preserving Mapping on Ideals Preserves Directed Suprema}
Tags: Order Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be ordered sets.
Let $f: S \to T$ be a mapping.
Let every filter $F$ in $\left({S, \preceq}\right)$, $f$ preserve the infimum on $F$.
Then $f$ preserves directed suprema.
\end{theorem}
\begin{proof}
This follows by mutatis mutandis of the proof of Infima Preserving Mapping on Filters Preserves Filtered Infima.
{{qed}}
\end{proof}
|
22417
|
\section{Suprema Preserving Mapping on Ideals is Increasing}
Tags: Increasing Mappings, Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered sets.
Let $f: S \to T$ be a mapping.
For every ideal $I$ in $\struct {S, \preceq}$, let $f$ preserve the supremum on $I$.
Then $f$ is increasing.
\end{theorem}
\begin{proof}
Let $x, y \in S$ such that:
:$x \preceq y$
By Supremum of Singleton:
:$\set x$ and $\set y$ admit suprema in $\struct {S, \preceq}$
By Supremum of Lower Closure of Set:
:$\set x^\preceq$ and $\set y^\preceq$ admit suprema in $\struct {S, \preceq}$
where $\set x^\preceq$ denotes the lower closure of $\set x$
By Lower Closure of Singleton:
:$x^\preceq$ and $y^\preceq$ admit suprema in $\struct {S, \preceq}$
By Lower Closure of Element is Ideal:
:$x^\preceq$ and $y^\preceq$ are ideals in $\struct {S, \preceq}$
By assumption and definition of mapping preserves the supremum on subset:
:$\map {f^\to} {x^\preceq}$ and $\map {f^\to} {y^\preceq}$ admit suprema in $\struct {T, \precsim}$
and:
:$\map \sup {\map {f^\to} {x^\preceq} } = \map f {\map \sup {x^\preceq} }$
and:
:$\map \sup {\map {f^\to} {y^\preceq} } = \map f {\map \sup {y^\preceq} }$
By Supremum of Lower Closure of Element:
:$\map \sup {x^\preceq} = x$ and $\map \sup {y^\preceq} = y$
By Lower Closure is Increasing:
:$x^\preceq \subseteq y^\preceq$
By Image of Subset under Relation is Subset of Image/Corollary 2:
:$\map {f^\to} {x^\preceq} \subseteq \map {f^\to} {y^\preceq}$
Thus by Supremum of Subset:
:$\map f x \precsim \map f y$
Thus by definition:
:$f$ is increasing.
{{qed}}
\end{proof}
|
22418
|
\section{Suprema and Infima of Combined Bounded Functions/Bounded Above}
Tags: Suprema and Infima of Combined Bounded Functions
\begin{theorem}
Let $f$ and $g$ be real functions.
Let $c$ be a constant.
Let both $f$ and $g$ be bounded above on $S \subseteq \R$.
Then:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$
where $\ds \map \sup {\map f x}$ is the supremum of $\map f x$.
\end{theorem}
\begin{proof}
First we show that:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
Let $T = \set {\map f x: x \in S}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\sup_{x \mathop \in S} } {\map f x + c}
| r = \map {\sup_{y \mathop \in T} } {y + c}
| c =
}}
{{eqn | r = c + \map {\sup_{y \mathop \in T} } y
| c = Supremum Plus Constant
}}
{{eqn | r = c + \map {\sup_{x \mathop \in S} } {\map f x}
| c =
}}
{{end-eqn}}
Next we show that $\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$:
Let:
:$\ds H = \map {\sup_{x \mathop \in S} } {\map f x}$
:$\ds K = \map {\sup_{x \mathop \in S} } {\map g x}$
Then:
:$\forall x \in S: \map f x + \map g x \le H + K$
Hence $H + K$ is an upper bound for $\set {\map f x + \map g x: x \in S}$.
The result follows.
{{qed}}
\end{proof}
|
22419
|
\section{Suprema and Infima of Combined Bounded Functions/Bounded Below}
Tags: Suprema and Infima of Combined Bounded Functions
\begin{theorem}
Let $f$ and $g$ be real functions.
Let $c$ be a constant.
Let both $f$ and $g$ be bounded below on $S \subseteq \R$.
Then:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$
where $\ds \map \inf {\map f x}$ is the infimum of $\map f x$.
\end{theorem}
\begin{proof}
First we show that:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
Let $T = \set {\map f x: x \in S}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\inf_{x \mathop \in S} } {\map f x + c}
| r = \map {\inf_{y \mathop \in T} } {y + c}
| c =
}}
{{eqn | r = c + \map {\sup_{y \mathop \in T} } y
| c = Infimum Plus Constant
}}
{{eqn | r = c + \map {\inf_{x \mathop \in S} } {\map f x}
| c =
}}
{{end-eqn}}
Next we show that:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$
Let:
:$\ds H = \map {\inf_{x \mathop \in S} } {\map f x}$
:$\ds K = \map {\inf_{x \mathop \in S} } {\map g x}$
Then:
:$\forall x \in S: \map f x + g \map x \ge H + K$
Hence $H + K$ is a lower bound for $\set {\map f x + \map g x: x \in S}$.
The result follows.
{{qed}}
\end{proof}
|
22420
|
\section{Supremum Inequality for Ordinals}
Tags: Ordinals
\begin{theorem}
Let $A \subseteq \On$ and $B \subseteq \On$ be ordinals.
Then:
:$\ds \forall x \in A: \exists y \in B: x \le y \implies \bigcup A \le \bigcup B$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x
| o = <
| r = \bigcup A
}}
{{eqn | ll= \leadsto
| q = \exists z
| l = x
| o = <
| r = z
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \land
| l = z < A
| o = <
| r = A
| c =
}}
{{eqn | ll= \leadsto
| q = \exists y \in B: \exists z
| l = x
| o = <
| r = z
| c = {{hypothesis}}
}}
{{eqn | lo= \land
| l = z
| o = \le
| r = y
| c =
}}
{{eqn | ll= \leadsto
| q = \exists y \in B
| l = x
| o = <
| r = y
}}
{{eqn | ll= \leadsto
| l = x
| o = <
| r = \bigcup B
| c = Union of Ordinals is Least Upper Bound
}}
{{end-eqn}}
Therefore:
:$\ds \bigcup A \subseteq \bigcup B$
and:
:$\ds \bigcup A \le \bigcup B$
{{qed}}
\end{proof}
|
22421
|
\section{Supremum Metric and L1 Metric on Closed Real Intervals are not Topologically Equivalent}
Tags: L1 Metric, Supremum Metric
\begin{theorem}
Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$.
Let $d_1$ be the $L^1$ metric on $S$:
:$\ds \forall f, g \in S: \map {d_1} {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$
Let $d_\infty$ be the supremum metric on $S$:
:$\ds \forall f, g \in S: \map {d_\infty} {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$
Then $d_1$ and $d_\infty$ are not topologically equivalent.
\end{theorem}
\begin{proof}
Let $f, g \in S$.
Then by definition of supremum metric:
:$\forall x \in \closedint a b: \size {\map f x - \map g x} \le \map {d_\infty} {f, g}$
Hence by ...
:$\map {d_1} {f, g} = \ds \int_a^b \size {\map f x - \map g x} \rd x \le \paren {b - a} \map {d_\infty} {f, g}$
{{finish|Find that link}}
and so:
:$\map {B_\epsilon} {f; d_\infty} \subseteq \map {b_{\paren {b - a} } } {f; d_1}$
Now let $f_0$ denote the constant mapping:
:$\forall x: \map {f_0} x = 0$
We show that $\map {B_1} {f_0; d_\infty}$ is not $d_1$-open.
{{AimForCont}} instead that $\map {B_1} {f_0; d_\infty}$ is $d_1$-open.
Then we should have:
:$\map {B_1} {f_0; d_1} \subseteq \map {B_1} {f_0; d_\infty}$
for some $\epsilon \in \R_{>0}$.
But for $\epsilon > 0$ there exists a continuous function on $\closedint a b$ such that:
:$\map {d_1} {f, f_0} < \epsilon$
yet:
:$\map {d_\infty} {f, f_0} = 1$
So:
:$f \in \map {B_1} {f_0; d_1}$
but:
:$f \notin \map {B_1} {f_0; d_\infty}$
which contradicts our deduction that $\map {B_1} {f_0; d_1} \subseteq \map {B_1} {f_0; d_\infty}$.
Hence it cannot be the case that $\map {B_1} {f_0; d_\infty}$ is $d_1$-open.
The result follows.
{{qed}}
\end{proof}
|
22422
|
\section{Supremum Metric is Metric}
Tags: Supremum Metric
\begin{theorem}
Let $S$ be a set.
Let $M = \struct {A', d'}$ be a metric space.
Let $A$ be the set of all bounded mappings $f: S \to M$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
\end{theorem}
\begin{proof}
We have that the supremum metric on $A \times A$ is defined as:
:$\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x}$
where $f$ and $g$ are bounded mappings.
First note that we have:
{{begin-eqn}}
{{eqn | l = \size {\map f x - \map g x}
| r = \size {\map f x + \paren {-\map g x} }
| c =
}}
{{eqn | o = \le
| r = \size {\map f x} + \size {\paren {-\map g x} }
| c = Triangle Inequality for Real Numbers
}}
{{eqn | r = \size {\map f x} + \size {\map g x}
| c = {{Defof|Absolute Value}}
}}
{{eqn | o = \le
| r = K + L
| c =
}}
{{end-eqn}}
and so the {{RHS}} exists.
\end{proof}
|
22423
|
\section{Supremum Metric on Bounded Continuous Mappings is Metric}
Tags: Supremum Metric
\begin{theorem}
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $A$ be the set of all continuous mappings $f: M_1 \to M_2$ which are also bounded.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
\end{theorem}
\begin{proof}
The set $A$ is a subset of the set $A'$ of all bounded mappings $f: M_1 \to M_2$.
Let $d': A' \times A' \to \R$ be the supremum metric on $A'$.
From Supremum Metric is Metric, $\struct {A', d'}$ is a metric space.
By definition, $A$ is a metric subspace of $A'$.
Hence the result.
{{qed}}
\end{proof}
|
22424
|
\section{Supremum Metric on Bounded Real-Valued Functions is Metric}
Tags: Supremum Metric on Bounded Real-Valued Functions is Metric, Supremum Metric
\begin{theorem}
Let $X$ be a set.
Let $A$ be the set of all bounded real-valued functions $f: X \to \R$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
\end{theorem}
\begin{proof}
We have that the supremum metric on $A \times A$ is defined as:
:$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in X} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert$
where $f$ and $g$ are bounded real-valued functions.
So:
:$\exists K, L \in \R: \left\vert{f \left({x}\right)}\right\vert \le K, \left\vert{g \left({x}\right)}\right\vert \le L$
for all $x \in X$.
First note that we have:
{{begin-eqn}}
{{eqn | l = \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert
| r = \left\vert{f \left({x}\right) + \left({- g \left({x}\right)}\right)}\right\vert
| c =
}}
{{eqn | o = \le
| r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{\left({- g \left({x}\right)}\right)}\right\vert
| c = Triangle Inequality for Real Numbers
}}
{{eqn | r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{g \left({x}\right)}\right\vert
| c = Definition of Absolute Value
}}
{{eqn | o = \le
| r = K + L
| c =
}}
{{end-eqn}}
and so the RHS exists.
\end{proof}
|
22425
|
\section{Supremum Metric on Bounded Real Functions on Closed Interval is Metric}
Tags: Supremum Metric, Supremum Metric on Bounded Real Functions on Closed Interval is Metric
\begin{theorem}
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $A$ be the set of all bounded real functions $f: \closedint a b \to \R$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
\end{theorem}
\begin{proof}
We have that the supremum metric on $A \times A$ is defined as:
:$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert$
where $f$ and $g$ are bounded real functions.
So:
:$\exists K, L \in \R: \left\vert{f \left({x}\right)}\right\vert \le K, \left\vert{g \left({x}\right)}\right\vert \le L$
for all $x \in \left[{a \,.\,.\, b}\right]$.
First note that we have:
{{begin-eqn}}
{{eqn | l = \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert
| r = \left\vert{f \left({x}\right) + \left({- g \left({x}\right)}\right)}\right\vert
| c =
}}
{{eqn | o = \le
| r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{\left({- g \left({x}\right)}\right)}\right\vert
| c = Triangle Inequality for Real Numbers
}}
{{eqn | r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{g \left({x}\right)}\right\vert
| c = Definition of Absolute Value
}}
{{eqn | o = \le
| r = K + L
| c =
}}
{{end-eqn}}
and so the RHS exists.
\end{proof}
|
22426
|
\section{Supremum Metric on Bounded Real Sequences is Metric}
Tags: Supremum Metric on Bounded Real Sequences is Metric, Supremum Metric
\begin{theorem}
Let $A$ be the set of all bounded real sequences.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
\end{theorem}
\begin{proof}
We have that the supremum metric on $A \times A$ is defined as:
:$\displaystyle \forall x, y \in A: d \left({x, y}\right) := \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert$
where $x = \left\langle{x_i}\right\rangle$ and $y = \left\langle{y_i}\right\rangle$ are bounded real sequences.
So:
:$\exists K, L \in \R: \left\vert{x_n}\right\vert \le K, \left\vert{y_n}\right\vert \le L$
for all $n \in \N$.
First note that we have:
{{begin-eqn}}
{{eqn | l = \left\vert{x_n - y_n}\right\vert
| r = \left\vert{x_n + \left({- y_n}\right)}\right\vert
| c =
}}
{{eqn | o = \le
| r = \left\vert{x_n}\right\vert + \left\vert{\left({- y_n}\right)}\right\vert
| c = Triangle Inequality for Real Numbers
}}
{{eqn | r = \left\vert{x_n}\right\vert + \left\vert{y_n}\right\vert
| c = Definition of Absolute Value
}}
{{eqn | o = \le
| r = K + L
| c =
}}
{{end-eqn}}
and so the RHS exists.
\end{proof}
|
22427
|
\section{Supremum Metric on Continuous Real Functions is Subspace of Bounded}
Tags: Supremum Metric
\begin{theorem}
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $\mathscr C \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$.
Let $\map {\mathscr B} {\closedint a b, \R}$ be the set of all bounded real functions $f: \closedint a b \to \R$.
Let $d$ be the supremum metric on $\map {\mathscr B} {\closedint a b, \R}$.
Then $\struct {\mathscr C \closedint a b, d_{\mathscr C} }$ is a subspace of $\struct {\map {\mathscr B} {\closedint a b, \R}, d}$.
\end{theorem}
\begin{proof}
Let $f \in \mathscr C \closedint a b$.
Then by Image of Closed Real Interval is Bounded, $f$ is bounded on $\closedint a b$.
Thus $f \in \map {\mathscr B} {\closedint a b, \R}$ and the result follows.
{{qed}}
\end{proof}
|
22428
|
\section{Supremum Metric on Differentiability Class is Metric}
Tags: Supremum Metric
\begin{theorem}
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $r \in \N$ be a natural number.
Let $A := \mathscr D^r \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$ which are of differentiability class $r$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
\end{theorem}
\begin{proof}
We have that the supremum metric on $A \times A$ is defined as:
:$\ds \forall f, g \in A: \map d {f, g} := \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {g^{\paren i} } x}$
where $f$ and $g$ are continuous functions on $\closedint a b$ which are of differentiability class $r$.
\end{proof}
|
22429
|
\section{Supremum Norm is Norm}
Tags: Supremum Norm, Examples of Norms, Supremum Norm is Norm, Norm Examples, Functional Analysis
\begin{theorem}
Let $S$ be a set.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $K \in \set {\R, \C}$.
Let $\BB$ be the set of bounded mappings $S \to X$.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\BB$.
Then $\norm {\, \cdot \,}_\infty$ is a norm on $\BB$.
{{MissingLinks|Add a link that establishes that $\BB$ is a vector space}}
\end{theorem}
\begin{proof}
First:
{{begin-eqn}}
{{eqn | l = \norm f_\infty
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| l = \sup_{x \mathop \in S} \norm {\map f x}
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in S
| l = \norm {\map f x}
| r = 0
| c = since $\norm {\, \cdot \,}$ is a norm, and hence non-negative
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in S
| l = \map f x
| r = 0
| c = since $\norm x = 0 \iff x = 0$
}}
{{eqn | ll= \leadstoandfrom
| l = f
| r = 0
| c =
}}
{{end-eqn}}
Now let $\lambda \in K, f \in \BB$
We have:
{{begin-eqn}}
{{eqn | l = \norm {\lambda f}_\infty
| r = \sup_{x \mathop \in S} \norm {\lambda \map f x}
| c =
}}
{{eqn | r = \size \lambda_K \sup_{x \mathop \in S} \norm {\map f x}
| c = Multiple of Supremum, and because $\norm {\, \cdot \,}$ is a norm
}}
{{eqn | r = \size \lambda_K \, \norm f_\infty
| c =
}}
{{end-eqn}}
Finally let $f, g \in \BB$.
We have:
{{begin-eqn}}
{{eqn | l = \norm {f + g}_\infty
| r = \sup_{x \mathop \in S} \norm {\map f x + \map g x}
| c =
}}
{{eqn | o = \le
| r = \sup_{x \mathop \in S} \paren {\norm {\lambda \map f x} + \norm {\lambda \map g x} }
| c = because $\norm {\, \cdot \,}$ is a norm
}}
{{eqn | o = \le
| r = \sup_{x \mathop \in S} \norm {\lambda \map f x} + \sup_{x \mathop \in S} \norm {\lambda \map g x}
| c = Supremum of Sum
}}
{{eqn | r = \norm f_\infty + \norm g_\infty
| c =
}}
{{end-eqn}}
Thus $\norm {\, \cdot \,}_\infty$ has the defining properties of a norm.
{{qed}}
Category:Examples of Norms
Category:Supremum Norm
Category:Supremum Norm is Norm
\end{proof}
|
22430
|
\section{Supremum Norm on Vector Space of Real Matrices is Norm}
Tags: Examples of Norms
\begin{theorem}
Supremum Norm forms a norm on the vector space of real matrices.
\end{theorem}
\begin{proof}
Let $M \in \R^{m \times n}: m, n \in \N_{>0}$ be a real matrix.
Denote the $\paren {i, j}$-th entry of $M$ by $a_{i j}$.
Note that the set of matrix elements of $M$ is a finite set of real numbers.
We have that:
:Real Numbers form Ordered Field
:Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements
Therefore, $M$ has the greatest element.
\end{proof}
|
22431
|
\section{Supremum Plus Constant}
Tags: Real Analysis, Analysis
\begin{theorem}
Let $S$ be a subset of the set of real numbers $\R$.
Let $S$ be bounded above.
Let $\xi \in \R$.
Then:
:$\ds \map {\sup_{x \mathop \in S} } {x + \xi} = \xi + \map {\sup_{x \mathop \in S} } x$
where $\sup$ denotes supremum.
\end{theorem}
\begin{proof}
Let $B = \sup S$.
Let $T = \set {x + \xi: x \in S}$.
Since $\forall x \in S: x \le B$ it follows that:
:$\forall x \in S: x + \xi \le B + \xi$
Hence $\xi + B$ is an upper bound for $T$.
If $C$ is the supremum for $T$ then $C \le \xi + B$.
On the other hand:
:$\forall y \in T: y \le C$
Therefore:
:$\forall y \in T: y - \xi \le C - \xi$
Since $S = \set {y - \xi: y \in T}$ it follows that $C - \xi$ is an upper bound for $S$ and so $B \le C - \xi$.
So we have shown that $C \le \xi + B$ and $C \ge \xi + B$, hence the result.
{{qed}}
\end{proof}
|
22432
|
\section{Supremum by Suprema of Directed Set in Simple Order Product}
Tags: Simple Order Product, Up-Complete Semilattices
\begin{theorem}
Let $\struct {S, \preceq}$ be an up-complete meet semilattice.
Let $\struct {S \times S, \precsim}$ be the simple order product of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.
Let $D$ be a directed subset of $S \times S$.
Then:
:$\sup D = \tuple {\map \sup {\map {\pr_1^\to} D}, \map \sup {\map {\pr_2^\to} D} }$
where
:$\pr_1$ denotes the first projection on $S \times S$
:$\pr_2$ denotes the second projection on $S \times S$
:$\map {\pr_1^\to} D$ denotes the image of $D$ under $\pr_1$
\end{theorem}
\begin{proof}
By Up-Complete Product:
:$\struct {S \times S, \precsim}$ is up-complete.
By definition of up-complete:
:$D$ admits a supremum.
By definition of Cartesian product:
:$\exists d_1, d_2 \in S: \sup D = \tuple {d_1, d_2}$
By Up-Complete Product/Lemma 2:
:$D_1 := \map {\pr_1^\to} D$ is directed
and
:$D_2 := \map {\pr_2^\to} D$ is directed.
By definition of up-complete:
:$D_1$ admits a supremum
and
:$D_2$ admits a supremum
We will prove that
:$d_2$ is upper bound for $D_2$
Let $x \in D_2$.
By definition of image of set:
:$\exists \tuple {a, b} \in D: \map {\pr_2} {a, b} = x$
By definition of second projection:
$b = x$
By definition of supremum:
:$\tuple {d_1, d_2}$ is upper bound for $D$.
By definition of upper bound:
:$\tuple {a, x} \precsim \tuple {d_1, d_2}$
Thus by definition of simple order product:
:$x \preceq d_2$
{{qed|lemma}}
Analogically we have that
:$d_1$ is upper bound for $D_1$
By definition of supremum:
:$\sup D_1 \preceq d_1$ and $\sup D_2 \preceq d_2$
By definition of simple order product:
:$\tuple {\sup D_1, \sup D_2} \precsim \sup D$
By Up-Complete Product/Lemma 1:
:$D_1 \times D_2$ is directed.
By definition of up-complete:
:$D_1 \times D_2$ admits a supremum.
By definition of subset:
:$D \subseteq D_1 \times D_2$
By Supremum of Subset:
:$\sup D \precsim \map \sup {D_1 \times D_2}$
By Supremum of Simple Order Product:
:$\sup D \precsim \tuple {\sup D_1, \sup D_2}$
Thus by definition of antisymmetry:
:$\sup D = \tuple {\sup D_1, \sup D_2}$
{{qed}}
\end{proof}
|
22433
|
\section{Supremum does not Precede Infimum}
Tags: Order Theory, Orderings, Infima, Suprema
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ admit both a supremum $M$ and an infimum $m$.
Then $m \preceq M$.
\end{theorem}
\begin{proof}
By definition of supremum:
:$\forall a \in T: a \preceq M$
By definition of infimum:
:$\forall a \in T: m \preceq a$
The result follows from transitivity of ordering.
{{qed}}
\end{proof}
|
22434
|
\section{Supremum in Ordered Subset}
Tags: Order Theory
\begin{theorem}
Let $L = \left({S, \preceq}\right)$ be an ordered set.
Let $R = \left({T, \preceq'}\right)$ be an ordered subset of $L$.
Let $X \subseteq T$ such that
:$X$ admits an supremum in $L$.
Then $\sup_L X \in T$ {{iff}}
:$X$ admits an supremum in $R$ and $\sup_R X = \sup_L X$
\end{theorem}
\begin{proof}
This follows by mutatis mutandis of the proof of Infimum in Ordered Subset.
{{qed}}
\end{proof}
|
22435
|
\section{Supremum is Coproduct in Order Category}
Tags: Poset Categories, Order Categories
\begin{theorem}
Let $\mathbf P$ be an order category with ordering $\preceq$.
Let $p, q \in P_0$, and suppose they have some supremum $r = \sup \left\{{p, q}\right\}$.
Then $r$ is the coproduct of $p$ and $q$ in $\mathbf P$.
\end{theorem}
\begin{proof}
Let $\mathbf P^{\text{op}}$ be the dual category of $\mathbf P$.
From Dual of Order Category, it is the order category corresponding to the dual ordering $\succeq$.
From Dual Pairs (Order Theory), it follows that in $\mathbf P^{\text{op}}$:
:$r = \inf \left\{{p, q}\right\}$
where $\inf$ denotes infimum.
By Infimum is Product in Order Category, $r$ is the product of $p$ and $q$ in $\mathbf P^{\text{op}}$.
By Dual Pairs (Category Theory), $r$ is the coproduct of $p$ and $q$ in $\mathbf P$.
{{qed}}
\end{proof}
|
22436
|
\section{Supremum is Dual to Infimum}
Tags: Order Theory, Infima, Suprema
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$ and $T \subseteq S$.
The following are dual statements:
:$a$ is a supremum for $T$
:$a$ is an infimum for $T$
\end{theorem}
\begin{proof}
By definition, $a$ is a supremum for $T$ {{iff}}:
:$a$ is an upper bound for $T$
:$a \preceq b$ for all upper bounds $b$ of $T$
The dual of this statement is:
:$a$ is a lower bound for $T$
:$b \preceq a$ for all lower bounds $b$ of $T$
by Dual Pairs (Order Theory).
By definition, this means $a$ is an infimum for $T$.
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}}
\end{proof}
|
22437
|
\section{Supremum is Increasing relative to Product Ordering}
Tags: Increasing Mappings, Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $I$ be a set.
Let $f, g: I \to S$.
Let $f \sqbrk I$ denote the image of $I$ under $f$.
Let:
:$\forall i \in I: \map f i \preceq \map g i$
That is, let $f \preceq g$ in the product ordering.
Let $f \sqbrk I$ and $g \sqbrk I$ admit suprema.
Then:
:$\sup f \sqbrk I \preceq \sup g \sqbrk I$
\end{theorem}
\begin{proof}
Let $x \in f \sqbrk I$.
Then:
:$\exists j \in I: \map f j = x$
Then:
:$\map f j \prec \map g j$
By the definition of supremum:
:$\sup g \sqbrk I$ is an upper bound of $g \sqbrk I$
Thus:
:$\map g j \preceq \sup g \sqbrk I$
Since $\preceq$ is transitive:
:$x = \map f j \preceq \sup g \sqbrk I$
Since this holds for all $x \in f \sqbrk I$, $\sup g \sqbrk I$ is an upper bound of $f \sqbrk I$.
Thus by the definition of supremum:
:$\sup f \sqbrk I \preceq \sup g \sqbrk I$
{{qed}}
Category:Order Theory
Category:Increasing Mappings
\end{proof}
|
22438
|
\section{Supremum is Unique}
Tags: Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a non-empty subset of $S$.
Then $T$ has at most one supremum in $S$.
\end{theorem}
\begin{proof}
Let $c$ and $c'$ both be suprema of $T$ in $S$.
From the definition of supremum, $c$ and $c'$ are upper bounds of $T$ in $S$.
By that definition:
:$c$ is an upper bound of $T$ in $S$ and $c'$ is a supremum of $T$ in $S$ implies that $c' \preceq c$
:$c'$ is an upper bound of $T$ in $S$ and $c$ is a supremum of $T$ in $S$ implies that $c \preceq c'$.
So:
:$c' \preceq c \land c \preceq c'$
and thus by the antisymmetry of the ordering $\preceq$:
:$c = c'$
{{Qed}}
\end{proof}
|
22439
|
\section{Supremum is not necessarily Greatest Element}
Tags: Suprema, Order Theory, Supremum is not necessarily Greatest Element
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ admit a supremum in $S$.
Then the supremum of $T$ in $S$ is not necessarily the greatest element of $T$.
\end{theorem}
\begin{proof}
Consider the subset $T$ of the set of real numbers $\R$:
:$T := \set {x \in \R: 1 \le x < 2}$
The number $2$ cannot be the greatest element of $T$ as $2 \notin T$.
However, $2$ is the supremum of $T$ in $S$.
Indeed, by definition:
:$\forall x \in T: x < 2$
So, let $x < 2$.
Then consider $y = \dfrac x 1 + \dfrac 2 1 = \dfrac {x + 2} 2$.
We have that $x < y < 2$ by Mediant is Between.
Thus $y \in T$ but $y > x$ and so $x$ cannot be the greatest element of $T$.
Neither can $y$ be the supremum of $T$ in $S$.
The conclusion is that there is no greatest element of $T$.
Hence the result.
{{qed}}
\end{proof}
|
22440
|
\section{Supremum of Absolute Value of Difference equals Difference between Supremum and Infimum}
Tags: Real Analysis
\begin{theorem}
Let $f$ be a real function.
Let $S$ be a subset of the domain of $f$.
Let $\ds \sup_{x \mathop \in S} \set {\map f x}$ and $\ds \inf_{x \mathop \in S} \set {\map f x}$ exist.
Then $\ds \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} }$ exists and:
:$\ds \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} } = \sup_{x \mathop \in S} \set {\map f x} - \inf_{x \mathop \in S} \set {\map f x}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sup_{x \mathop \in S} \set {\map f x} - \inf_{x \mathop \in S} \set {\map f x}
| r = \sup_{x \mathop \in S} \set {\map f x} + \sup_{x \mathop \in S} \set {-\map f x}
| c = Negative of Infimum is Supremum of Negatives
}}
{{eqn | r = \sup_{x, y \mathop \in S} \set {\map f x + \paren {-\map f y} }
| c = Supremum of Sum equals Sum of Suprema
}}
{{eqn | r = \sup_{x, y \mathop \in S} \set {\map f x - \map f y}
}}
{{eqn | r = \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} }
| c = Supremum of Absolute Value of Difference equals Supremum of Difference
}}
{{end-eqn}}
{{qed}}
Category:Real Analysis
\end{proof}
|
22441
|
\section{Supremum of Absolute Value of Difference equals Supremum of Difference}
Tags: Suprema, Real Analysis, Absolute Value Function
\begin{theorem}
Let $S$ be a non-empty real set.
Let $\ds \sup_{x, y \mathop \in S} \paren {x - y}$ exist.
Then $\ds \sup_{x, y \mathop \in S} \size {x - y}$ exists and:
:$\ds \sup_{x, y \mathop \in S} \size {x - y} = \sup_{x, y \mathop \in S} \paren {x - y}$
\end{theorem}
\begin{proof}
Consider the set $\set {x - y: x, y \in S, x - y \le 0}$.
There is a number $x'$ in $S$ as $S$ is non-empty.
Therefore, $0 \in \set {x - y: x, y \in S, x - y \le 0}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, and $x - y \le 0$.
Also, $0$ is an upper bound for $\set {x - y: x, y \in S, x - y \le 0}$ by definition.
Accordingly:
:$\ds \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$
Consider the set $\left\{{x - y: x, y \in S, x - y \ge 0}\right\}$.
There is a number $x'$ in $S$ as $S$ is non-empty.
Therefore, $0 \in \left\{{x - y: x, y \in S, x - y \ge 0}\right\}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, $x - y \ge 0$.
Accordingly:
:$\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$
{{improve|I can't immediately think of how it would be done, but it would be good if we could devise a neater and more compact notation that what is used here. All the complicated mathematics is being done in the underscript, which makes it not easy to follow. (Improved Dec. 2016.)}}
{{begin-eqn}}
{{eqn | l = \sup_{x, y \mathop \in S} \paren {x - y}
| r = \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \paren {x - y}
| c = as ($x - y \ge 0$ or $x - y \le 0$) is true
}}
{{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} }
| c = by Supremum of Set Equals Maximum of Suprema of Subsets
}}
{{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}, 0}
| c = as $\ds \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$
}}
{{eqn | r = \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}
| c = as $\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$
}}
{{eqn | r = \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}
| c = as $\size {x − y} = x − y$ since $x − y \ge 0$
}}
{{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y} }
| c = as the two arguments of max are equal
}}
{{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{y, x \mathop \in S, y − x \mathop \ge 0} \size {y - x} }
| c = by renaming variables $x \leftrightarrow y$
}}
{{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \size {x - y} }
| c =
}}
{{eqn | r = \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \size {x - y}
| c = by Supremum of Set Equals Maximum of Suprema of Subsets
}}
{{eqn | r = \sup_{x, y \mathop \in S} \size {x - y}
| c = as ($x - y \ge 0$ or $x - y \le 0$) is true
}}
{{end-eqn}}
{{qed}}
Category:Suprema
Category:Absolute Value Function
\end{proof}
|
22442
|
\section{Supremum of Bounded Above Set of Reals is in Closure}
Tags: Set Closures, Real Analysis, Analysis
\begin{theorem}
Let $\R$ be the real number line under the Euclidean metric.
Let $H \subseteq \R$ be a bounded above subset of $\R$ such that $H \ne \O$.
Let $u = \sup H$ be the supremum of $H$.
Then:
:$u \in \map \cl H$
where $\map \cl H$ denotes the closure of $H$ in $\R$.
\end{theorem}
\begin{proof}
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {B_\epsilon} u$ be the open $\epsilon$-ball of $u$ in $\R$.
From Distance from Subset of Real Numbers:
:$\map d {u, H} = 0$
Thus by definition of distance from subset:
:$\exists x \in H: \map d {u, x} < \epsilon$
Thus $x \in \map {B_\epsilon} u$.
As $x \in H$ and $x \in \map {B_\epsilon} u$, from the definition of intersection:
:$x \in H \cap \map {B_\epsilon} u$
The result follows from Condition for Point being in Closure.
{{qed}}
\end{proof}
|
22443
|
\section{Supremum of Elements of Sublattice not necessarily Same as for Lattice}
Tags: Sublattices, Suprema
\begin{theorem}
Let $\struct {S, \preceq}$ be a '''lattice'''.
Let $\struct {T, \preceq_T}$ be a '''sublattice''' of $S$.
Let $a, b \in T$.
Then it is not necessarily the case that:
:$\sup_S \set {a, b}$
is the same as:
:$\sup_T \set {a, b}$
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Let $\powerset G$ denote the power set of $G$.
Let $\struct {\powerset G, \subseteq}$ be the complete lattice formed by $\powerset G$ and $\subseteq$.
From Power Set is Complete Lattice, $\struct {\powerset G, \subseteq}$ indeed forms a complete lattice.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by $\mathbb G$ and $\subseteq$.
From Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ indeed forms a complete lattice.
By definition, $\struct {\mathbb G, \subseteq}$ is a sublattice of $\struct {\powerset G, \subseteq}$.
Let $H, K \in \mathbb G$.
We have that:
:$H \cup K$ is the supremum of $H$ and $K$ in $\struct {\powerset G, \subseteq}$.
But from Union of Subgroups, $H \cup K$ is not a subgroup of $G$ element of $\mathbb G$.
However, from Supremum of Subgroups in Lattice, we do have that the subset product $H K$ in $G$ is a subgroup of $G$ such that:
:$\sup_{\mathbb G} \set {H, K} = H K$
{{qed}}
\end{proof}
|
22444
|
\section{Supremum of Empty Set is Smallest Element}
Tags: Empty Set, Order Theory, Suprema
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Then:
:the supremum of the empty set exists {{iff}} the smallest element of $S$ exists
in which case:
:$\map \sup \O$ is the smallest element of $S$
\end{theorem}
\begin{proof}
Observe that, vacuously, any $s \in S$ is an upper bound for $\O$.
\end{proof}
|
22445
|
\section{Supremum of Function is less than Supremum of Greater Function}
Tags: Real Analysis, Suprema
\begin{theorem}
Let $f$ and $g$ be real functions.
Let $S$ be a subset of $\Dom f \cap \Dom g$.
Let $\map f x \le \map g x$ for every $x \in S$.
Let $\ds \sup_{x \mathop \in S} \map g x$ exist.
Then $\ds \sup_{x \mathop \in S} \map f x$ exists and:
:$\ds \sup_{x \mathop \in S} \map f x \le \sup_{x \mathop \in S} \map g x$.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \sup g
| r = \map \sup {f + \paren {g - f} }
}}
{{eqn | r = \sup f + \sup \left({g - f}\right)
| c = Supremum of Sum equals Sum of Suprema
}}
{{end-eqn}}
Supremum of Sum equals Sum of Suprema also gives that $\sup f$ and $\sup \paren {g - f}$ exist.
We have:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map g x
| o = \ge
| r = \map f x
| c =
}}
{{eqn | ll= \leadsto
| q = \forall x \in S
| l = \map g x - \map f x
| o = \ge
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| q = \forall x \in S
| l = \map \sup {g - f}
| o = \ge
| r = \map g x - \map f x \ge 0
| c = as $\map \sup {g - f}$ is an upper bound for $\set {\map g x - \map f x: x \in S}$
}}
{{eqn | ll= \leadsto
| l = \map \sup {g - f}
| o = \ge
| r = 0
}}
{{eqn | ll= \leadsto
| l = \sup f + \map \sup {g - f}
| o = \ge
| r = \sup f
}}
{{eqn | ll= \leadsto
| l = \sup g
| o = \ge
| r = \sup f
| c = as $\sup g = \sup f + \map \sup {g - f}$
}}
{{end-eqn}}
{{qed}}
Category:Real Analysis
Category:Suprema
\end{proof}
|
22446
|
\section{Supremum of Ideals is Increasing}
Tags: Supremum, Increasing Mappings
\begin{theorem}
Let $L = \left({S, \preceq}\right)$ be an up-complete ordered set.
Let $\mathit{Ids}\left({L}\right)$ be the set of all ideals in $L$.
Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an ordered set where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$
Let $f: \mathit{Ids}\left({L}\right) \to S$ be a mapping such that
:$\forall I \in \mathit{Ids}\left({L}\right): f\left({I}\right) = \sup I$
Then $f$ is an increasing mapping.
\end{theorem}
\begin{proof}
Let $I, J \in \mathit{Ids}\left({L}\right)$ such that
:$I \precsim J$
By definition of $\precsim$:
:$I \subseteq J$
By definition of up-complete:
:$I$ and $J$ admit suprema in $L$.
By Supremum of Subset:
:$\sup I \preceq \sup J$
Thus by definition of $f$:
:$f\left({I}\right) \preceq f\left({J}\right)$
Hence $f$ is an increasing mapping.
{{qed}}
\end{proof}
|
22447
|
\section{Supremum of Ideals is Upper Adjoint}
Tags: Galois Connections, Continuous Lattices
\begin{theorem}
Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below continuous join semilattice.
Let $\mathit{Ids}\left({L}\right)$ be the set of all ideals in $L$.
Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an ordered set where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$
Let $f: \mathit{Ids}\left({L}\right) \to S$ be a mapping such that
:$\forall I \in \mathit{Ids}\left({L}\right): f\left({I}\right) = \sup I$
Then $f$ is an upper adjoint of Galois connection.
\end{theorem}
\begin{proof}
Define $d: S \to \mathit{Ids}\left({L}\right)$
:$\forall t \in S: d\left({t}\right) = \inf \left({f^{-1}\left[{t^\succeq}\right]}\right)$
where
:$t^\succeq$ denotes the upper closure of $t$,
:$f^{-1}\left[{t^\succeq}\right]$ denotes the image of $t^\succeq$ over $f^{-1}$.
We will prove that
:$\forall t \in S: d\left({t}\right) = \min \left({f^{-1}\left[{t^\succeq}\right]}\right)$
Let $t \in S$.
By Continuous iff For Every Element There Exists Ideal Element Precedes Supremum:
:there exists an ideal $J$ in $L$ such that
::$t \preceq \sup J$ and for every ideal $K$ in $L$: $t \preceq \sup K \implies J \subseteq K$
We will prove that
:$\forall K \in \mathit{Ids}\left({L}\right): K$ is lower bound for $f^{-1}\left[{t^\succeq}\right] \implies K \precsim J$
Let $K \in \mathit{Ids}\left({L}\right)$ such that
:$K$ is lower bound for $f^{-1}\left[{t^\succeq}\right]$
By definition of $f$:
:$t \preceq f\left({J}\right)$
By definition of upper closure of element:
:$f\left({J}\right) \in t^\succeq$
By definition of image of set:
:$J \in f^{-1}\left[{t^\succeq}\right]$
Thus by definition of lower bound:
:$K \precsim J$
{{qed|lemma}}
We will prove that
:$J$ is lower bound for $f^{-1}\left[{t^\succeq}\right]$
Let $K \in \mathit{Ids}\left({L}\right)$ such that
:$K \in f^{-1}\left[{t^\succeq}\right]$
By definition of image of set:
:$f\left({K}\right) \in t^\succeq$
By definition of upper closure of element:
:$t \preceq f\left({K}\right)$
By definition of $f$:
:$t \preceq \sup K$
Then
:$J \subseteq K$
Thus by definition of $\precsim$:
:$J \precsim K$
{{qed|lemma}}
By definition of supremum:
:$t \preceq \sup \left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right)$
By definition of $f$:
:$t \preceq f\left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right)$
By definition of upper closure of element:
:$f\left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right) \in t^\succeq$
By definition of image of set:
:$\inf \left({f^{-1}\left[{t^\succeq}\right]}\right) \in f^{-1}\left[{t^\succeq}\right]$
Thus by definition of smallest element:
:$d\left({t}\right) = \min \left({f^{-1}\left[{t^\succeq}\right]}\right)$
{{qed|lemma}}
By Supremum of Ideals is Increasing:
:$f$ is an increasing mapping.
By Galois Connection is Expressed by Minimum:
:$\left({f, d}\right)$ is a Galois connection.
Hence $f$ is an upper adjoint of Galois connection.
{{qed}}
\end{proof}
|
22448
|
\section{Supremum of Ideals is Upper Adjoint implies Lattice is Continuous}
Tags: Galois Connections, Continuous Lattices
\begin{theorem}
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below up-complete lattice.
Let $\map {\mathit {Ids} } L$ be the set of all ideals in $L$.
Let $P = \struct {\map {\mathit {Ids} } L, \precsim}$ be an ordered set where $\mathord \precsim = \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
Let $f: \map {\mathit {Ids} } L \to S$ be a mapping such that
:$\forall I \in \map {\mathit {Ids} } L: f \sqbrk I = \sup I$
Let $f$ be an upper adjoint of a Galois connection.
Then $L$ is continuous.
\end{theorem}
\begin{proof}
We will prove that
:$\forall x \in S: \exists I \in \map {\mathit {Ids} } L: x \preceq \sup I \land \forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$
Let $x \in S$.
Define $I := \map \inf {f^{-1} \sqbrk {x^\succeq} }$.
By definition of $P$:
:$I \in \map {\mathit {Ids} } L$
We will prove that
:$\forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$
Let $J \in \map {\mathit {Ids} } L$ such that
:$x \preceq \sup J$
By definition of $f$:
:$x \preceq f \sqbrk J$
By definition of upper closure of element:
:$f \sqbrk J \in x^\succeq$
By definition of image of set:
:$J \in f^{-1} \sqbrk {x^\succeq}$
By definition of infimum:
:$I \precsim J$
Hence by definition of $\precsim$:
:$I \subseteq J$
{{qed|lemma}}
By definition of upper adjoint of a Galois connection:
:there exists a mapping $d: S \to \map {\mathit {Ids} } L$: $\struct {f, d}$ is a Galois connection.
By Galois Connection is Expressed by Minimum
:$\map d x = \map \min {f^{-1} \sqbrk {x^\succeq} }$
By definition of smallest element:
:$I \in f^{-1} \sqbrk {x^\succeq}$
By definition of image of set:
:$f \sqbrk I \in x^\succeq$
By definition of upper closure of element:
:$x \preceq f \sqbrk I$
Thus by definition of $f$:
:$x \preceq \sup I$
Hence
:$\exists I \in \map {\mathit {Ids} } L: x \preceq \sup I \land \forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$
{{qed|lemma}}
Hence by Continuous iff For Every Element There Exists Ideal Element Precedes Supremum:
:$L$ is continuous.
{{qed}}
\end{proof}
|
22449
|
\section{Supremum of Lower Closure of Set}
Tags: Lower Closures, Order Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$.
Let $L = T^\preceq$ be the lower closure of $T$ in $S$.
Let $s \in S$
Then $s$ is the supremum of $T$ {{iff}} it is the supremum of $L$.
\end{theorem}
\begin{proof}
By Supremum and Infimum are Unique we need only show that $s$ is a supremum of $L$ {{iff}} it is a supremum of $T$.
\end{proof}
|
22450
|
\section{Supremum of Lower Sums Never Greater than Upper Sum}
Tags: Real Analysis
\begin{theorem}
Let $\closedint a b$ be a closed real interval .
Let $f$ be a bounded real function defined on $\closedint a b$.
Let $S$ be a finite subdivision of $\closedint a b$.
Let $\map U S$ be the upper sum of $f$ on $\closedint a b$ with respect to $S$.
Let $\map L P$ be the lower sum of $f$ on $\closedint a b$ with respect to a finite subdivision $P$.
Then:
:$\sup_P \map L P \le \map U S$
\end{theorem}
\begin{proof}
From Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, $\map U S$ is an upper bound for the real set:
:$T = \leftset {\map L P: P}$ is a finite subdivision of $\rightset {\closedint a b}$
Since $\sup_P \map L P$ is the supremum of $T$:
:$\sup_P \map L P \le \map U S$
Hence the result.
{{qed}}
Category:Real Analysis
\end{proof}
|
22451
|
\section{Supremum of Meet Image of Directed Set}
Tags: Up-Complete Semilattices
\begin{theorem}
Let $\struct {S, \preceq}$ be an up-complete meet semilattice.
Let $f: S \times S \to S$ be a mapping such that:
:$\forall \tuple {x, y} \in S \times S: \map f {x, y} = x \wedge y$
Let $D$ be directed subset of $S \times S$ in the simple order product $\struct {S \times S, \precsim}$ of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.
Then:
:$\map \sup {\map {f^\to} D} = \sup \set {x \wedge y: x \in \map {\pr_1^\to} D, y \in \map {\pr_2^\to} D}$
where
:$\pr_1$ denotes the first projection on $S \times S$
:$\pr_2$ denotes the second projection on $S \times S$
:$\map {\pr_1^\to} D$ denotes the image of $D$ under $\pr_1$.
\end{theorem}
\begin{proof}
By definition of image of set:
:$\map {f^\to} D = \set {x \wedge y: \tuple {x, y} \in D}$
By definition of subset:
:$\map {f^\to} D \subseteq \set {x \wedge y: x \in \map {\pr_1^\to} D, y \in \map {\pr_2^\to} D}$
By Up-Complete Product/Lemma 2:
:$D_1 := \map {\pr_1^\to} D$ is directed
and
:$D_2 := \map {\pr_2^\to} D$ is directed.
By Meet of Directed Subsets is Directed:
:$\set {x \wedge y: x \in D_1, y \in D_2}$ is directed.
By definition of up-complete:
:$\set {x \wedge y: x \in D_1, y \in D_2}$ admits a supremum.
By Meet is Increasing:
:$f$ is increasing mapping.
By Image of Directed Subset under Increasing Mapping is Directed:
:$\map {f^\to} D$ is directed.
By definition of up-complete:
:$\map {f^\to} D$ admits a supremum.
By Supremum of Subset:
:$\map \sup {\map {f^\to} D} \preceq \sup \set {x \wedge y: x \in D_1, y \in D_2}$
We will prove that
:$\set {x \wedge y: x \in D_1, y \in D_2} \subseteq \paren {\map {f^\to} D}^\preceq$
where
:$\paren {\map {f^\to} D}^\preceq$ denotes the lower closure of $\map {f^\to} D$.
Let $z \in \set {x \wedge y: x \in D_1, y \in D_2}$.
Then
:$\exists x \in D_1, y \in D_2: z = x \wedge y$
By definition of image of set:
:$\exists \tuple {a, b} \in D: \map {\pr_1} {a, b} = x$
and
:$\exists \tuple {c, d} \in D: \map {\pr_2} {c, d} = y$:
By definition of first projection and second projection:
:$a = x$ and $d = y$
By definition of directed subset:
:$\exists \tuple {g, h} \in D: \tuple {x, b} \precsim \tuple {g, h} \land \tuple {c, y} \precsim \tuple {g, h}$
By definition of simple order product:
:$x \preceq g$ and $y \preceq h$
By Meet Semilattice is Ordered Structure:
:$x \wedge y \preceq g \wedge h \in \set {x \wedge y: \tuple {x, y} \in D}$
Thus by definition of lower closure:
:$z \in \paren {\map {f^\to} D}^\preceq$
{{qed|lemma}}
By Supremum of Lower Closure of Set:
:$\paren {\map {f^\to} D}^\preceq$ admits a supremum
and
:$\map \sup {\map {f^\to} D}^\preceq = \map \sup {\map {f^\to} D}$
By Supremum of Subset:
:$\sup \set {x \wedge y: x \in D_1, y \in D_2} \preceq \map \sup {\map {f^\to} D}$
Thus by definition of antisymmetry:
:$\map \sup {\map {f^\to} D} = \sup \set {x \wedge y: x \in D_1, y \in D_2}$
{{qed}}
\end{proof}
|
22452
|
\section{Supremum of Power Set}
Tags: Power Set, Orderings, Order Theory
\begin{theorem}
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the relation $\subseteq$.
(From Subset Relation on Power Set is Partial Ordering, this is an ordered set.)
Then the supremum of $\struct {\powerset S, \subseteq}$ is the set $S$.
\end{theorem}
\begin{proof}
By the definition of the power set:
:$\forall X \in \powerset S: X \subseteq S$
The result then follows from the definition of supremum.
{{qed}}
Category:Power Set
Category:Order Theory
\end{proof}
|
22453
|
\section{Supremum of Product}
Tags: Max and Min Operations, Suprema
\begin{theorem}
Let $\struct {G, \circ, \preceq}$ be an ordered group.
Suppose that subsets $A$ and $B$ of $G$ admit suprema in $G$.
Then:
:$\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$
where $\circ_\PP$ denotes subset product.
\end{theorem}
\begin{proof}
Let $a \in A$, $b \in B$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
| o = \preceq
| r = \sup A \circ b
| c = {{Defof|Supremum of Set}}
}}
{{eqn | o = \preceq
| r = \sup A \circ \sup B
| c = {{Defof|Supremum of Set}}
}}
{{end-eqn}}
Hence $\sup A \circ \sup B$ is an upper bound for $A \circ_\PP B$.
Suppose that $u$ is an upper bound for $A \circ_\PP B$.
Then:
{{begin-eqn}}
{{eqn | q = \forall b \in B: \forall a \in A
| l = a \circ b
| o = \preceq
| r = u
}}
{{eqn | ll= \leadsto
| q = \forall b \in B: \forall a \in A
| l = a
| o = \preceq
| r = u \circ b^{-1}
}}
{{eqn | ll= \leadsto
| q = \forall b \in B
| l = \sup A
| o = \preceq
| r = u \circ b^{-1}
| c = {{Defof|Supremum of Set}}
}}
{{eqn | ll= \leadsto
| q = \forall b \in B
| l = b
| o = \preceq
| r = \paren {\sup A}^{-1} \circ u
}}
{{eqn | ll= \leadsto
| l = \sup B
| o = \preceq
| r = \paren {\sup A}^{-1} \circ u
| c = {{Defof|Supremum of Set}}
}}
{{eqn | ll= \leadsto
| l = \sup A \circ \sup B
| o = \preceq
| r = u
}}
{{end-eqn}}
Therefore:
:$\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$
{{qed}}
\end{proof}
|
22454
|
\section{Supremum of Set of Integers equals Greatest Element}
Tags: Integers
\begin{theorem}
Let $S \subset \Z$ be a non-empty subset of the set of integers.
Let $S$ be bounded above in the set of real numbers $\R$.
Then $S$ has a greatest element, and it is equal to the supremum $\sup S$.
\end{theorem}
\begin{proof}
By Set of Integers Bounded Above by Real Number has Greatest Element, $S$ has a greatest element, say $n \in S$.
By Greatest Element is Supremum, $n$ is the supremum of $S$.
{{qed}}
\end{proof}
|
22455
|
\section{Supremum of Set of Integers is Integer}
Tags: Integers
\begin{theorem}
Let $S \subset \Z$ be a non-empty subset of the set of integers.
Let $S$ be bounded above in the set of real numbers.
Then its supremum $\sup S$ is an integer.
\end{theorem}
\begin{proof}
By Supremum of Set of Integers equals Greatest Element, $S$ has a greatest element $n \in \Z$, that is equals to the supremum of $S$.
{{qed}}
\end{proof}
|
22456
|
\section{Supremum of Set of Real Numbers is at least Supremum of Subset}
Tags: Supremum of Set of Real Numbers is at least Supremum of Subset, Supremum of Subset is Less Than or Equal to Supremum of Set, Real Analysis
\begin{theorem}
Let $S$ be a set of real numbers.
Let $S$ have a supremum.
Let $T$ be a non-empty subset of $S$.
Then $\sup T$ exists and:
:$\sup T \le \sup S$
\end{theorem}
\begin{proof}
The number $\sup S$ is an upper bound for $S$.
Therefore, $\sup S$ is an upper bound for $T$ as $T$ is a non-empty subset of $S$.
Accordingly, $T$ has a supremum by the continuum property.
The number $\sup S$ is an upper bound for $T$.
Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the least upper bound of $T$.
{{qed}}
Category:Real Analysis
239707
239703
2015-12-01T20:29:22Z
Prime.mover
59
239707
wikitext
text/x-wiki
\end{proof}
|
22457
|
\section{Supremum of Simple Order Product}
Tags: Simple Order Product, Order Theory, Suprema
\begin{theorem}
Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be ordered sets.
Let $\struct {S_1 \times S_2, \precsim}$ be the simple order product of $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$.
Let $X_1$ be a non-empty subset of $S_1$, $X_2$ be a non-empty subset of $S_2$ such that
:$X_1$ and $X_2$ admit suprema.
Then:
:$X_1 \times X_2$ admits a supremum
and:
:$\map \sup {X_1 \times X_2} = \tuple {\sup X_1, \sup X_2}$
\end{theorem}
\begin{proof}
We will prove that:
:$\tuple {\sup X_1, \sup X_2}$ is upper bound for $X_1 \times X_2$
Let $\tuple {a, b} \in X_1 \times X_2$.
By definition of Cartesian product:
:$a \in X_1$ and $b \in X_2$
By definitions of supremum and upper bound:
:$a \preceq_1 \sup X_1$ and $b \preceq_2 \sup X_2$
Thus by definition of simple order product:
:$\tuple {a, b} \precsim \tuple {\sup X_1, \sup X_2}$
{{qed|lemma}}
We will prove that:
:$\forall \tuple {a, b} \in S1 \times S_2: \tuple {a, b}$ is upper bound for $X_1 \times X_2 \implies \tuple {\sup X_1, \sup X_2} \precsim \tuple {a, b}$
Let $\tuple {a, b} \in S1 \times S_2$ such that:
:$\tuple {a, b}$ is upper bound for $X_1 \times X_2$
We will prove as sublemma that:
:$a$ is upper bound for $X_1$
Let $c \in X_1$.
By definition of non-empty set:
:$\exists d: d \in X_2$
By definition of Cartesian product:
:$\tuple {c, d} \in X_1 \times X_2$
By definition of upper bound:
:$\tuple {c, d} \precsim \tuple {a, b}$
Thus by definition of simple order product:
:$c \preceq_1 a$
This ends the proof of sublemma.
Analogically we have that:
:$b$ is upper bound for $X_2$
By definition of supremum:
:$\sup X_1 \preceq_1 a$ and $\sup X_2 \preceq_2 b$
Thus by definition of simple order product:
:$\tuple {\sup X_1, \sup X_2} \precsim \tuple {a, b}$
{{qed|lemma}}
Thus by definition:
:$X_1 \times X_2$ admits a supremum
and:
:$\map \sup {X_1 \times X_2} = \tuple {\sup X_1, \sup X_2}$
{{qed}}
\end{proof}
|
22458
|
\section{Supremum of Singleton}
Tags: Singletons, Order Theory, Suprema
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Then for all $a \in S$:
:$\sup \set a = a$
where $\sup$ denotes supremum.
\end{theorem}
\begin{proof}
Since $a \preceq a$, $a$ is an upper bound of $\set a$.
Let $b$ be another upper bound of $\set a$.
Then necessarily $a \preceq b$.
It follows that indeed:
:$\sup \set a = a$
as desired.
{{qed}}
\end{proof}
|
22459
|
\section{Supremum of Subgroups in Lattice}
Tags: Lattice Theory, Subgroups, Suprema, Complete Lattices
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by $\mathbb G$ and $\subseteq$.
Let $H, K \in \mathbb G$.
Let either $H$ or $K$ be normal in $G$.
Then:
:$\sup \set {H, K} = H \circ K$
where $H \circ K$ denotes subset product.
\end{theorem}
\begin{proof}
Recall that Set of Subgroups forms Complete Lattice.
Let $L = \sup \set {H, K}$.
Let either $H$ or $K$ be normal in $G$.
Since $L$ contains $H$ and $K$, then $L$ contains $H \circ K$.
The smallest subgroup of $G$ containing $H$ and $K$ is:
:$\gen {H, K}$
the subgroup generated by $H$ and $K$.
From Subset Product with Normal Subgroup as Generator:
:$\gen {H, K} = H \circ K$
when either $H$ or $K$ is normal.
The result follows.
{{qed}}
\end{proof}
|
22460
|
\section{Supremum of Subset}
Tags: Orderings, Order Theory
\begin{theorem}
Let $\left({U, \preceq}\right)$ be an ordered set.
Let $S \subseteq U$.
Let $T \subseteq S$.
Let $S$ admit a supremum (in $U$).
If $T$ also admits a supremum (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.
\end{theorem}
\begin{proof}
Let $B = \sup \left({S}\right)$.
Then $B$ is an upper bound for $S$.
As $T \subseteq S$, it follows by the definition of a subset that $x \in T \implies x \in S$.
Because $x \in S \implies x \preceq B$ (as $B$ is an upper bound for $S$) it follows that $x \in T \implies x \preceq B$.
So $B$ is an upper bound for $T$.
Therefore $B$ succeeds the supremum of $T$ in $S$.
Hence the result.
{{qed}}
Category:Order Theory
\end{proof}
|
22461
|
\section{Supremum of Subset of Bounded Above Set of Real Numbers}
Tags: Boundedness
\begin{theorem}
Let $A$ and $B$ be sets of real numbers such that $A \subseteq B$.
Let $B$ be bounded above.
Then:
:$\sup A \le \sup B$
where $\sup$ denotes the supremum.
\end{theorem}
\begin{proof}
Let $B$ be bounded above.
By the Continuum Property, $B$ admits a supremum.
By Subset of Bounded Above Set is Bounded Above, $A$ is also bounded above.
Hence also by the Continuum Property, $A$ also admits a supremum.
{{AimForCont}} $\sup A > \sup B$.
Then:
:$\exists y \in A: y > \sup B$
Thus by definition of supremum, $y \notin B$.
That is:
:$A \nsubseteq B$
which contradicts our initial assumption that $A \subseteq B$.
Hence the result by Proof by Contradiction.
{{qed}}
\end{proof}
|
22462
|
\section{Supremum of Subset of Real Numbers May or May Not be in Subset}
Tags: Suprema
\begin{theorem}
Let $S \subset \R$ be a proper subset of the set $\R$ of real numbers.
Let $S$ admit a supremum $M$.
Then $M$ may or may not be an element of $S$.
\end{theorem}
\begin{proof}
Consider the subset $S$ of the real numbers $\R$ defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
It is seen that:
:$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$
and hence $\sup S = 1$.
Thus $\sup S \in S$.
Consider the subset $T$ of the real numbers $\R$ defined as:
:$T = \set {-\dfrac 1 n: n \in \Z_{>0} }$
It is seen that:
:$T = \set {-1, -\dfrac 1 2, -\dfrac 1 3, \ldots}$
and hence $\sup T = 0$.
Thus $\sup T \notin T$.
{{Qed}}
\end{proof}
|
22463
|
\section{Supremum of Subset of Real Numbers is Arbitrarily Close}
Tags: Real Analysis
\begin{theorem}
Let $A \subseteq \R$ be a subset of the real numbers.
Let $b$ be a supremum of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A: b − x < \epsilon$
\end{theorem}
\begin{proof}
Note that $A$ is non-empty as the empty set does not admit a supremum (in $\R$).
Suppose $\epsilon \in \R_{>0}$ such that:
:$\forall x \in A: b − x \ge \epsilon$
Then:
:$\forall x \in A: b − \epsilon \ge x$
and so $b − \epsilon$ would be an upper bound of $A$ which is less than $b$.
But since $b$ is a supremum of $A$ there can be no such $b − \epsilon$.
From that contradiction it follows that:
:$\exists x \in A: b − x < \epsilon$
{{qed}}
\end{proof}
|
22464
|
\section{Supremum of Subset of Union Equals Supremum of Union}
Tags: Real Analysis
\begin{theorem}
Let $S$ be a non-empty real set.
Let $S$ have a supremum.
Let $\set {S_i: i \in \set {1, 2, \ldots, n} }$, $n \in \N_{>0}$, be a set of non-empty subsets of $S$.
Let $\bigcup S_i = S$.
Then there exists a $j$ in $\set {1, 2, \ldots, n}$ such that:
:$\sup S_j = \sup S$
\end{theorem}
\begin{proof}
If $S$ equals $S_j$ for a $j$ in $\set {1, 2, \ldots, n}$, it is trivially true that $\sup S = \sup S_j$.
Now assume that $S$ is unequal to $S_i$ for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$.
By Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S \ge \sup S_i$ for every $i$ in $\set{1, 2, \ldots, n}$.
There are two alternatives; either:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
or:
:$\sup S = \sup S_j$ for at least one $j$ in $\set {1, 2, \ldots, n}$.
Suppose that:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
Let $\epsilon = \sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$.
We note that $\epsilon > 0$.
By Supremum of Subset of Real Numbers is Arbitrarily Close, $S$ has an element $x$ that satisfies:
:$x > \sup S - \epsilon$
We have:
{{begin-eqn}}
{{eqn | o = >
| l = x
| r = \sup S - \epsilon
}}
{{eqn | r = \sup S - \paren {\sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n} }
| c = definition of $\epsilon$
}}
{{eqn | r = \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}
}}
{{end-eqn}}
Therefore:
:$x > \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$
This means that $x > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.
However, $x$ must be an element of $S_j$ for some $j$ in $\set {1, 2, \ldots, n}$ as $x \in S$ and $S = \bigcup S_i$.
Accordingly, it is not true that $\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.
We just concluded that the alternative:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
is not true.
Therefore, the other alternative:
:$\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$
is true.
{{qed}}
Category:Real Analysis
\end{proof}
|
22465
|
\section{Supremum of Sum equals Sum of Suprema}
Tags: Real Analysis
\begin{theorem}
Let $A$ and $B$ be non-empty sets of real numbers.
Let $A + B$ be $\set {x + y: x \in A, y \in B}$.
Let either $A$ and $B$ have suprema or $A + B$ have a supremum.
Then all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist and:
:$\sup \paren {A + B} = \sup A + \sup B$
\end{theorem}
\begin{proof}
Assume first that $A$ and $B$ have suprema.
We have:
:$x \le \sup A$ for an arbitrary $x$ in $A$
:$y \le \sup B$ for an arbitrary $y$ in $B$
Adding these inequalities, we get:
:$x + y \le \sup A + \sup B$
The number $x + y$ is an arbitrary element of $A + B$ as $x$ and $y$ are arbitrary elements of $A$ and $B$ respectively.
Therefore, $\sup A + \sup B$ is an upper bound for $A + B$.
$A + B$ is non-empty as $A$ and $B$ are non-empty.
Accordingly, $A + B$ has a supremum by the Continuum Property.
Next, assume that $\sup \paren {A + B}$ has a supremum.
We need to prove that $A$ and $B$ have suprema.
Let $y$ be a point in $B$.
We have:
{{begin-eqn}}
{{eqn | l = x + y
| o = \le
| r = \sup \paren {A + B}
| c = for every $x$ in $A$ as $x + y$ is a point in $A + B$
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \le
| r = \sup \paren {A + B} - y
}}
{{end-eqn}}
Therefore, $A$ has an upper bound as $x$ is an arbitrary point in $A$.
Also, we have that $A$ is non-empty.
Accordingly, $A$ has a supremum by the Continuum Property.
A similar argument gives that $B$ has a supremum.
So, we have shown that all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist.
We proceed to show that $\sup \paren {A + B} = \sup A + \sup B$.
We have $\sup \paren {A + B} \le \sup A + \sup B$ as $\sup A + \sup B$ is an upper bound for $A + B$.
Accordingly, either:
:$\sup \paren {A + B} < \sup A + \sup B$
or:
:$\sup \paren {A + B} = \sup A + \sup B$.
{{AimForCont}}:
:$\sup \paren {A + B} < \sup A + \sup B$.
Let $\epsilon = \sup A + \sup B - \sup \paren {A + B}$.
We note that $\epsilon > 0$.
Since $\sup A$ is the least upper bound of $A$, there is an element $x$ in $A$ such that:
:$x > \sup A - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close
Since $\sup B$ is the least upper bound of $B$, there is an element $y$ in $B$ such that:
:$y > \sup B - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close
Adding these inequalities, we get:
{{begin-eqn}}
{{eqn | l = x + y
| o = >
| r = \sup A - \frac \epsilon 2 + \sup B - \frac \epsilon 2
}}
{{eqn | ll= \leadstoandfrom
| l = x + y
| o = >
| r = \sup A + \sup B - \epsilon
}}
{{eqn | ll= \leadstoandfrom
| l = x + y
| o = >
| r = \sup A + \sup B - \paren {\sup A + \sup B - \sup \paren {A + B} }
| c = definition of $\epsilon$
}}
{{eqn | ll= \leadstoandfrom
| l = x + y
| o = >
| r = \sup \paren {A + B}
}}
{{end-eqn}}
which is impossible since the number $x + y$ is an element of $A + B$ as $x \in A$ and $y \in B$.
We have found that:
:$\sup \paren {A + B} < \sup A + \sup B$ is not true.
Therefore:
:$\sup \paren {A + B} = \sup A + \sup B$ as $\sup \paren {A + B} \le \sup A + \sup B$.
{{qed}}
Category:Real Analysis
\end{proof}
|
22466
|
\section{Supremum of Suprema}
Tags: Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $\mathbb T \subseteq \powerset S$, where $\powerset S$ is the power set of $S$.
Suppose all $T \in \mathbb T$ admit a supremum $\sup T$ in $S$.
Then:
:$\sup \bigcup \mathbb T = \sup {\set {\sup T: T \in \mathbb T} }$
if one of these two quantities exists (in $S$).
\end{theorem}
\begin{proof}
Suppose that $s = \sup \bigcup \mathbb T \in S$.
By Set is Subset of Union, $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$.
Hence by Supremum of Subset:
:$\forall T \in \mathbb T: \sup T \preceq s$
Suppose now that $a \in S$ satisfies:
:$\forall T \in \mathbb T: \sup T \preceq a$
Then by transitivity of $\preceq$:
:$\forall t \in T: t \preceq a$
Since this holds for any $T \in \mathbb T$, also:
:$\forall t \in \bigcup \mathbb T: t \preceq a$
Hence $s \preceq a$, by definition of supremum.
That is, $s = \sup {\set {\sup T: T \in \mathbb T} }$.
{{qed|lemma}}
Suppose now that $r = \sup {\set {\sup T: T \in \mathbb T} } \in S$.
By definition of supremum, for all $T \in \mathbb T$ and $t \in T$:
:$t \preceq \sup T$
By transitivity of $\preceq$:
:$\forall T \in \mathbb T: \forall t \in T: t \preceq r$
Hence for all $t \in \bigcup \mathbb T$:
:$t \preceq r$
Suppose that $a \in S$ satisfies:
:$\forall t \in \bigcup \mathbb T: t \preceq a$
In particular, for any $T \in \mathbb T$, since $T \subseteq \bigcup \mathbb T$:
:$\sup T \preceq a$
and therefore by definition of supremum, also:
:$r \preceq a$
That is, $r = \sup \bigcup \mathbb T$.
{{qed}}
Category:Order Theory
\end{proof}
|
22467
|
\section{Supremum of Union of Bounded Above Sets of Real Numbers}
Tags: Set Union, Boundedness, Suprema
\begin{theorem}
Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ both be bounded above.
Then:
:$\map \sup {A \cup B} = \max \set {\sup A, \sup B}$
where $\sup$ denotes the supremum.
\end{theorem}
\begin{proof}
Let $A$ and $B$ both be bounded above.
By the Continuum Property, $A$ and $B$ both admit a supremum.
Let $x \in A \cup B$.
Then either $x \le \sup A$ or $x \le \sup B$ by definition of supremum.
Hence:
:$x \le \max \set {\sup A, \sup B}$
and so $\max \set {\sup A, \sup B}$ is certainly an upper bound of $A \cup B$.
It remains to be shown that $\max \set {\sup A, \sup B}$ is the smallest upper bound.
{{AimForCont}} there exists $m \in \R$ such that:
:$m < \max \set {\sup A, \sup B}$
and:
:$\forall x \in A \cup B: x \le m$
{{WLOG}}, let $\sup A \ge \sup B$.
Then:
:$\max \set {\sup A, \sup B} = \sup A$
and so:
:$m < \sup A$
But then by definition of supremum:
:$\exists a \in A: a > m$
and so $m$ is not an upper bound of $A \cup B$.
This contradicts our assumption that $m$ is a supremum of $A \cup B$.
It follows by Proof by Contradiction that $\max \set {\sup A, \sup B}$ is the supremum of $A \cup B$.
The same argument shows, mutatis mutandis, that if $\sup A \le \sup B$, $\max \set {\sup A, \sup B}$ is the supremum of $A \cup B$.
{{qed}}
\end{proof}
|
22468
|
\section{Surgery for Rings}
Tags: Ring Theory, Ring Isomorphisms, Ring Monomorphisms
\begin{theorem}
Let $R$ and $S$ be commutative rings with unity, and $\phi: R \to S$ a ring monomorphism.
Then there is a ring $T$ isomorphic to $S$ that contains $R$ as a subring.
\end{theorem}
\begin{proof}
Let $T$ be the disjoint union $T = R \cup \paren {S \setminus \Img \phi}$.
Define $\theta : T \to S$ as follows:
:If $x \in R$, then $\map \theta x = \map \phi x$
:If $x \in \paren {S \setminus \Img \phi}$ then $\map \theta x = x$
We claim that $\theta$ is an isomorphism.
''Injectivity'': Let $\map \theta x = \map \theta y$.
Since $\theta \sqbrk R = \phi \sqbrk R$, we have $\theta \sqbrk R \cap \paren {S \setminus \Img \phi} = \O$.
Therefore either $x, y \in R$ or $x, y \in \paren {S \setminus \Img \phi}$.
If $x, y \in R$ then $\map \phi x = \map \phi y$, so $\theta$ is injective because $\phi$ is.
If $x, y \in \paren {S \setminus \Img \phi}$ then $\theta$ acts as the identity and $x = y$.
''Surjectivity'': Under $\theta$, $R$ surjects onto $\Img \phi = \phi \sqbrk R$ by definition.
Moreover:
:$\theta \sqbrk {S \setminus \Img \phi} = I \sqbrk {S \setminus \Img \phi} = \paren {S \setminus \Img \phi}$
where $I$ denotes the identity mapping.
Therefore $\theta$ is surjective everywhere.
Now we endow $T$ with addition and multiplication: for $x, y \in T$, let
:$x + y = \map {\theta^{-1} } {\map \theta x + \map \theta y}$
:$x y = \map {\theta^{-1} } {\map \theta x \, \map \theta y}$
so trivially we have
:$\map \theta {x + y} = \map \theta x + \map \theta y$
:$\map \theta {x y} = \map \theta x \, \map \theta y$
Therefore $T$ is the isomorphic image of a ring, and an isomorphism is precisely a map that preserves the ring axioms.
Thus $T$ is a ring isomorphic to $S$ containing $R$ as a subring.
{{qed}}
{{proof wanted|I wonder whether this page might be no more than an application of the Embedding Theorem.}}
Category:Ring Isomorphisms
Category:Ring Monomorphisms
\end{proof}
|
22469
|
\section{Surjection Induced by Powerset is Induced by Surjection}
Tags: Power Set, Induced Mappings, Direct Image Mappings, Relation Theory, Named Theorems, Mappings, Surjections, Relations
\begin{theorem}
Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping $\RR$.
Let $\RR^\to$ be a surjection.
Let $X = \Preimg \RR$, that is, the preimage of $\RR$.
Then $\RR {\restriction_X} \subseteq X \times T$, that is, the restriction of $\RR$ to $X$, is a surjection.
\end{theorem}
\begin{proof}
Let $X$ be the preimage of $\RR$.
Suppose $\RR {\restriction_X} \subseteq X \times T$ is a mapping, but not a surjection.
Then:
:$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in \RR$
Because no element of $S$ relates to $y$, no subset of $S$ contains any element of $S$ that relates to the subset $\set y \subseteq T$.
Thus:
:$\exists \set y \in \powerset T: \neg \exists X \in \powerset S: \map {\RR^\to} X = \set y$
So we see that $\RR^\to: \powerset S \to \powerset T$ is not a surjection.
Now, suppose $\RR {\restriction_X} \subseteq X \times T$ is not even a mapping.
This could happen, according to the definition of a mapping, in one of two ways:
:$(1): \quad \exists x \in X: \neg \exists y \in T: \tuple {x, y} \in \RR$
:$(2): \quad \exists x \in S: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR: y_1 \ne y_2$
Because $X$ is already the preimage of $\RR$, the first of these cannot happen here.
For the second, it can be seen that neither $\set {y_1}$ nor $\set {y_2}$ can be in $\Rng {\map {\RR^\to} {\powerset S} }$.
Therefore $\RR^\to: \powerset S \to \powerset T$ can not be a surjection.
{{questionable|See talk page}}
Thus, by the Rule of Transposition, the result follows.
{{qed}}
Category:Surjections
Category:Direct Image Mappings
\end{proof}
|
22470
|
\section{Surjection by Free Module}
Tags: Homological Algebra
\begin{theorem}
Let $A$ be a ring.
Let $M$ be a left $A$-module.
Then there exists a free $A$-module $F$ and a surjective $A$-module homomorphism $f : F \to M$.
\end{theorem}
\begin{proof}
Let $F = A^{\paren M}$ be the free $A$-module on the set $M$.
Let $c : M \to A^{\paren M}$ be the canonical mapping.
Let $f : F \to M$ be the $A$-module homomorphism induced the by the Universal Property of Free Modules applied to the identity $\operatorname {id}_M$ of $M$.
We have
:$f \circ c = \operatorname {id}_M$
Thus $f$ is a split epimorphism in the category of sets.
By Split Epimorphism is Epic and Surjection iff Epimorphism in Category of Sets $f$ is surjective.
{{qed}}
Category:Homological Algebra
\end{proof}
|
22471
|
\section{Surjection from Aleph to Ordinal}
Tags: Aleph Function, Aleph Mapping
\begin{theorem}
Let $x$ and $y$ be ordinals.
Suppose that:
:$0 < y < \aleph_{x+1}$
Then there is a surjection:
:$f : \aleph_x \to y$
\end{theorem}
\begin{proof}
:$y < \aleph_{x+1}$, then $y < \aleph_x \lor y \sim \aleph_x$ by Ordinal Less than Successor Aleph.
In either case, $\left|{ y }\right| \le \aleph_x$ by Ordinal in Aleph iff Cardinal in Aleph and Equivalent Sets have Equal Cardinal Numbers.
The existence of the surjection follows from Surjection iff Cardinal Inequality.
{{qed}}
\end{proof}
|
22472
|
\section{Surjection from Class to Proper Class}
Tags: Gödel-Bernays Class Theory, Class Mappings
\begin{theorem}
Let $A$ be a class.
Let $\mathrm P$ be a proper class.
Let $f: A \to \mathrm P$ be a surjection.
Then $A$ is proper.
\end{theorem}
\begin{proof}
{{NotZFC}}
{{AimForCont}} $A$ is not proper.
Then $A$ must be a set.
By the Axiom of Powers, $\powerset A$ is also a set.
Let $g: f \sqbrk A \to \powerset A$ be defined as:
:$\map g {\map f a} = f^{-1} \sqbrk {\set {\map f a} }$
It should be noted that:
:$\forall a \in A: \map {\paren {g \circ f} } a \ne \O$:
Suppose that $\map g {\map f a} = \map g {\map f b}$.
Let $x \in f^{-1} \sqbrk {\set {\map f a} }$.
Then by the definition of a singleton:
:$x \in f^{-1} \sqbrk {\set {\map f a} } \implies \map f x \in \set {\map f a} \implies \map f x = \map f a$
By the same argument:
:$x \in f^{-1} \sqbrk {\set {\map f b} } \implies \map f x = \map f b$
Hhence by the properties of equality:
:$\map f a = \map f b$
It has been shown that $g$ is an injection.
Because $f$ is a surjection, it follows by definition of surjection that:
:$f \sqbrk A = \mathrm P$
But this contradicts Injection from Proper Class to Class.
Thus by contradiction $A$ must be proper.
Hence the result.
{{qed}}
Category:Gödel-Bernays Class Theory
Category:Class Mappings
\end{proof}
|
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