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21773
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\section{Stabilizer of Polynomial}
Tags: Group Theory, Polynomial Theory, Stabilizers
\begin{theorem}
Let $n \in \Z: n > 0$.
Let $\map f {x_1, x_2, \ldots, x_n}$ be a polynomial in $n$ variables $x_1, x_2, \ldots, x_n$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi, \rho \in S_n$.
Let the group action $\pi * f$ be defined as the permutation on the polynomial $f$ by $\pi$.
Then the stabilizer of $f$ is the set of permutations on $n$ letters which fix $f$.
\end{theorem}
\begin{proof}
Follows directly from the definition of the stabilizer of $f$.
{{qed}}
\end{proof}
|
21774
|
\section{Stabilizer of Subgroup Action is Identity}
Tags: Subgroup Action, Examples of Group Actions, Stabilizers
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\struct {H, \circ}$ be a subgroup of $G$.
Let $*: H \times G \to G$ be the subgroup action defined for all $h \in H, g \in G$ as:
:$\forall h \in H, g \in G: h * g := h \circ g$
The stabilizer of $x \in G$ is $\set e$:
:$\Stab x = \set e$
\end{theorem}
\begin{proof}
From Subgroup Action is Group Action we have that $*$ is a group action.
Let $x \in G$.
Then:
{{begin-eqn}}
{{eqn | l = \Stab x
| r = \set {h \in H: h * x = x}
| c = {{Defof|Stabilizer}}
}}
{{eqn | r = \set {h \in H: h \circ x = x}
| c = Definition of $*$
}}
{{eqn | r = \set {h \in H: h = x \circ x^{-1} }
| c =
}}
{{eqn | r = \set {h \in H: h = e}
| c =
}}
{{eqn | r = \set e
| c =
}}
{{end-eqn}}
Hence the result, by definition of right coset.
{{qed}}
\end{proof}
|
21775
|
\section{Stabilizer of Subset Product Action on Power Set}
Tags: Subset Product Action, Group Actions, Stabilizers
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ be the power set of $\struct {G, \circ}$.
Let $*: G \times \powerset G \to \powerset G$ be the subset product action on $\powerset G$ defined as:
:$\forall g \in G: \forall S \in \powerset G: g * S = g \circ S$
where $g \circ S$ is the subset product $\set g \circ S$.
Then the stabilizer of $S$ in $\powerset G$ is the set:
:$\Stab S = S$
\end{theorem}
\begin{proof}
From the definition of stabilizer:
:$\Stab S = \set {g \in G: g * S = S}$
The result follows from the definition of the group action $*$ given.
{{qed}}
\end{proof}
|
21776
|
\section{Stabilizer of Subspace stabilizes Orthogonal Complement}
Tags: Hilbert Spaces
\begin{theorem}
Let $H$ be a finite-dimensional real or complex Hilbert space (that is, inner product space).
Let $t: H \to H$ be a normal operator on $H$.
Let $t$ stabilize a subspace $V$.
Then $t$ also stabilizes its orthogonal complement $V^\perp$.
\end{theorem}
\begin{proof}
Let $p: H \to V$ be the orthogonal projection of $H$ onto $V$.
Then the orthogonal projection of $H$ onto $V^\perp$ is $\mathbf 1 - p$, where $\mathbf 1$ is the identity map of $H$.
The fact that $t$ stabilizes $V$ can be expressed as:
:$\paren {\mathbf 1 - p} t p = 0$
or:
:$p t p = t p$
The goal is to show that:
:$p t \paren {\mathbf 1 - p} = 0$
We have that $\tuple {a, b} \mapsto \map \tr {a b^*}$ is an inner product on the space of endomorphisms of $H$.
Here, $b^*$ denotes the adjoint operator of $b$.
Thus it will suffice to show that $\map \tr {x x^*} = 0$ for $x = p t \paren {\mathbf 1 - p}$.
This follows from a direct computation, using properties of the trace and orthogonal projections:
{{begin-eqn}}
{{eqn | l = x x^*
| r = p t \paren {\mathbf 1 - p}^2 t^* p
| c =
}}
{{eqn | r = p t \paren {\mathbf 1 - p} t^* p
| c =
}}
{{eqn | r = p t t^* p - p t p t^* p
| c =
}}
{{eqn | ll= \leadsto
| l = \map \tr {x x^*}
| r = \map \tr {p t t^* p} - \map \tr {p t p t^* p}
| c =
}}
{{eqn | r = \map \tr {p^2 t t^*} - \map \tr {p^2 t p t^*}
| c =
}}
{{eqn | r = \map \tr {p t t^*} - \map \tr {\paren {p t p} t^*}
| c =
}}
{{eqn | r = \map \tr {p t t^*} - \map \tr {t p t^*}
| c =
}}
{{eqn | r = \map \tr {p t t^*} - \map \tr {p t^* t}
| c =
}}
{{eqn | r = \map \tr {p \paren {t t^* - t^*t} }
| c =
}}
{{eqn | r = \map \tr 0
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
{{qed}}
Category:Hilbert Spaces
\end{proof}
|
21777
|
\section{Stabilizers of Elements in Same Orbit are Conjugate Subgroups}
Tags: Group Actions, Stabilizers
\begin{theorem}
Let $G$ be a group acting on a set $X$.
Let:
:$y, z \in \Orb x$
where $\Orb x$ denotes the orbit of some $x \in X$.
Then their stabilizers $\Stab y$ and $\Stab z$ are conjugate subgroups.
\end{theorem}
\begin{proof}
From Stabilizer is Subgroup we have that both $\Stab y$ and $\Stab z$ are subgroups of $G$.
From definition of orbits:
:$\exists h_1, h_2 \in G: y = h_1 * x, z = h_2 * x$
Then $y = h_1 * \paren {h_2^{-1} * z} = h_1 h_2^{-1} * z$.
Thus:
{{begin-eqn}}
{{eqn | l = \Stab y
| r = \set {g \in G: g * y = y}
| c = {{Defof|Stabilizer}}
}}
{{eqn | r = \set {g \in G: g * \paren {h_1 h_2^{-1} * z} = h_1 h_2^{-1} * z}
}}
{{eqn | r = \set {g \in G: h_1^{-1} h_2 * \paren {g h_1 h_2^{-1} * z} = z}
}}
{{eqn | r = \set {g \in G: \paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } * z = z}
}}
{{eqn | r = \paren {h_1 h_2^{-1} } \set {\paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } \in G: \paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } * z = z} \paren {h_1 h_2^{-1} }^{-1}
}}
{{eqn | r = \paren {h_1 h_2^{-1} } \Stab z \paren {h_1 h_2^{-1} }^{-1}
| c = {{Defof|Stabilizer}}
}}
{{end-eqn}}
This shows that $\Stab y$ and $\Stab z$ are conjugate.
Hence the result.
{{qed}}
\end{proof}
|
21778
|
\section{Standard Bounded Metric is Metric}
Tags: Standard Bounded Metric is Metric, Minimum of 1 and Metric forms Metric, Examples of Metric Spaces
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $\bar d: A^2 \to \R$ be the standard bounded metric of $d$:
:$\forall \tuple {x, y} \in A^2: \map {\bar d} {x, y} = \min \set {1, \map d {x, y} }$
Then $\bar d$ is a metric for $A$.
\end{theorem}
\begin{proof}
It is to be demonstrated that $\bar d$ satisfies all the metric space axioms.
\end{proof}
|
21779
|
\section{Standard Bounded Metric is Metric/Topological Equivalence}
Tags: Minimum of 1 and Metric forms Metric, Standard Bounded Metric is Metric
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $\bar d: A^2 \to \R$ be the standard bounded metric of $d$:
:$\forall \tuple {x, y} \in A^2: \map {\bar d} {x, y} = \min \set {1, \map d {x, y} }$
$\bar d$ is topologically equivalent to $d$.
\end{theorem}
\begin{proof}
That $\bar d$ forms a metric on $M$ is demonstrated in Standard Bounded Metric is Metric.
We have that:
:$\forall x, y \in A^2: \map {\bar d} {x, y} \le \map d {x, y}$
Hence:
:$\map {B_\epsilon} {x; d} \subseteq \map {B_\epsilon} {x; \bar d}$
where $\map {B_\epsilon} {x; d}$ denotes the open $\epsilon$-ball of $x$ in $\struct {A, d}$.
Hence:
:if $U$ is $\bar d$-open, the $U$ is $d$-open
where $U$ is a subset of $A$.
Let $U$ be $d$-open.
Let $x \in U$.
Then $\map {B_\epsilon} {x; d} \subseteq U$ for some $\epsilon \in \R_{>0}$.
Let us take $\epsilon < 1$.
Then:
:$\map {B_\epsilon} {x; \bar d} = \map {B_\epsilon} {x; d} \subseteq U$
demonstrating that $U$ is $\bar d$-open.
The result follows.
{{qed}}
\end{proof}
|
21780
|
\section{Standard Continuous Uniform Distribution in terms of Exponential Distribution}
Tags: Exponential Distribution, Continuous Uniform Distribution
\begin{theorem}
Let $X$ and $Y$ be independent random variables.
Let $\beta$ be a strictly positive real number.
Let $X$ and $Y$ be random samples from the exponential distribution with parameter $\beta$.
Then:
:$\dfrac X {X + Y} \sim \operatorname U \openint 0 1$
where $\operatorname U \openint 0 1$ is the uniform distribution on $\openint 0 1$.
\end{theorem}
\begin{proof}
Note that the support of $\operatorname U \openint 0 1$ is $\openint 0 1$.
It is therefore sufficient to show that for $0 < z < 1$:
:$\map \Pr {\dfrac X {X + Y} \le z} = z$
Note that if $x, y > 0$ then:
:$0 < \dfrac x {x + y} < 1$
Note also that:
:$\dfrac x {x + y} \le z$
with $0 < z < 1$ is equivalent to:
:$x \le z x + z y$
which is in turn equivalent to:
:$\paren {1 - z} x \le z y$
That is:
:$x \le \dfrac z {1 - z} y$
We therefore have:
:$\map \Pr {\dfrac X {X + Y} \le z} = \map \Pr {X \le \dfrac z {1 - z} Y}$
Let $f_{X, Y}$ be the joint probability density function of $X$ and $Y$.
Let $f_X$ and $f_Y$ be the probability density function of $X$ and $Y$ respectively.
From Condition for Independence from Joint Probability Density Function, we have for each $x, y \in \R_{> 0}$:
:$\map {f_{X, Y} } {x, y} = \map {f_X} x \map {f_Y} y$
We therefore have:
{{begin-eqn}}
{{eqn | l = \map \Pr {\dfrac X {X + Y} \le z}
| r = \map \Pr {X \le \frac z {1 - z} Y}
}}
{{eqn | r = \int_0^\infty \int_0^{\frac z {1 - z} y} \map {f_{X, Y} } {x, y} \rd x \rd y
}}
{{eqn | r = \int_0^\infty \int_0^{\frac z {1 - z} y} \map {f_X} x \map {f_Y} y \rd x \rd y
}}
{{eqn | r = \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \paren {\int_0^{\frac z {1 - z} y} \frac 1 \beta \map \exp {-\frac x \beta} \rd x} \rd y
| c = {{Defof|Exponential Distribution}}
}}
{{eqn | r = \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \intlimits {-\map \exp {-\frac x \beta} } 0 {\frac z {1 - z} y} \rd y
| c = Primitive of Exponential of $a x$
}}
{{eqn | r = \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \paren {1 - \map \exp {-\frac z {\beta \paren {1 - z} } y} } \rd y
| c = Exponential of Zero
}}
{{eqn | r = 1 - \frac 1 \beta \int_0^\infty \map \exp {-y \paren {\frac z {\beta \paren {1 - z} } + \frac 1 \beta} } \rd y
}}
{{eqn | r = 1 - \frac 1 \beta \int_0^\infty \map \exp {-\frac y {\beta \paren {1 - z} } } \rd y
}}
{{eqn | r = 1 + \frac 1 \beta \intlimits {\beta \paren {1 - z} \map \exp {-\frac y {\beta \paren {1 - z} } } } 0 \infty
| c = Primitive of Exponential of $a x$
}}
{{eqn | r = 1 + \frac 1 \beta \paren {0 - \beta \paren {1 - z} }
| c = Exponential Tends to Zero and Infinity, Exponential of Zero
}}
{{eqn | r = 1 - \paren {1 - z}
}}
{{eqn | r = z
}}
{{end-eqn}}
{{qed}}
Category:Continuous Uniform Distribution
Category:Exponential Distribution
\end{proof}
|
21781
|
\section{Standard Discrete Metric induces Discrete Topology}
Tags: Discrete Metric, Discrete Metrics, Metrizable Topologies, Discrete Topology
\begin{theorem}
Let $M = \struct {A, d}$ be the (standard) discrete metric space on $A$.
Then $d$ induces the discrete topology on $A$.
Thus the discrete topology is metrizable.
\end{theorem}
\begin{proof}
Let $a \in A$.
From Subset of Standard Discrete Metric Space is Open, a set $U \subseteq A$ is open in $M$.
So, in particular, $\set a$ is open in $\struct {A, d}$.
This holds for all $a \in A$.
From Metric Induces Topology it follows that $\set a$ is an open set in $\struct {A, \tau_{A, d} }$.
The result follows from Basis for Discrete Topology.
{{qed}}
\end{proof}
|
21782
|
\section{Standard Discrete Metric is Metric}
Tags: Discrete Metrics, Metric Spaces, Examples of Metrics, Discrete Metric
\begin{theorem}
The standard discrete metric is a metric.
\end{theorem}
\begin{proof}
Let $d: S \times S \to \R$ denote the standard discrete metric on the underlying set $S$ of some space $\struct {S, d}$.
By definition:
:$\forall x, y \in S: \map d {x, y} = \begin {cases}
0 & : x = y \\
1 & : x \ne y
\end {cases}$
\end{proof}
|
21783
|
\section{Standard Discrete Metric is not Topologically Equivalent to p-Product Metrics}
Tags: Standard Discrete Metric, Discrete Metrics, P-Product Metrics
\begin{theorem}
For $n \in \N$, let $\R^n$ be an Euclidean space.
Let $p \in \R_{\ge 1}$.
Let $d_p$ be the $p$-product metric on $\R^n$.
Let $d_0$ be the standard discrete metric on $\R^n$.
Then $d_p$ and $d_0$ are not topologically equivalent.
\end{theorem}
\begin{proof}
From Open Ball in Standard Discrete Metric Space it is seen that singletons are open sets in $\struct {\R^n, d_0}$.
However, this is not the case in the $\struct {\R^n, d_p}$.
{{qed}}
\end{proof}
|
21784
|
\section{Standard Gaussian Random Variable as Transformation of Gaussian Random Variable}
Tags: Gaussian Distribution
\begin{theorem}
Let $\mu$ be a real number.
Let $\sigma$ be a positive real number.
Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the Gaussian distribution with parameters $\mu$ and $\sigma^2$.
Then:
:$\dfrac {X - \mu} \sigma \sim \Gaussian 0 1$
where $\Gaussian 0 1$ is the standard Gaussian distribution.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \frac {X - \mu} \sigma
| r = \frac 1 \sigma X - \frac \mu \sigma
}}
{{eqn | o = \sim
| r = \Gaussian {\frac \mu \sigma - \frac \mu \sigma} {\paren {\frac 1 \sigma}^2 \sigma^2}
| c = Linear Transformation of Gaussian Random Variable
}}
{{eqn | r = \Gaussian 0 1
}}
{{end-eqn}}
{{qed}}
Category:Gaussian Distribution
\end{proof}
|
21785
|
\section{Standard Machinery}
Tags: Measure Theory, Proof Techniques
\begin{theorem}
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^1_{\overline \R} } \mu$ be the space of $\mu$-integrable functions.
Let $\map P {f_1, \ldots, f_n}$ be a proposition, where the variables $f_i$ denote $\mu$-measurable functions $f_i: X \to \overline \R$.
Let every occurrence of an $f_i$ be of the form:
:$\ds \int \map \Phi {f_i} \rd \mu$
for a suitable index set $I$ and multilinear mapping $\Phi: \map {\LL^1_{\overline \R} } \mu^I \to \map {\LL^1_{\overline \R} } \mu$.
Denote with $\map \chi \Sigma$ the set of characteristic functions of elements of $\Sigma$, that is:
:$\map \chi \Sigma := \set {\chi_E: X \to \R: E \in \Sigma}$
Then the following are equivalent:
:$(A): \quad \forall f_1, \ldots, f_n \in \map \chi \Sigma: \map P {f_1, \ldots, f_n}$
:$(B): \quad \forall f_1, \ldots, f_n \in \map {\LL^1_{\overline \R} } \mu: \map P {f_1, \ldots, f_n}$
that is, it suffices to verify $P$ holds for characteristic functions.
{{WIP|Not completely valid as stated; need to restrict to $E$ with $\map \mu E < +\infty$ and impose more conditions on $P$}}
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Proof Techniques
Category:Measure Theory
\end{proof}
|
21786
|
\section{Star Convex Set is Path-Connected}
Tags: Path-Connected Spaces, Connectedness, Vector Spaces, Path-Connected Sets
\begin{theorem}
Let $A$ be a star convex subset of a vector space $V$ over $\R$ or $\C$.
Then $A$ is path-connected.
\end{theorem}
\begin{proof}
Let $x_1, x_2 \in A$.
Let $a \in A$ be a star center of $A$.
By definition of star convex set, it follows that for all $t \in \left[{0 \,.\,.\, 1}\right]$, we have $t x_1 + \left({1 - t}\right) a, t x_2 + \left({1 - t}\right) a \in A$.
Define two paths $\gamma_1, \gamma_2: t \in \left[{0 \,.\,.\, 1}\right] \to A$ by $\gamma_1 \left({t}\right) = t x_1 + \left({1 - t}\right) a$, and $\gamma_2 \left({t}\right) = t a + \left({1 - t}\right) x_2$.
As $\gamma_2 \left({t}\right) = \left({1 - t}\right) x_2 + \left({1 - \left({1 - t}\right) }\right) a$, and $\left({1 - t}\right) \in \left[{0 \,.\,.\, 1}\right]$, it follows that $\gamma_2 \left({t}\right) \in A$.
Note that $\gamma_1 \left({0}\right) = x_1$, $\gamma_1 \left({1}\right) = \gamma_2 \left({0}\right) = a$, and $\gamma_2 \left({1}\right) = x_2$.
Define $\gamma: \left[{0 \,.\,.\, 1}\right] \to A$ as the concatenation $\gamma_1 * \gamma_2$.
Then $\gamma$ is a path in $A$ joining $x_1$ and $x_2$, so $A$ is path-connected.
{{qed}}
\end{proof}
|
21787
|
\section{State Code Function is Primitive Recursive}
Tags: Primitive Recursive Functions, URM Programs
\begin{theorem}
Let $k \in \N^*$.
Let $e = \gamma \left({P}\right)$ be the code number of a URM program $P$.
Let $\left({n_1, n_2, \ldots, n_k}\right)$ be the input of $P$.
Let $S_k: \N^{k+2} \to \N$ be the function defined as:
:$S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$ is the state code for $P$ at stage $t$ of computation of $P$.
If $e$ does not code a URM Program then $S_k = 0$.
Also, if $P$ terminates at stage $t_0$, then we put:
:$\forall t > t_0: S_k \left({e, n_1, n_2, \ldots, n_k, t}\right) = S_k \left({e, n_1, n_2, \ldots, n_k, t_0}\right)$.
Then for all $k \ge 1$, the function $S_k$ is primitive recursive.
\end{theorem}
\begin{proof}
It can easily be seen that $S_k$ is a total function.
Suppose $e = \gamma \left({P}\right)$ for some URM program $P$.
At stage $0$, we are about to carry out instruction $1$ with the input $\left({n_1, n_2, \ldots, n_k}\right)$.
So we have:
:$S_k \left({e, n_1, n_2, \ldots, n_k, 0}\right) = \begin{cases}
2^1 3^{n_1} 5^{n_2} \cdots p_{k+1}^{n_k} & : e \in \operatorname{Prog} \\
0 & : \text{otherwise}
\end{cases}$
where $\operatorname{Prog}$ is the set of code numbers of all URM programs.
We see that $S_k \left({e, n_1, n_2, \ldots, n_k, 0}\right)$ does not actually depend upon the actual program being run, beyond the fact that it matters whether it actually ''is'' a program or not.
Now $\operatorname{Prog}$ is a primitive recursive set.
So from results about primitive recursive functions, the relations defining the cases are primitive recursive.
We can also deduce from various results about primitive recursive functions that the functions given by the formulas $2^1 3^{n_1} 5^{n_2} \cdots p_{k+1}^{n_k}$ and $0$ are primitive recursive.
In particular, we use the results:
* Multiplication is Primitive Recursive;
* Exponentiation is Primitive Recursive;
* Prime Enumeration Function is Primitive Recursive.
So from Definition by Cases is Primitive Recursive, $S_k \left({e, n_1, n_2, \ldots, n_k, 0}\right)$ is primitive recursive.
Now we want to be able to express $S_k \left({e, n_1, n_2, \ldots, n_k, t+1}\right)$ in terms of $e, \left({n_1, n_2, \ldots, n_k}\right), t$ and $S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$ using known primitive recursive functions.
We need to consider a number of cases:
#$e$ does not code a URM program;
#$e = \gamma \left({P}\right)$ and the computation halts at stage $t$ or earlier;
#$e = \gamma \left({P}\right)$ and the instruction to be carried out at stage $t$ is a <tt>Zero</tt> instruction;
#$e = \gamma \left({P}\right)$ and the instruction to be carried out at stage $t$ is a <tt>Successor</tt> instruction;
#$e = \gamma \left({P}\right)$ and the instruction to be carried out at stage $t$ is a <tt>Copy</tt> instruction;
#$e = \gamma \left({P}\right)$ and the instruction to be carried out at stage $t$ is a <tt>Jump</tt> instruction.
These cases are clearly mutually exclusive and exhaustive.
First we need to check that each case corresponds to a primitive recursive relation.
* The set $\operatorname{Prog}$ is primitive recursive so its complement is also primitive recursive.
So 1. is a primitive recursive relation.
* So we have that $e$ codes a URM program. Call that program $P$.
From the definition of state code, we see that if a computation halts at stage $t$ or earlier, then the number of the instruction to be carried out at stage $t$ is greater than the number of instructions in $P$.
From the definition of the code number of $P$, the number of instructions in $P$ is $\operatorname{len} \left({e}\right)$ where $\operatorname{len} \left({e}\right)$ is the length of $e$, which is primitive recursive.
Now let $r = S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$.
Let $\left({r}\right)_j$ be defined as the prime exponent function.
By the definition of the state code, the number of the instruction to be carried out at stage $t$ is $\left({r}\right)_1$, which is primitive recursive.
So 2. can be expressed as:
:$e \in \operatorname{Prog} \text { and } \left({S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)}\right)_1 > \operatorname{len} \left({e}\right)$
Both $\operatorname{Prog}$ and $\left({r}\right)_1$ are primitive recursive, so from Set Operations on Primitive Recursive Relations, 2. is a primitive recursive relation.
* So, let the number of the instruction to be carried out at stage $t$ be $a = \left({S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)}\right)_1$.
From the definition of the code number of $P$, the code number of this instruction is $\left({e}\right)_a$.
Now from Set of Codes for URM Instructions is Primitive Recursive, each of the sets $\operatorname{Zinstr}$, $\operatorname{Sinstr}$, $\operatorname{Cinstr}$ and $\operatorname{Jinstr}$ are primitive recursive.
So each of 3. to 6. above can be expressed as:
:$e \in \operatorname{Prog} \text { and } a \le \operatorname{len} \left({e}\right) \text { and } \left({e}\right)_a \in \operatorname{Instr}$
and is a primitive recursive relation.
So relations 1. to 6. are all primitive recursive.
Now we need to show how, in each case, $S_k \left({e, n_1, n_2, \ldots, n_k, t+1}\right)$ can be obtained from $e, \left({n_1, n_2, \ldots, n_k}\right), t$ and $S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$ using known primitive recursive functions.
First, if $e$ does not code a URM program then $S_k \left({e, n_1, n_2, \ldots, n_k, t+1}\right) = 0$, which is primitive recursive.
Second, we have adopted the convention that if $P$ has halted, then $S_k$ does not change.
So if $P$ halts at or before stage $t$, we have that $S_k \left({e, n_1, n_2, \ldots, n_k, t+1}\right) = S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$
Next, we look at the individual commands.
As an example we will investigate the <tt>Successor</tt> command. The others are treated similarly.
Suppose the instruction to be carried out at stage $t$ is a <tt>Successor</tt> command.
We know that the code number $c$ is given by $c = \left({e}\right)_a$ where $a = \left({S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)}\right)_1$.
Suppose the instruction is $S \left({n}\right)$. Then $c = 6 n$.
So $n = \operatorname{quot} \left({6, n}\right)$ which is recursive from Quotient is Primitive Recursive.
This instruction adds $1$ to the number in $R_n$.
This increases the exponent $p_{n+1}$ in the state code by $1$.
This is achieved by multiplying $S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$ by $p \left({n+1}\right)$, where $p \left({n+1}\right)$ is the prime enumeration function which is primitive recursive.
Since the instruction to be carried out at stage $t$ is a <tt>Successor</tt> the instruction number at stage $t+1$ is $a+1$ so the factor $2^a$ in $S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$ is replaced by $2^{a+1}$.
So:
:$S_k \left({e, n_1, n_2, \ldots, n_k, t+1}\right) = 2 \times p_{n+1} \times S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$
where $n = \operatorname{quot} \left({6, n}\right)$, $c = \left({e}\right)_a$ and $a = \left({S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)}\right)_1$.
This is the required expression for $S_k \left({e, n_1, n_2, \ldots, n_k, t+1}\right)$ obtained by substitution from primitive recursive functions.
The proofs for $\operatorname{Zinstr}$, $\operatorname{Cinstr}$ and $\operatorname{Jinstr}$ are along the same lines.
In each case, the value of $S_k \left({e, n_1, n_2, \ldots, n_k, t+1}\right)$ can be obtained by substitution from primitive recursive functions (but I'd hate to have to do the calculations on my fingers).
Thus by Definition by Cases is Primitive Recursive, $S_k \left({e, n_1, n_2, \ldots, n_k, t+1}\right)$ is primitive recursive.
Hence $S_k$ is defined by primitive recursion from functions known to be primitive recursive.
Hence the result.
{{qed}}
\end{proof}
|
21788
|
\section{Steiner's Calculus Problem}
Tags: Euler's Number
\begin{theorem}
Let $f: \R_{>0} \to \R$ be the real function defined as:
:$\forall x \in \R_{>0}: \map f x = x^{1/x}$
Then $\map f x$ reaches its maximum at $x = e$ where $e$ is Euler's number .
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map {f'} x
| r = \frac \d {\d x} x^{1/x}
}}
{{eqn | r = \frac \d {\d x} e^{\ln x / x}
}}
{{eqn | r = e^{\ln x / x} \paren {\frac 1 {x^2} - \frac {\ln x} {x^2} }
}}
{{eqn | r = \frac {x^{1/x} } {x^2} \paren {1 - \ln x}
}}
{{end-eqn}}
$\dfrac {x^{1/x} } {x^2}$ is always greater than $0$.
Therefore:
:$\map {f'} x > 0$ for $\ln x < 1$
:$\map {f'} x = 0$ for $\ln x = 1$
:$\map {f'} x < 0$ for $\ln x > 1$
By Derivative at Maximum or Minimum, maximum is obtained when $\ln x = 1$,
that is, when $x = e$.
{{qed}}
{{Namedfor|Jakob Steiner|cat = Steiner}}
\end{proof}
|
21789
|
\section{Steiner-Lehmus Theorem}
Tags: Triangles, Steiner-Lehmus Theorem
\begin{theorem}
Let $ABC$ be a triangle.
Denote the lengths of the angle bisectors through the vertices $A$ and $B$ by $\omega_\alpha$ and $\omega_\beta$.
Let $\omega_\alpha = \omega_\beta$.
Then $ABC$ is an isosceles triangle.
:250px
\end{theorem}
\begin{proof}
:250px
Let $a$, $b$, and $c$ be the sides opposite $A$, $B$ and $C$ respectively.
By Length of Angle Bisector, $\omega_\alpha, \omega_\beta$ are given by:
:$\omega_\alpha^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$
:$\omega_\beta^2 = \dfrac {a c} {\paren {a + c}^2} \paren {\paren {a + c}^2 - b^2}$
Equating $\omega_\alpha^2$ with $\omega_\beta^2$:
{{begin-eqn}}
{{eqn | l = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}
| r = \dfrac {a c} {\paren {a + c}^2} \paren {\paren {a + c}^2 - b^2}
}}
{{eqn | ll= \leadsto
| l = b c \paren {a + c}^2 \paren {\paren {b + c}^2 - a^2}
| r = a c \paren {b + c}^2 \paren {\paren {a + c}^2 - b^2}
}}
{{eqn | ll= \leadsto
| l = b c \paren {a + c}^2 \paren {b + c + a} \paren {b + c - a}
| r = a c \paren {b + c}^2 \paren {a + c + b} \paren {a + c - b}
}}
{{eqn | ll= \leadsto
| l = b \paren {a + c}^2 \paren {b + c - a}
| r = a \paren {b + c}^2 \paren {a + c - b}
| c = as $a + b + c > 0$
}}
{{eqn | ll= \leadsto
| l = a^2 b^2 + 2 a b^2 c + b^2 c^2 + a^2 b c + 2 a b c^2 + b c^3 - a^3 b - 2 a^2 b c - a b c^2
| r = a^2 b^2 + 2 a^2 b c + a^2 c^2 + a b^2 c + 2 a b c^2 + a c^3 - a b^3 - 2 a b^2 c - a b c^2
}}
{{eqn | ll= \leadsto
| l = 2 a b^2 c + b^2 c^2 - a^2 b c + b c^3 - a^3 b
| r = 2 a^2 b c + a^2 c^2 - a b^2 c + a c^3 - a b^3
}}
{{eqn | ll= \leadsto
| l = 3 a b^2 c - 3 a^2 b c + b^2 c^2 + b c^3 - a^3 b - a^2 c^2 - a c^3 + a b^3
| r = 0
}}
{{eqn | ll= \leadsto
| l = 3 a b c \paren {b - a} + c^2 \paren {b^2 - a^2} + c^3 \paren {b - a} + a b \paren {b^2 - a^2}
| r = 0
}}
{{eqn | ll= \leadsto
| l = \paren {b - a} \paren {3 a b c + a b \paren {a + b} + c^3 + c^2 \paren {a + b} }
| r = 0
}}
{{eqn | ll= \leadsto
| l = b - a
| r = 0
| c = as $a, b, c > 0$
}}
{{eqn | ll= \leadsto
| l = a
| r = b
}}
{{end-eqn}}
Therefore $ABC$ is an isosceles triangle.
{{qed}}
\end{proof}
|
21790
|
\section{Step Function satisfies Dirichlet Conditions}
Tags: Step Functions, Dirichlet Conditions
\begin{theorem}
Let $\alpha, \beta \in \R$ be a real numbers such that $\alpha < \beta$.
Let $\map f x$ be a step function defined on the interval $\openint \alpha \beta$.
Then $f$ satisfies the Dirichlet conditions.
\end{theorem}
\begin{proof}
Recall the definition of step function:
:{{Definition:Step Function}}
Recall the Dirichlet conditions:
{{:Definition:Dirichlet Conditions}}
We inspect the Dirichlet conditions in turn.
\end{proof}
|
21791
|
\section{Stewart's Theorem}
Tags: Triangles, Euclidean Geometry, Named Theorems
\begin{theorem}
Let $\triangle ABC$ be a triangle with sides $a, b, c$.
Let $CP$ be a cevian from $C$ to $P$.
:400px
Then:
:$a^2 \cdot AP + b^2 \cdot PB = c \paren {CP^2 + AP \cdot PB}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | n = 1
| l = b^2
| r = AP^2 + CP^2 - 2 AP \cdot CP \cdot \map \cos {\angle APC}
| c = Law of Cosines
}}
{{eqn | n = 2
| l = a^2
| r = PB^2 + CP^2 - 2 CP \cdot PB \cdot \map \cos {\angle BPC}
| c = Law of Cosines
}}
{{eqn | r = PB^2 + CP^2 + 2 CP \cdot PB \cdot \map \cos {\angle APC}
| c = Cosine of Supplementary Angle
}}
{{eqn | n = 3
| ll= \leadsto \quad
| l = b^2 \cdot PB
| r = AP^2 \cdot PB + CP^2 \cdot PB - 2 PB \cdot AP \cdot CP \cdot \map \cos {\angle APC}
| c = $(1) \ \times PB$
}}
{{eqn | n = 4
| l = a^2 \cdot AP
| r = PB^2 \cdot AP + CP^2 \cdot AP + 2 AP \cdot CP \cdot PB \cdot \map \cos {\angle APC}
| c = $(2) \ \times AP$
}}
{{eqn | ll= \leadsto \quad
| l = a^2 \cdot AP + b^2 \cdot PB
| r = AP^2 \cdot PB + PB^2 \cdot AP + CP^2 \cdot PB + CP^2 \cdot AP
| c = $(3) \ + \ (4)$
}}
{{eqn | r = CP^2 \left({PB + AP}\right) + AP \cdot PB \paren {PB + AP}
}}
{{eqn | r = c \paren {CP^2 + AP \cdot PB}
| c = as $PB + AP = c$
}}
{{end-eqn}}
{{qed}}
{{namedfor|Matthew Stewart|cat = Stewart}}
It is also known as Apollonius's Theorem after {{AuthorRef|Apollonius of Perga}}.
Category:Triangles
\end{proof}
|
21792
|
\section{Stieltjes Function of Measure is Stieltjes Function}
Tags: Measure Theory
\begin{theorem}
Let $\mu$ be a measure on $\R$ with the Borel $\sigma$-algebra $\map \BB \R$.
Suppose that for every $n \in \N$:
:$\map \mu {\closedint {-n} n} < +\infty$
Then $F_\mu: \R \to \overline \R$, the Stieltjes function of $\mu$, is a Stieltjes function.
\end{theorem}
\begin{proof}
By definition, $F_\mu$ is a Stieltjes function {{iff}} it is increasing and left-continuous.
\end{proof}
|
21793
|
\section{Stieltjes Function of Measure of Finite Stieltjes Function}
Tags: Measure Theory
\begin{theorem}
Let $f: \R \to \R$ be a finite Stieltjes function.
Let $\mu_f$ be the measure of $f$.
Let $f_{\mu_f}$ be the Stieltjes function of $\mu_f$.
Then $f_{\mu_f} = f$.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Measure Theory
\end{proof}
|
21794
|
\section{Stirling's Formula}
Tags: Factorials, Special Functions, Asymptotics, Analysis, Named Theorems, Stirling's Formula
\begin{theorem}
The factorial function can be approximated by the formula:
:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
where $\sim$ denotes asymptotically equal.
\end{theorem}
\begin{proof}
Let $a_n = \dfrac {n!} {\sqrt{2n} \left({\frac n e}\right)^n}$.
\end{proof}
|
21795
|
\section{Stirling's Formula/Proof 2/Lemma 2}
Tags: Stirling's Formula
\begin{theorem}
The sequence $\sequence {d_n}$ defined as:
:$d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$
is decreasing.
\end{theorem}
\begin{proof}
The proof strategy is to demonstrate that the sign of $d_n - d_{n + 1}$ is positive.
{{begin-eqn}}
{{eqn | l = d_n - d_{n + 1}
| r = \map \ln {n!} - \paren {n + \frac 1 2} \ln n + n
| c =
}}
{{eqn | o =
| ro= -
| r = \paren {\map \ln {\paren {n + 1}!} - \paren {n + 1 + \frac 1 2} \map \ln {n + 1} + n + 1}
| c =
}}
{{eqn | r = -\map \ln {n + 1} - \paren {n + \frac 1 2} \ln n + \paren {n + \frac 3 2} \map \ln {n + 1} - 1
| c = (as $\map \ln {\paren {n + 1}!} = \map \ln {n + 1} + \map \ln {n!}$)
}}
{{eqn | r = \paren {n + \frac 1 2} \map \ln {\frac {n + 1} n} - 1
| c =
}}
{{eqn | n = 1
| r = \frac {2 n + 1} 2 \map \ln {\frac {1 + \paren {2 n + 1}^{-1} } {1 - \paren {2 n + 1}^{-1} } } - 1
| c =
}}
{{end-eqn}}
Let:
:$\map f x := \dfrac 1 {2 x} \map \ln {\dfrac {1 + x} {1 - x} } - 1$
for $\size x < 1$.
Then from Lemma 1:
:$(2): \quad \ds \map f x = \sum_{k \mathop = 1}^\infty \frac {x^{2 n} } {2 n + 1}$
Thus $\map f x > 0$ for $\size x < 1$.
Putting $x = \dfrac 1 {2 n + 1}$ it can be seen that $(1)$ is $\map f {\dfrac 1 {2 n + 1} }$.
As $-1 < \dfrac 1 {2 n + 1} < 1$ it can be seen that $(2)$ can be applied and so:
:$\forall n \in \N: d_n - d_{n + 1} \ge 0$
Thus $\sequence {d_n}$ is a decreasing sequence.
{{qed}}
\end{proof}
|
21796
|
\section{Stirling's Formula/Proof 2/Lemma 5}
Tags: Stirling's Formula, Sine Function, Integral Calculus
\begin{theorem}
:$\dfrac {n!} {n^n \sqrt n e^{-n} } \to \sqrt {2 \pi}$ as $n \to \infty$
\end{theorem}
\begin{proof}
By previous work done in Stirling's Formula: Proof 2 it is noted that $\sequence {\dfrac {n!} {n^n \sqrt n e^{-n} } }_{n \mathop \in \N}$ is a convergent sequence.
Let $\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$.
Let $\ds I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$.
Then:
{{begin-eqn}}
{{eqn | l = \frac {I_{2 n} } {I_{2 n + 1} }
| r = \frac {\paren {2 n}!} {\paren {2^n n!}^2} \frac \pi 2 \cdot \frac {\paren {2 n + 1}!} {\paren {2^n n!}^2}
| c = Definite Integral from $0$ to $\dfrac \pi 2$ of Even Power of $\sin x$ and Corollary 2
}}
{{eqn | r = \frac \pi 2 \paren {2 n + 1} \frac {\paren {\paren {2 n}!}^2} {\paren {2^n n!}^4}
| c = extracting $\paren {2 n + 1}$ as a factor and rearranging
}}
{{eqn | o = \sim
| r = \frac \pi 2 \paren {2 n + 1} \frac {\paren {C \paren {2 n}^{2 n + 1/2} e^{-2 n} }^2} {\paren {2^n C n^{n + 1/2} e^{-n} }^4}
| c = entering $C$ into top and bottom
}}
{{eqn | r = \frac \pi 2 \frac {\paren {2 n + 1} } {C^2} \frac {2^{4 n} 2 \paren {n^{4 n + 1} } e^{-4 n} } {2^{4 n} \paren {n^{4 n + 2} } e^{-4 n} }
| c = rearranging
}}
{{eqn | r = \frac {\pi \paren {2 n + 1} } {n C^2}
| c = much simplification
}}
{{eqn | o = \to
| r = \frac {2 \pi} {C^2}
| rr= \text {as $n \to \infty$}
}}
{{end-eqn}}
From Lemma 4:
:$\ds \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$
from which:
{{begin-eqn}}
{{eqn | l = \frac {2 \pi} {C^2}
| r = 1
}}
{{eqn | ll= \leadsto
| l = C^2
| r = 2 \pi
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
21797
|
\section{Stirling's Formula/Refinement}
Tags: Stirling's Formula
\begin{theorem}
A refinement of Stirling's Formula is:
:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {12 n} }$
where $\sim$ denotes asymptotically equal.
\end{theorem}
\begin{proof}
From Limit of Error in Stirling's Formula:
:$e^{1 / \paren {12 n + 1} } \le \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } \le e^{1 / 12 n}$
{{Finish}}
\end{proof}
|
21798
|
\section{Stirling Number of n with n-m is Polynomial in n of Degree 2m/Unsigned First Kind}
Tags: Stirling Numbers
\begin{theorem}
Let $m \in \Z_{\ge 0}$.
The unsigned Stirling number of the first kind $\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.
\end{theorem}
\begin{proof}
The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.
\end{proof}
|
21799
|
\section{Stirling Number of the Second Kind of 0}
Tags: Stirling Numbers, Examples of Stirling Numbers of the Second Kind
\begin{theorem}
:$\ds {0 \brace n} = \delta_{0 n}$
where:
:$\ds {0 \brace n}$ denotes a Stirling number of the second kind
:$\delta_{0 n}$ denotes the Kronecker delta.
\end{theorem}
\begin{proof}
By definition of Stirling numbers of the second kind:
$\ds x^{\underline 0} = \sum_k {0 \brace k} x^k$
Thus we have:
{{begin-eqn}}
{{eqn | l = x^0
| r = 1
| c = {{Defof|Integer Power}}
}}
{{eqn | r = x^{\underline 0}
| c = Number to Power of Zero Falling is One
}}
{{end-eqn}}
Thus, in the expression:
:$\ds x^0 = \sum_k {0 \brace k} x^{\underline k}$
we have:
:$\ds {0 \brace 0} = 1$
and for all $k \in \Z_{>0}$:
:$\ds {0 \brace k} = 0$
That is:
:$\ds {0 \brace k} = \delta_{0 k}$
{{qed}}
\end{proof}
|
21800
|
\section{Stirling Number of the Second Kind of Number with Greater}
Tags: Stirling Numbers, Stirling Number of the Second Kind of Number with Greater
\begin{theorem}
Let $n, k \in \Z_{\ge 0}$.
Let $k > n$.
Let $\ds {n \brace k}$ denote a Stirling number of the second kind.
Then:
:$\ds {n \brace k} = 0$
\end{theorem}
\begin{proof}
By definition, the Stirling numbers of the second kind are defined as the coefficients $\displaystyle \left\{ {n \atop k}\right\}$ which satisfy the equation:
:$\displaystyle x^n = \sum_k \left\{ {n \atop k}\right\} x^{\underline k}$
where $x^{\underline k}$ denotes the $k$th falling factorial of $x$.
Both of the expressions on the {{LHS}} and {{RHS}} are polynomials in $x$ of degree $n$.
Hence the coefficient $\displaystyle \left\{ {n \atop k}\right\}$ of $x^{\underline k}$ where $k > n$ is $0$.
{{qed}}
\end{proof}
|
21801
|
\section{Stirling Number of the Second Kind of Number with Greater/Proof 1}
Tags: Stirling Numbers, Stirling Number of the Second Kind of Number with Greater
\begin{theorem}
Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Stirling Number of the Second Kind of Number with Greater}}
\end{theorem}
\begin{proof}
By definition, the Stirling numbers of the second kind are defined as the coefficients $\ds {n \brace k}$ which satisfy the equation:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
where $x^{\underline k}$ denotes the $k$th falling factorial of $x$.
Both of the expressions on the {{LHS}} and {{RHS}} are polynomials in $x$ of degree $n$.
Hence the coefficient $\ds {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$.
{{qed}}
Category:Stirling Number of the Second Kind of Number with Greater
\end{proof}
|
21802
|
\section{Stirling Number of the Second Kind of Number with Greater/Proof 2}
Tags: Stirling Numbers, Stirling Number of the Second Kind of Number with Greater
\begin{theorem}
Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Stirling Number of the Second Kind of Number with Greater}}
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds k > n \implies {n \brace k} = 0$
\end{proof}
|
21803
|
\section{Stirling Number of the Second Kind of Number with Self}
Tags: Stirling Numbers
\begin{theorem}
:$\ds {n \brace n} = 1$
where $\ds {n \brace n}$ denotes a Stirling number of the second kind.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds {n \brace n} = 1$
\end{proof}
|
21804
|
\section{Stirling Number of the Second Kind of n+1 with 0}
Tags: Stirling Numbers, Examples of Stirling Numbers of the Second Kind
\begin{theorem}
Let $n \in \Z_{\ge 0}$.
Then:
:$\ds {n + 1 \brace 0} = 0$
where $\ds {n + 1 \brace 0}$ denotes a Stirling number of the second kind.
\end{theorem}
\begin{proof}
We are given that $k = 0$.
So by definition of unsigned Stirling number of the first kind:
:$\ds {n \brace k} = \delta_{n k}$
where $\delta_{n k}$ is the Kronecker delta.
Thus
{{begin-eqn}}
{{eqn | l = n
| o = \ge
| r = 0
| c = by hypothesis
}}
{{eqn | ll= \leadsto
| l = n + 1
| o = >
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = n + 1
| o = \ne
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = \delta_{\paren {n + 1} 0}
| r = 0
| c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
21805
|
\section{Stirling Number of the Second Kind of n+1 with 1}
Tags: Stirling Numbers, Examples of Stirling Numbers of the Second Kind
\begin{theorem}
Let $n \in \Z_{\ge 0}$.
Then:
:$\ds {n + 1 \brace 1} = 1$
where $\ds {n + 1 \brace 1}$ denotes a Stirling number of the second kind.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n + 1 \brace 1} = 1$
\end{proof}
|
21806
|
\section{Stirling Number of the Second Kind of n+1 with 2}
Tags: Stirling Numbers, Examples of Stirling Numbers of the Second Kind
\begin{theorem}
Let $n \in \Z_{\ge 0}$.
Then:
:$\ds {n + 1 \brace 2} = 2^n - 1$
where $\ds {n + 1 \brace 2}$ denotes a Stirling number of the second kind.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n + 1 \brace 2} = 2^n - 1$
\end{proof}
|
21807
|
\section{Stirling Number of the Second Kind of n with n-1}
Tags: Stirling Numbers, Examples of Stirling Numbers of the Second Kind
\begin{theorem}
Let $n \in \Z_{> 0}$ be an integer greater than $0$.
Then:
:$\ds {n \brace n - 1} = \binom n 2$
where:
:$\ds {n \brace n - 1}$ denotes a Stirling number of the second kind
:$\dbinom n 2$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds {n \brace n - 1} = \binom n 2$
\end{proof}
|
21808
|
\section{Stirling Number of the Second Kind of n with n-2}
Tags: Examples of Stirling Numbers of the Second Kind
\begin{theorem}
Let $n \in \Z_{\ge 2}$ be an integer greater than or equal to $2$.
Then:
:$\ds {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$
where:
:$\ds {n \brace n - 2}$ denotes an Stirling number of the second kind
:$\dbinom n 4$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
\end{proof}
|
21809
|
\section{Stirling Number of the Second Kind of n with n-3}
Tags: Examples of Stirling Numbers of the Second Kind
\begin{theorem}
Let $n \in \Z_{\ge 3}$ be an integer greater than or equal to $3$.
Then:
:$\ds {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$
where:
:$\ds {n \brace n - 3}$ denotes an Stirling number of the second kind
:$\dbinom n 6$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
\end{proof}
|
21810
|
\section{Stolz-Cesàro Theorem}
Tags: Limits of Sequences, Analysis
\begin{theorem}
Let $\sequence {a_n}$ be a sequence.
{{explain|Domain of $\sequence {a_n}$ -- $\R$ presumably but could it be $\C$?}}
Let $\sequence {b_n}$ be a sequence of (strictly) positive real numbers such that:
:$\ds \sum_{i \mathop = 0}^\infty b_n = \infty$
If:
:$\ds \lim_{n \mathop \to \infty} \dfrac {a_n} {b_n} = L \in \R$
then also:
:$\ds \lim_{n \mathop \to \infty} \dfrac {a_1 + a_2 + \cdots + a_n} {b_1 + b_2 + \cdots + b_n} = L$
\end{theorem}
\begin{proof}
Define the following sums:
:$\ds A_n = \sum_{i \mathop = 1}^n a_i$
:$\ds B_n = \sum_{i \mathop = 1}^n b_i$
Let $\epsilon > 0$ and $\mu = \dfrac {\epsilon} 2$.
By the definition of convergent sequences, there exists $k \in \N$ such that:
:$\forall n > k: \paren {L - \mu} b_n < a_n < \paren {L + \mu} b_n$
Rewrite the sum as:
{{begin-eqn}}
{{eqn | l = A_n = a_1 + a_2 + \cdots + a_k + a_{k + 1} + \cdots + a_n
| o =
}}
{{eqn | ll= \leadsto
| l = a_1 + a_2 + \cdots + a_k + \paren {L - \mu} \paren {b_{k + 1} + \cdots + b_n}
| o = <
| m = a_1 + a_2 + \cdots + a_n
| mo= <
| r = a_1 + a_2 + \cdots + a_k + \paren {L + \mu} \paren {b_{k + 1} + \cdots + b_n}
}}
{{eqn | ll= \leadsto
| l = A_k + \paren {L - \mu} \paren {B_n - B_k}
| o = <
| m = A_n
| mo= <
| r = A_k + \paren {L + \mu} \paren {B_n - B_k}
}}
{{end-eqn}}
Divide above by $B_n$:
:$\dfrac {A_k + \paren {L - \mu} B_k} {B_n} + \paren {L - \mu} < \dfrac {A_n} {B_n} < \paren {L + \mu} + \dfrac {A_k + \paren {L + \mu} B_k} {B_n}$
Let $k$ be fixed.
From Reciprocal of Null Sequence and Combination Theorem for Sequences, the sequence $\sequence {\dfrac {A_k + \paren {L \pm \epsilon} B_k} {B_n} }$ converges to zero.
By the definition of convergent sequences, there exists $N > k > 0$ such that:
:$\size {\dfrac {A_k + \paren {L \pm \mu} B_k} {B_n} } < \mu$ for all $n > N$
Substitute the above into the inequality and obtain:
{{begin-eqn}}
{{eqn | o =
| r = L - 2 \mu < \frac {A_n} {B_n} < L + 2 \mu
}}
{{eqn | o = \leadstoandfrom
| r = \size {\frac {A_n} {B_n} - L} < \epsilon
| c = for all $n > N$
}}
{{end-eqn}}
Hence by the definition of convergent sequences the result follows.
{{qed}}
\end{proof}
|
21811
|
\section{Stone Space is Topological Space}
Tags: Stone Spaces, Stone Space
\begin{theorem}
Let $\struct {B, \preceq, \wedge, \vee}$ be a non-empty Boolean algebra.
Let $\struct {U, \tau}$ be the Stone space of $B$.
Then $\struct {U, \tau}$ is a topological space.
\end{theorem}
\begin{proof}
{{explain|generally, why the following purports to be a proof. Suggest the specific basis axioms are invoked.}}
$U$ is non-empty by the Ultrafilter Lemma.
The topology of the Stone space is defined as the topology generated by the basis $Q$ consisting of all sets of the form
:$\set {x \in U: b \in x}$
for some $b \in B$.
We must show that $Q$ is in fact a basis.
$Q$ is trivially a subset of the power set of $U$.
Suppose that $x \in U$.
Then by the definition of ultrafilter, $x \ne \O$.
Thus $x$ has some element $b$. Then $x \in \set {x \in U: b \in x} \in Q$.
So $Q$ covers $U$.
Let $P, Q \in Q$. Then for some $b, c \in B$,
:$P = \set {x \in U: b \in x}$ and
:$Q = \set {y \in U: c \in y}$.
Let $z \in P \cap Q$.
Then by the definitions of $P$ and $Q$,
:$b \in z$ and $c \in z$. So $z$ is an ultrafilter on $B$ containing $b$ and $c$.
Since $z$ is a filter containing $b$ and $c$, it also contains $b \wedge c$.
Let $R = \set {w \in U: b \wedge c \in w}$.
$R$ is clearly an element of $Q$ containing $z$.
Suppose that $v \in R$.
Then $b \wedge c \in v$.
Since $v$ is a filter,
$b \in v$ and $c \in v$.
Thus $v \in P \cap Q$.
Since this holds for each $v \in R$, $R \subseteq P \cap Q$
{{qed}}
Category:Stone Spaces
\end{proof}
|
21812
|
\section{Straight Line has Zero Curvature}
Tags: Straight Lines, Curvature
\begin{theorem}
A straight lines has zero curvature.
\end{theorem}
\begin{proof}
From Equation of Straight Line in Plane: Slope-Intercept Form, a straight line has the equation:
:$y = m x + c$
Differentiating twice {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | l = \dfrac {\d y} {\d x}
| r = m
| c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
| l = \dfrac {\d^2 y} {\d x^2}
| r = 0
| c =
}}
{{end-eqn}}
By definition, the curvature of a curve is defined as:
:$\kappa = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$
But we have that:
:$y'' := \dfrac {\d^2 y} {\d x^2} = 0$
and so, as in general $y' := \dfrac {\d y} {\d x} = m \ne 0$:
:$\kappa = \dfrac 0 {\paren {1 + m^2}^{3/2} }$
the curvature is zero.
{{qed}}
\end{proof}
|
21813
|
\section{Straight Lines which make Equal Angles with Perpendicular to Straight Line are Equal}
Tags: Angles, Lines
\begin{theorem}
Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Let $E, F$ be points on $AB$ such that $\angle DCE = \angle DCF$.
Then $CE = CF$.
\end{theorem}
\begin{proof}
$\triangle CDE$ and $\triangle CDF$ are right triangle where $CE$ and $CF$ are the hypotenuses.
We have:
:$\angle CDE = \angle CDF$ as both are right angles.
:$\angle DCE = \angle DCF$ by hypothesis.
:$CD$ is common.
Thus by Triangle Angle-Side-Angle Equality, $\triangle CDE$ and $\triangle CDF$ are congruent.
Hence the result.
{{qed}}
\end{proof}
|
21814
|
\section{Strict Lower Closure in Restricted Ordering}
Tags: Lower Closures, Order Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\prec T} = T \cap t^{\prec S}$
where:
: $t^{\prec T}$ is the strict lower closure of $t$ in $\left({T, \preceq \restriction_T}\right)$
: $t^{\prec S}$ is the strict lower closure of $t$ in $\left({S, \preceq}\right)$.
\end{theorem}
\begin{proof}
Let $t \in T$, and suppose that $t' \in t^{\prec T}$.
By definition of strict lower closure, this is equivalent to:
:$t' \preceq \restriction_T t \land t \ne t'$
By definition of $\preceq \restriction_T$, the first condition comes down to:
:$t' \preceq t \land t' \in T$
as it is assumed that $t \in T$.
In conclusion, $t' \in t^{\prec T}$ is equivalent to:
:$t' \in T \land t' \preceq t \land t \ne t'$
These last two conjuncts precisely express that $t' \in t^{\prec S}$.
By definition of set intersection, it also holds that:
:$t' \in T \cap t^{\prec S}$
{{iff}} $t' \in T$ and $t' \in t^{\prec S}$.
Thus, it follows that the following are equivalent:
:$t' \in t^{\prec T}$
:$t' \in T \cap t^{\prec S}$
and hence the result follows, by definition of set equality.
{{qed}}
Category:Lower Closures
\end{proof}
|
21815
|
\section{Strict Lower Closure is Dual to Strict Upper Closure}
Tags: Lower Closures, Upper Closures, Order Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a, b \in S$.
The following are dual statements:
:$b \in a^\prec$, the strict lower closure of $a$
:$b \in a^\succ$, the strict upper closure of $a$
\end{theorem}
\begin{proof}
By definition of strict lower closure:
:$b \in a^\prec$
{{iff}}
:$b$ strictly precedes $a$
The dual of this statement is:
:$b$ strictly succeeds $a$
by Dual Pairs (Order Theory).
By definition of strict upper closure, this means:
: $b \in a^\succ$
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}}
\end{proof}
|
21816
|
\section{Strict Lower Closure is Lower Set}
Tags: Lower Sets, Strict Lower Closure is Lower Set, Lower Closures, Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $p \in S$.
Then $p^\prec$, the strict lower closure of $p$, is a lower set.
\end{theorem}
\begin{proof}
Let $l \in {\dot\downarrow}p$.
Let $s \in S$ with $s \preceq l$.
Then by the definition of strict down-set, $l \prec p$.
Thus by Extended Transitivity, $s \prec p$.
So by the definition of strict down-set, $s \in {\dot\downarrow} p$.
Since this holds for all such $l$ and $s$, ${\dot\downarrow} p$ is a lower set.
{{qed}}
\end{proof}
|
21817
|
\section{Strict Lower Closure of Sum with One}
Tags: Naturally Ordered Semigroup
\begin{theorem}
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Then:
:$\forall n \in \struct {S, \circ, \preceq}: \paren {n \circ 1}^\prec = n^\prec \cup \set n$
where $n^\prec$ is defined as the strict lower closure of $n$, that is, the set of elements strictly preceding $n$.
\end{theorem}
\begin{proof}
First note that as $\struct {S, \circ, \preceq}$ is well-ordered and hence totally ordered, the Trichotomy Law applies.
Thus:
{{begin-eqn}}
{{eqn | q = \forall m \in S
| o =
| r = m \notin n^\prec
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \neg \ m \prec n
| c = {{Defof|Strict Lower Closure of Element}}
}}
{{eqn | o = \leadstoandfrom
| r = m = n \lor n \prec m
| c = Trichotomy Law
}}
{{eqn | o = \leadstoandfrom
| r = n \preceq m
| c = {{Defof|Strictly Precede}}
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | q = \forall p \in S
| o =
| r = p \notin \paren {n \circ 1}^\prec
| c =
}}
{{eqn | o = \leadstoandfrom
| r = n \circ 1 \preceq p
| c = from the above
}}
{{eqn | o = \leadstoandfrom
| r = n \prec p
| c = Sum with One is Immediate Successor in Naturally Ordered Semigroup
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | q = \forall p \in S
| o =
| r = p \notin n^\prec \cup \set n
| c =
}}
{{eqn | o = \leadstoandfrom
| r = n \preceq p \land n \ne p
| c = from the above
}}
{{eqn | o = \leadstoandfrom
| r = n \prec p
| c = {{Defof|Strictly Precede}}
}}
{{end-eqn}}
So:
:$p \notin n^\prec \cup \set n \iff p \notin \paren {n \circ 1}^\prec$
Thus:
:$\relcomp S {\paren {n \circ 1}^\prec} = \relcomp S {n^\prec \cup \set n}$
from the definition of relative complement.
So:
{{begin-eqn}}
{{eqn | l = \paren {n \circ 1}^\prec
| r = \relcomp S {\relcomp S {\paren {n \circ 1}^\prec} }
| c = Relative Complement of Relative Complement
}}
{{eqn | r = \relcomp S {\relcomp S {n^\prec \cup \set n} }
| c = from above
}}
{{eqn | r = n^\prec \cup \set n
| c = Relative Complement of Relative Complement
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21818
|
\section{Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup}
Tags: Ordered Semigroups, Order Theory, Semigroups
\begin{theorem}
Let $\struct {S, \circ, \preceq}$ be a totally ordered semigroup.
If either:
:$x \circ z \prec y \circ z$
or
:$z \circ x \prec z \circ y$
then $x \prec y$.
\end{theorem}
\begin{proof}
Let $x \circ z \prec y \circ z$.
{{AimForCont}} $x \succeq y$.
As $\struct {S, \circ, \preceq}$ is an ordered semigroup, $\preceq$ is compatible with $\circ$.
Hence we have:
:$x \succeq y \implies x \circ z \succeq y \circ z$
which contradicts $x \circ z \prec y \circ z$.
We have that $\preceq$ is a total ordering, and that it is not the case that $x \succeq y$.
Hence by the Trichotomy Law:
:$x \prec y$
Similarly for $z \circ x \prec z \circ y$.
{{qed}}
\end{proof}
|
21819
|
\section{Strict Ordering Preserved under Product with Cancellable Element}
Tags: Ordered Semigroups, Semigroups, Orderings, Total Orderings, Order Theory
\begin{theorem}
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y, z \in S$ be such that:
:$(1): \quad z$ is cancellable for $\circ$
:$(2): \quad x \prec y$
Then:
:$x \circ z \prec y \circ z$
:$z \circ x \prec z \circ y$
\end{theorem}
\begin{proof}
Let $z$ be cancellable and $x \prec y$.
Then by the definition of ordered semigroup:
:$x \circ z \preceq y \circ z$
From the fact that $z$ is cancellable:
:$x \circ z = y \circ z \iff x = y$
Thus as $x \circ z \ne y \circ z$ it follows from Strictly Precedes is Strict Ordering that:
:$x \circ z \prec y \circ z$
Similarly, $z \circ x \prec z \circ y$ follows from $z \circ x \preceq z \circ y$.
{{qed}}
\end{proof}
|
21820
|
\section{Strict Ordering Preserved under Product with Invertible Element}
Tags: Ordered Semigroups, Order Theory, Semigroups
\begin{theorem}
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $z \in S$ be invertible.
Suppose that either $x \circ z \prec y \circ z$ or $z \circ x \prec z \circ y$.
Then $x \prec y$.
\end{theorem}
\begin{proof}
Suppose $x \circ z \prec y \circ z$.
By Invertible Element of Monoid is Cancellable, $z^{-1}$ is cancellable.
Then from Strict Ordering Preserved under Product with Cancellable Element:
:$x = \paren {x \circ z} \circ z^{-1} \prec \paren {y \circ z} \circ z^{-1} = y$
Likewise, if $z \circ x \prec z \circ y$:
:$x = z^{-1} \circ \paren {z \circ x} \prec z^{-1} \circ \paren {z \circ y} = y$
{{qed}}
\end{proof}
|
21821
|
\section{Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible}
Tags: Naturally Ordered Semigroup
\begin{theorem}
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Then $\prec$ is strongly compatible with $\circ$:
:$\forall m, n, p \in S: m \prec n \iff m \circ p \prec n \circ p$
\end{theorem}
\begin{proof}
By {{NOSAxiom|2}}, all $n \in S$ are cancellable.
Hence from Strict Ordering Preserved under Product with Cancellable Element:
:$\forall m, n, p \in S: m \prec n \implies m \circ p \prec n \circ p$
By {{NOSAxiom|1}}, $\preceq$ is a total ordering.
Therefore, the contrapositive of:
:$\forall m, n, p \in S: m \circ p \prec n \circ p \implies m \prec n$
is:
:$\forall m, n, p \in S: m \preceq n \implies m \circ p \preceq n \circ p$
which we know to be true by virtue of {{NOSAxiom|2}}.
The result follows.
{{qed}}
\end{proof}
|
21822
|
\section{Strict Ordering on Integers is Asymmetric}
Tags: Orderings on Integers
\begin{theorem}
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Then:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | lo= \implies
| l = \eqclass {c, d} {}
| o = \nless
| r = \eqclass {a, b} {}
| c =
}}
{{end-eqn}}
That is, strict ordering on the integers is asymmetric.
\end{theorem}
\begin{proof}
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.
To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \prec b$ denote that the natural number $a$ is less than the natural number $b$.
We have:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \prec
| r = b + c
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | ll= \leadsto
| l = b + c
| o = \nprec
| r = a + d
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {c, d} {}
| o = \nless
| r = \eqclass {a, b} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21823
|
\section{Strict Ordering on Integers is Transitive}
Tags: Orderings on Integers
\begin{theorem}
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Then:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | lo= \land
| l = \eqclass {c, d} {}
| o = <
| r = \eqclass {e, f} {}
| c =
}}
{{eqn | ll= \implies
| l = \eqclass {a, b} {}
| o = <
| r = \eqclass {e, f} {}
| c =
}}
{{end-eqn}}
That is, strict ordering on the integers is transitive.
\end{theorem}
\begin{proof}
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.
To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \prec b$ denote that the natural number $a$ is less than the natural number $b$.
We have:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | n = 1
| ll= \leadsto
| l = a + d
| o = \prec
| r = b + c
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | l = \eqclass {c, d} {}
| o = <
| r = \eqclass {e, f} {}
| c =
}}
{{eqn | n = 2
| ll= \leadsto
| l = c + f
| o = \prec
| r = d + e
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | ll= \leadsto
| l = a + d + f
| o = \prec
| r = b + c + f
| c = adding $f$ to both sides of $(1)$
}}
{{eqn | ll= \leadsto
| l = a + d + f
| o = \prec
| r = b + d + e
| c = from $(2)$: $b + \paren {c + f} \prec b + \paren {d + e}$
}}
{{eqn | ll= \leadsto
| l = a + f
| o = \prec
| r = b + e
| c = subtracting $d$ from both sides
}}
{{eqn | ll= \leadsto
| l = \eqclass {a, b} {}
| o = <
| r = \eqclass {e, f} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21824
|
\section{Strict Ordering on Integers is Trichotomy}
Tags: Orderings on Integers
\begin{theorem}
Let $\eqclass {a, b} {}$ and $\eqclass {c, d} {}$ be integers, as defined by the formal definition of integers.
Then exactly one of the following holds:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | l = \eqclass {a, b} {}
| o = =
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | l = \eqclass {a, b} {}
| o = >
| r = \eqclass {c, d} {}
| c =
}}
{{end-eqn}}
That is, strict ordering is a trichotomy.
\end{theorem}
\begin{proof}
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.
To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \preccurlyeq b$ denote that the natural number $a$ is less than or equal to the natural number $b$.
We have:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = \eqclass {c, d} {}
| o = \nless
| r = \eqclass {a, b} {}
| c = Strict Ordering on Integers is Asymmetric
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \prec
| r = b + c
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \ne
| r = b + c
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {a, b} {}
| o = \ne
| r = \eqclass {c, d} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = >
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = \eqclass {c, d} {}
| o = \ngtr
| r = \eqclass {a, b} {}
| c = Strict Ordering on Integers is Asymmetric
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = >
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \succ
| r = b + c
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \ne
| r = b + c
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {a, b} {}
| o = \ne
| r = \eqclass {c, d} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = =
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = a + d
| o = =
| r = b + c
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \nprec
| r = b + c
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | lo= \land
| l = a + d
| o = \nsucc
| r = b + c
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {a, b} {}
| o = \nless
| r = \eqclass {c, d} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | lo= \land
| l = \eqclass {a, b} {}
| o = \ngtr
| r = \eqclass {c, d} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{end-eqn}}
This demonstrates that $<$, $=$ and $>$ are mutually exclusive.
Now:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \prec
| r = b + c
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | ll= \leadsto
| l = b + c
| o = \preccurlyeq
| r = a + d
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {c, d} {}
| o = \le
| r = \eqclass {a, b} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = >
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \succ
| r = b + c
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | ll= \leadsto
| l = b + c
| o = \succcurlyeq
| r = a + d
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {c, d} {}
| o = \ge
| r = \eqclass {a, b} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = \ne
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \ne
| r = b + c
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | ll= \leadsto
| l = a + d
| o = \prec
| r = b + c
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | lo= \lor
| l = a + d
| o = \succ
| r = b + c
| c = {{Defof|Ordering on Natural Numbers}}
}}
{{eqn | ll= \leadsto
| l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{eqn | lo= \land
| l = \eqclass {a, b} {}
| o = \>
| r = \eqclass {c, d} {}
| c = {{Defof|Strict Ordering on Integers}}
}}
{{end-eqn}}
demonstrating that either $<$, $=$ or $>$ must hold.
{{qed}}
\end{proof}
|
21825
|
\section{Strict Ordering on Integers is Well-Defined}
Tags: Orderings on Integers
\begin{theorem}
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Let:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| r = \eqclass {a', b'} {}
| c =
}}
{{eqn | l = \eqclass {c, d} {}
| r = \eqclass {c', d'} {}
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {}
| o = <
| r = \eqclass {c, d} {}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \eqclass {a', b'} {}
| o = <
| r = \eqclass {c', d'} {}
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
This is a direct application of the Extension Theorem for Total Orderings.
{{qed}}
\end{proof}
|
21826
|
\section{Strict Positivity Property induces Total Ordering}
Tags: Integral Domains, Strict Positivity Property induces Total Ordering, Ordered Integral Domains, Total Orderings
\begin{theorem}
Let $\struct {D, +, \times}$ be an integral domain whose zero is $0_D$.
Let $D$ be endowed with a (strict) positivity property $P: D \to \set {\T, \F}$.
Then there exists a total ordering $\le$ on $\struct {D, +, \times}$ induced by $P$ which is compatible with the ring structure of $\struct {D, +, \times}$.
\end{theorem}
\begin{proof}
By definition of the strict positivity property:
{{:Definition:Strict Positivity Property}}
Let us define a relation $<$ on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Setting $a = 0$:
:$\forall b \in D: 0 < b \iff \map P b$
demonstrating that (strictly) positive elements of $D$ are those which are greater than zero.
From Relation Induced by Strict Positivity Property is Compatible with Addition we have that $<$ is compatible with $+$.
From Relation Induced by Strict Positivity Property is Transitive we have that $<$ is transitive.
From Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive we have that $<$ is asymmetric and antireflexive.
Thus by definition, $<$ is a strict ordering.
Let the relation $\le$ be defined as the reflexive closure of $<$.
From Reflexive Closure of Strict Ordering is Ordering we have that $\le$ is an ordering on $D$.
From Relation Induced by Strict Positivity Property is Trichotomy, and from the Trichotomy Law (Ordering), we have that $\le$ is a total ordering.
Hence the result.
{{qed}}
\end{proof}
|
21827
|
\section{Strict Upper Closure in Restricted Ordering}
Tags: Upper Closures, Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\succ T} = T \cap t^{\succ S}$
where:
:$t^{\succ T}$ is the strict upper closure of $t$ in $\struct {T, \preceq \restriction_T}$
:$t^{\succ S}$ is the strict upper closure of $t$ in $\struct {S, \preceq}$.
\end{theorem}
\begin{proof}
Let $t \in T$, and suppose that $t' \in t^{\succ T}$.
By definition of strict upper closure, this is equivalent to:
:$t \preceq \restriction_T t' \land t \ne t'$
By definition of $\preceq \restriction_T$, the first condition comes down to:
:$t \preceq t' \land t' \in T$
as it is assumed that $t \in T$.
In conclusion, $t' \in t^{\succ T}$ is equivalent to:
:$t' \in T \land t \preceq t' \land t \ne t'$
These last two conjuncts precisely express that $t' \in t^{\succ S}$.
By definition of set intersection, it also holds that:
:$t' \in T \cap t^{\succ S}$
{{iff}} $t' \in T$ and $t' \in t^{\succ S}$.
Thus, it follows that the following are equivalent:
:$t' \in t^{\succ T}$
:$t' \in T \cap t^{\succ S}$
and hence the result follows, by definition of set equality.
{{qed}}
Category:Upper Closures
\end{proof}
|
21828
|
\section{Strict Upper Closure is Upper Set}
Tags: Upper Closures, Upper Sets, Order Theory
\begin{theorem}
Let $(S, \preceq)$ be an ordered set.
Let $p \in S$.
Then $p^\succ$, the strict upper closure of $p$, is an upper set.
\end{theorem}
\begin{proof}
Let $u \in p^\succ$.
Let $s \in S$ with $u \preceq s$.
Then by the definition of strict upper closure:
: $p \prec u$
Thus by Extended Transitivity:
: $p \prec s$
So by the definition of strict upper closure:
: $s \in p^\succ$
Since this holds for all such $u$ and $s$, $p^\succ$ is an upper set.
{{qed}}
\end{proof}
|
21829
|
\section{Strict Weak Ordering Induces Partition}
Tags: Order Theory, Equivalence Relations, Orderings, Relations
\begin{theorem}
Let $\struct {S, \prec}$ be a relational structure such that $\prec$ is a strict weak ordering on $S$.
Then $S$ can be partitioned into equivalence classes whose equivalence relation is "is non-comparable".
That is, each of the partitions $A$ of $S$ is a relational structure $\struct {\mathbb S, <}$ such that:
:$\mathbb S$ is the set of these partitions of $S$;
:$<$ is the strict total ordering on $\mathbb S$ '''induced by''' $\prec$.
\end{theorem}
\begin{proof}
From the definition of strict weak ordering, we define the symbol $\Bumpeq$ as:
:$a \Bumpeq b := \neg a \prec b \land \neg b \prec a$
that is, $a \Bumpeq b$ means "$a$ and $b$ are non-comparable".
Checking in turn each of the criteria for equivalence:
\end{proof}
|
21830
|
\section{Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping}
Tags: Well-Orderings, Ordinals, Order Theory
\begin{theorem}
Let $S$ be a set.
Let $\left({S, \prec}\right)$ be a strict well-ordering.
Then there exists a unique ordinal $x$ and unique mapping $f$ such that $f: x \to S$ is an order isomorphism.
\end{theorem}
\begin{proof}
The existence of $x$ and $f$ follows from Woset is Isomorphic to Unique Ordinal.
The uniqueness of $x$ follows from Woset is Isomorphic to Unique Ordinal.
The uniqueness of $f$ follows from Order Isomorphism between Wosets is Unique.
{{qed}}
\end{proof}
|
21831
|
\section{Strictly Decreasing Mapping is Decreasing}
Tags: Orderings, Decreasing Mappings, Order Theory, Mappings, Mapping Theory
\begin{theorem}
A mapping that is strictly decreasing is a decreasing mapping.
\end{theorem}
\begin{proof}
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be strictly decreasing.
From Strictly Precedes is Strict Ordering:
:$x \preceq_1 y \iff x = y \lor x \prec_1 y$
So:
{{begin-eqn}}
{{eqn | l = x
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi y
| r = \map \phi x
| c = {{Defof|Mapping}}
}}
{{eqn | ll= \leadsto
| l = \map \phi y
| o = \preceq_2
| r = \map \phi x
| c = as $\preceq_2$, being an ordering, is reflexive
}}
{{end-eqn}}
This leaves us with:
{{begin-eqn}}
{{eqn | l = x
| o = \prec_1
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi y
| o = \prec_2
| r = \map \phi x
| c = {{Defof|Strictly Decreasing Mapping}}
}}
{{eqn | ll= \leadsto
| l = \map \phi y
| o = \preceq_2
| r = \map \phi x
| c = Strictly Precedes is Strict Ordering
}}
{{end-eqn}}
{{Qed}}
\end{proof}
|
21832
|
\section{Strictly Increasing Infinite Sequence of Integers is Cofinal in Natural Numbers}
Tags: Definitions: Integers, Integers
\begin{theorem}
Let $S = \left\langle{x_n}\right\rangle$ be an infinite sequence of integers which is strictly increasing.
Then $S$ is a cofinal subset of $\left({\Z, \le}\right)$ where $\le$ is the usual ordering on the integers.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Integers
\end{proof}
|
21833
|
\section{Strictly Increasing Mapping Between Wosets Implies Order Isomorphism}
Tags: Well-Orderings
\begin{theorem}
Let $J$ and $E$ be well-ordered sets.
Let there exist a mapping $k: J \to E$ which is strictly increasing.
Then $J$ is order isomorphic to $E$ or an initial segment of $E$.
\end{theorem}
\begin{proof}
If the sets considered are empty or singletons, the theorem holds vacuously or trivially.
Suppose $J, E$ both have at least two elements.
Let $e_0 = \min E$, the smallest element of $E$.
Define the mapping:
:$h: J \to E$:
:$\map h \alpha = \begin {cases} \map \min {E \setminus h \sqbrk {S_\alpha} } & : h \sqbrk {S_\alpha} \ne E \\ e_0 & : h \sqbrk {S_\alpha} = E \end {cases}$
where $S_\alpha$ is the initial segment determined by $\alpha$ and $h \sqbrk {S_\alpha}$ is the image of $S_\alpha$ under $h$.
By the Principle of Recursive Definition for Well-Ordered Sets, this construction is well-defined and uniquely determined.
Observe that:
{{begin-eqn}}
{{eqn | l = h \sqbrk {S_\alpha}
| r = \set {\map h x \in E: \exists x \in J: \map h x = \map \min {E \setminus h \sqbrk {S_\alpha} } }
| c = {{Defof|Image of Subset under Mapping|image of a subset}}
}}
{{eqn | r = \set {\map h x \in E: \exists x \in J: \map h x \prec \map h \alpha}
}}
{{eqn | r = S_{\map h \alpha}
| c = {{Defof|Initial Segment|initial segment}}
}}
{{end-eqn}}
This equality also holds if $\map h \alpha = e_0$, by Initial Segment Determined by Smallest Element is Empty.
We claim that $\map h \alpha \preceq \map k \alpha$ for all $\alpha \in J$.
{{AimForCont}} there is some $a \in J$ such that $\map h a \not \preceq \map k a$.
Then $\map k a \prec \map h a$ by the trichotomy law.
Because $\map h a$ has an element preceding it, $\map h a \ne e_0$.
Thus $\map k a \prec \map \min {E \setminus h \sqbrk {S_a} }$ by the construction of $k$.
Then $\map k a \in h \sqbrk {S_a}$, because it precedes the smallest element that isn't in $h \sqbrk {S_a}$.
Recall that $h \sqbrk {S_a} = S_{\map h a}$.
Then $\map k a \in S_{\map h a}$.
This implies that $\map h a \prec \map k a$, contradicting the assumption that $\map h a \not \preceq \map k a$
From this contradiction we can conclude:
:$\map h \alpha \preceq \map k \alpha$ for all $\alpha \in J$.
{{AimForCont}} there is some $\alpha \in J$ such that $h \sqbrk {S_\alpha} = E$.
Recall that $h \sqbrk {S_\alpha} = S_{\map h \alpha}$.
Thus, were such an $\alpha$ to exist, then it would succeed all elements in $E$.
It particular, it would also succeed $\map k \alpha$.
But we showed above that $\map h \alpha \preceq \map k \alpha$.
From this contradiction we see that there cannot be any $\alpha \in J$ with $h \sqbrk {S_\alpha} = E$.
Thus the definition of $h$ can be simplified:
:$h: J \to E$:
:$\map h \alpha = \map \min {E \setminus h \sqbrk {S_\alpha} }$
Then the hypotheses of Characterization of Strictly Increasing Mapping on Woset are satisfied.
Thus $h$ is a strictly increasing mapping and its image is $E$ or an initial segment of $E$.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $h$ is also injective.
From Injection to Image is Bijection, $h$ is also bijective to its image.
We conclude that there is an order isomorphism from $J$ to $E$, or from $J$ to an initial segment of $E$.
{{qed}}
\end{proof}
|
21834
|
\section{Strictly Increasing Mapping is Increasing}
Tags: Orderings, Increasing Mappings, Ordering Theory, Mappings, Order Theory, Mapping Theory
\begin{theorem}
A mapping that is strictly increasing is an increasing mapping.
\end{theorem}
\begin{proof}
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be strictly increasing.
From Strictly Precedes is Strict Ordering:
:$x \preceq_1 y \implies x = y \lor x \prec_1 y$
So:
{{begin-eqn}}
{{eqn | l = x
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| r = \map \phi y
| c = {{Defof|Mapping}}
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| o = \preceq_2
| r = \map \phi y
| c = as $\preceq_2$, being an ordering, is reflexive
}}
{{end-eqn}}
This leaves us with:
{{begin-eqn}}
{{eqn | l = x
| o = \prec_1
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| o = \prec_2
| r = \map \phi y
| c = {{Defof|Strictly Increasing Mapping}}
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| o = \preceq_2
| r = \map \phi y
| c = Strictly Precedes is Strict Ordering
}}
{{end-eqn}}
{{Qed}}
\end{proof}
|
21835
|
\section{Strictly Increasing Mapping on Well-Ordered Class}
Tags: Set Theory, Increasing Mappings, Order Theory, Class Theory, Class Mappings
\begin{theorem}
Let $\struct {S, \prec}$ be a strictly well-ordered class.
Let $\struct {T, <}$ be a strictly ordered class.
Let $f$ be a mapping from $S$ to $T$.
For each $i \in S$ such that $i$ is not maximal in $S$, let:
: $\map f i < \map f {\map \Succ i}$
where $\map \Succ i$ is the immediate successor element of $i$.
Let:
:$\forall i, j \in S: i \preceq j \implies \map f i \le \map f j$
Then for each $i, j \in S$ such that $i \prec j$:
:$\map f i < \map f j$
{{MissingLinks|In the above, link to definitions of increasing mapping and strictly increasing mapping, in order to correlate with page title.}}
\end{theorem}
\begin{proof}
{{NotZFC}}
By Non-Maximal Element of Well-Ordered Class has Immediate Successor, $\map \Succ i$ is guaranteed to exist.
Let $i \prec j$.
Let $S_i := \set {q \in S: i \prec q}$.
{{explain|Important to specify what the domain of $q$ actually is -- presumably $S$.}}
Then $\map \Succ i$ is the minimal element of $S_i$.
By supposition, $j \in S_i$.
Thus:
:$j \nprec \map \Succ i$
Since Well-Ordering is Total Ordering:
:$\map \Succ i \preceq j$
Thus by supposition:
:$\map f {\map \Succ i} \le \map f j$
Since $\map f i < \map f {\map \Succ i}$:
:$\map f i < \map f j$
by transitivity.
{{qed}}
Category:Class Mappings
Category:Increasing Mappings
\end{proof}
|
21836
|
\section{Strictly Increasing Sequence induces Partition}
Tags: Set Theory, Sequences
\begin{theorem}
Let $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ be a strictly increasing finite sequence of natural numbers.
Let:
:$\forall k \in \closedint 1 n: A_k := \closedint {r_{k - 1} + 1} {r_k}$
Then:
:$\set {A_k: k \in \closedint 1 n}$ is a partition of $\closedint {r_0 + 1} {r_n}$.
\end{theorem}
\begin{proof}
First we show that the elements of $\set {A_k: k \in \closedint 1 n}$ are disjoint.
Let $j \in \closedint 1 n$.
We have that:
:$\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is strictly increasing
and:
:$0 \le j - 1 < j \le n$
Thus by Sum with One is Immediate Successor in Naturally Ordered Semigroup:
:$r_0 \le r_{j - 1} < r_j \le r_n \implies \paren {r_0 + 1} \le \paren {r_{j - 1} + 1} \le r_j \le r_n$
So:
:$\O \subset A_j \subseteq \closedint {r_0 + 1} {r_n}$
Because $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is strictly increasing:
:$1 \le j < k \le n \implies A_j \cap A_k = \O$
So the elements of $\set {A_k: k \in \closedint 1 n}$ are disjoint, as we were to show.
It remains to be established that:
:if $m \in \closedint {r_0 + 1} {r_n}$
then:
:$m \in A_k$ for some $k \in \closedint 1 n$.
Let $m \in \closedint {r_0 + 1} {r_n}$.
Consider the set:
:$J = \set {j \in \closedint 0 n: m \le r_j}$
Clearly $j \ne \O$ as $n \in J$.
Let $k$ be the smallest element of $J$.
Then because $r_0 < m$:
:$k \ne 0$
Thus:
:$k \in \closedint 1 n$
By its definition:
:$r_{k - 1} < m \le r_k$
Thus, by Sum with One is Immediate Successor in Naturally Ordered Semigroup:
:$r_{k - 1} + 1 \le m \le r_k$
Therefore $m \in A_k$.
{{qed}}
{{improve|A better link instead of Sum with One is Immediate Successor in Naturally Ordered Semigroup}}
\end{proof}
|
21837
|
\section{Strictly Increasing Sequence of Natural Numbers}
Tags: Sequences
\begin{theorem}
Let $\N_{>0}$ be the set of natural numbers without zero:
: $\N_{>0} = \left\{{1, 2, 3, \ldots}\right\}$
Let $\left \langle {n_r} \right \rangle$ be strictly increasing sequence in $\N_{>0}$.
Then:
: $\forall r \in \N_{>0}: n_r \ge r$
\end{theorem}
\begin{proof}
This is to be proved by induction on $r$.
For all $r \in \N_{>0}$, let $P \left({r}\right)$ be the proposition $n_r \ge r$.
\end{proof}
|
21838
|
\section{Strictly Increasing Sequence on Ordered Set}
Tags: Sequences, Proofs by Contraposition
\begin{theorem}
Let <math>\left({S; \preceq}\right)</math> be a totally ordered set.
Let <math>\left \langle {r_k} \right \rangle_{p \le k \le q}</math> be a sequence of elements of <math>\left({S; \preceq}\right)</math>.
Then <math>\left \langle {r_k} \right \rangle_{p \le k \le q}</math> is strictly increasing iff:
:<math>\forall k \in \left[{p + 1 \, . \, . \, q}\right]: r_{k - 1} \prec r_k</math>
\end{theorem}
\begin{proof}
Let $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ be strictly increasing.
Because $\forall k \in \N_{>0}: k - 1 < k$, it follows directly that:
:$\forall k \in \closedint {p + 1} q: r_{k - 1} \prec r_k$
For the other direction, we use a Proof by Contraposition.
To that end, suppose $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ is ''not'' strictly increasing.
Let $K$ be the set of all $k \in \closedint p q$ such that:
:$\exists j \in \closedint p q$ such that $j < k$ and $r_k \preceq r_j$
The set $K$ is not empty because $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ is not strictly increasing.
As $K \subset \N$ and the latter is well-ordered, then so is $K$.
Thus $K$ has a minimal element $m$.
Thus there exists $j \in \closedint p q$ such that:
:$j < m$
and:
:$r_m \preceq r_j$
Because $m - 1 < m$:
:$j \le m - 1$
and so:
:$m - 1 \notin K$
So:
:$r_m \preceq r_j\prec r_{m - 1}$
Since orderings are transitive, it follows
:$r_m \preceq r_{m - 1}$
From Rule of Transposition it follows that
:$\forall k \in \closedint {p + 1} q: r_{k - 1} \prec r_k \implies \sequence {r_k}_{p \mathop \le k \mathop \le q}$ is strictly increasing.
The result follows.
{{Qed}}
\end{proof}
|
21839
|
\section{Strictly Maximal Element is Maximal Element}
Tags: Maximal Elements
\begin{theorem}
Let $\struct {S, \RR}$ be a relational structure.
Let $T \subseteq S$ be a subset of $S$.
Let $m \in T$ be a strictly maximal element of $T$ under $\RR$.
Then $m$ is a maximal element of $T$ under $\RR$.
\end{theorem}
\begin{proof}
Let $m \in T$ be a strictly maximal element of $T$ under $\RR$.
Then by definition:
:$\forall x \in T: \tuple {m, x} \notin \RR$
{{AimForCont}} $m$ is not a maximal element of $T$ under $\RR$.
Then:
:$\exists y \in T: \tuple {m, y} \in \RR$
such that $y \ne m$.
But this contradicts the assertion that $\tuple {m, y} \notin \RR$.
Hence it cannot be the case that $m$ is not a maximal element of $T$ under $\RR$.
That is:
:$m$ is a maximal element of $T$ under $\RR$.
{{qed}}
Category:Maximal Elements
\end{proof}
|
21840
|
\section{Strictly Minimal Element is Minimal Element}
Tags: Minimal Elements
\begin{theorem}
Let $\struct {S, \RR}$ be a relational structure.
Let $T \subseteq S$ be a subset of $S$.
Let $m \in T$ be a strictly minimal element of $T$ under $\RR$.
Then $m$ is a minimal element of $T$ under $\RR$.
\end{theorem}
\begin{proof}
Let $m \in T$ be a strictly minimal element of $T$ under $\RR$.
Then by definition:
:$\forall x \in T: \tuple {x, m} \notin \RR$
{{AimForCont}} $m$ is not a minimal element of $T$ under $\RR$.
Then:
:$\exists y \in T: \tuple {y, m} \in \RR$
such that $y \ne m$.
But this contradicts the assertion that $\tuple {y, m} \notin \RR$.
Hence it cannot be the case that $m$ is not a minimal element of $T$ under $\RR$.
That is:
:$m$ is a minimal element of $T$ under $\RR$.
{{qed}}
Category:Minimal Elements
\end{proof}
|
21841
|
\section{Strictly Monotone Mapping is Monotone}
Tags: Orderings, Monotone Mappings, Order Theory, Mappings, Mapping Theory
\begin{theorem}
A mapping that is strictly monotone is a monotone mapping.
\end{theorem}
\begin{proof}
If $\phi$ is strictly monotone, then it is either strictly increasing or strictly decreasing.
If $\phi$ is strictly increasing, then by Strictly Increasing Mapping is Increasing, $\phi$ is increasing.
If $\phi$ is strictly decreasing, then by Strictly Decreasing Mapping is Decreasing, $\phi$ is decreasing.
Thus $\phi$ is monotone by definition.
{{qed}}
\end{proof}
|
21842
|
\section{Strictly Monotone Mapping with Totally Ordered Domain is Injective}
Tags: Orderings, Injections, Monotone Mappings, Order Theory, Mappings
\begin{theorem}
Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a strictly monotone mapping.
Then $\phi$ is injective.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x, y
| o = \in
| r = S
| c =
}}
{{eqn | lo= \land
| l = x
| o = \ne
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \prec_1
| r = y
}}
{{eqn | lo= \lor
| l = y
| o = \prec_1
| r = x
| c = Trichotomy Law
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| o = \prec_2
| r = \map \phi y
}}
{{eqn | lo= \lor
| l = \map \phi y
| o = \prec_2
| r = \map \phi x
| c = $\phi$ is strictly monotone
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| o = \ne
| r = \map \phi y
| c = Definition of $\prec_2$
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
21843
|
\section{Strictly Monotone Real Function is Bijective}
Tags: Bijections, Monotone Mappings, Real Analysis, Analysis
\begin{theorem}
Let $f$ be a real function which is defined on $I \subseteq \R$.
Let $f$ be strictly monotone on $I$.
Let the image of $f$ be $J$.
Then $f: I \to J$ is a bijection.
\end{theorem}
\begin{proof}
From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $f$ is an injection.
From Surjection by Restriction of Codomain, $f: I \to J$ is a surjection.
{{qed}}
\end{proof}
|
21844
|
\section{Strictly Order-Preserving and Order-Reversing Mapping on Strictly Totally Ordered Set is Injection}
Tags: Increasing Mappings, Total Orderings, Strict Orderings
\begin{theorem}
Let $\struct {S, \prec_1}$ and $\struct {T, \prec_2}$ be strictly totally ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\pi: S \to T$ be a mapping with the property that:
:$\forall x, y \in S: x \prec_1 y \iff \map \pi x \prec_2 \map \pi y$
Then $\pi$ is an injection.
\end{theorem}
\begin{proof}
{{AimForCont}} $\pi$ is not an injection.
Hence:
:$\exists x, y \in S: \map \pi x = \map \pi y$
As $S$ is strictly totally ordered:
:$x \prec_1 y$ or $y \prec_1 x$
{{WLOG}}, let $x \prec_1 y$.
Then we have:
:$\map \pi x = \map \pi y$
But {{hypothesis}}:
:$\map \pi x \prec_2 \map \pi y$
Because $\prec_2$ is a strict ordering, it follows that:
:$\map \pi x \ne \map \pi y$
It follows by Proof by Contradiction that it cannot be the case that $\pi$ is not an injection.
Hence the result.
{{qed}}
\end{proof}
|
21845
|
\section{Strictly Positive Integer Power Function Strictly Succeeds Each Element}
Tags: Ring Theory
\begin{theorem}
Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity.
Let $\struct {R, \le}$ be a directed set with no upper bound.
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
:$\forall x \in R: \map f x = \circ^n x$
Then the image of $f$ has elements strictly succeeding each elements of $R$.
\end{theorem}
\begin{proof}
Let $b \in R$.
By Directed Set has Strict Successors iff Unbounded Above:
:$\exists c \in R: b < c$
:$\exists d \in R: 1 < d$
By the definition of a directed set:
:$\exists e \in R: d \le e, c \le e$
$\struct {R, +, \circ, \le}$ is an ordered ring, so $\le$ is by definition a transitive relation.
Hence by transitivity:
:$b < e$
and:
:$1 < e$
By Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element:
:$e \le \map f e$
Thus by transitivity:
:$b < \map f e$
{{qed}}
Category:Ring Theory
\end{proof}
|
21846
|
\section{Strictly Positive Integer Power Function is Unbounded Above}
Tags: Real Numbers, Real Analysis
\begin{theorem}
Let $\R$ be the real numbers with the usual ordering.
Let $n \in \N_{>0}$.
Let $f: \R \to \R$ be defined by:
:$\map f x = x^n$
Then $f$ is unbounded above.
\end{theorem}
\begin{proof}
If $n = 1$, then $f$ is the identity function.
By the Archimedean Principle, the real numbers are unbounded above.
Thus by definition of the identity function:
$f$ is unbounded above.
{{qed|lemma}}
Let $n \ge 2$.
{{AimForCont}} that $f$ is bounded above by $b \in \R$.
{{WLOG}} suppose that $b > 0$.
Then by the definition of an upper bound:
:$\forall x \in \R: x^n \le b$
Let $x > b$.
Then:
:$\dfrac {x^n - b} {x - b} \le 0$
By the Mean Value Theorem, there exists a point $p$ between $b$ and $x$ such that:
:$\map {f'} p = \dfrac {x^n - b} {x - b}$
By Derivative of Power:
:$\map {f'} p = n p^{n - 1}$
Therefore $n p^{n - 1} \le 0$.
By:
:Power of Strictly Positive Real Number is Positive
and:
:Strictly Positive Real Numbers are Closed under Multiplication it follows that:
:$p > 0 \implies n p^{n - 1} > 0$
But this is impossible because it was previously established that $p \ge b > 0$.
From this contradiction it follows that there can be no such $b$.
Hence the result.
{{qed}}
\end{proof}
|
21847
|
\section{Strictly Positive Integer Power Function is Unbounded Above/General Case}
Tags: Totally Ordered Rings, Rings with Unity
\begin{theorem}
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring with unity.
Suppose that $R$ has no upper bound.
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
:$\map f x = \circ^n x$
Then the image of $f$ is unbounded above in $R$.
\end{theorem}
\begin{proof}
Let $1_R$ be the unity of $R$.
Let $b \in R$.
We will show that $b$ is not an upper bound of the image of $f$.
Since $R$ is totally ordered and unbounded above, there is an element $c \in R$ such that $b < c$ and $1_R < c$.
By Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element, $c \le \map f c$.
Thus by transitivity of ordering, $b < \map f c$.
{{qed}}
Category:Totally Ordered Rings
Category:Rings with Unity
\end{proof}
|
21848
|
\section{Strictly Positive Integers have same Cardinality as Natural Numbers}
Tags: Natural Numbers, Integers, Cardinality
\begin{theorem}
Let $\Z_{>0} := \set {1, 2, 3, \ldots}$ denote the set of strictly positive integers.
Let $\N := \set {0, 1, 2, \ldots}$ denote the set of natural numbers.
Then $\Z_{>0}$ has the same cardinality as $\N$.
\end{theorem}
\begin{proof}
Consider the mapping $f: \N \to \Z_{>0}$ defined as:
:$\forall x \in \N: \map f x = x + 1$
Then $f$ is trivially seen to be a bijection.
The result follows by definition of cardinality.
\end{proof}
|
21849
|
\section{Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element}
Tags: Ordered Rings
\begin{theorem}
Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity.
Let $x \in R$ with $x > 1$ and $x > 0$.
Let $n \in \N_{>0}$.
Then:
:$\circ^n x \ge x$
\end{theorem}
\begin{proof}
The proof proceeds by induction:
If $n = 1$, then $\circ^n x = x$.
So:
:$\circ^n x \ge x$
Suppose that $\circ^n x \ge x$.
Then since $x > 1$:
:$\circ^n x > 1$
By Product of Positive Element and Element Greater than One:
:$x \circ \paren {\circ^n x} > x$
Hence:
:$\circ^{n + 1} x \ge x$
{{qed}}
Category:Ordered Rings
\end{proof}
|
21850
|
\section{Strictly Positive Rational Numbers are Closed under Addition}
Tags: Addition, Rational Addition, Rational Numbers
\begin{theorem}
Let $\Q_{>0}$ denote the set of strictly positive rational numbers:
:$\Q_{>0} := \set {x \in \Q: x > 0}$
where $\Q$ denotes the set of rational numbers.
The algebraic structure $\struct {\Q_{>0}, +}$ is closed in the sense that:
:$\forall a, b \in \Q_{>0}: a + b \in \Q_{>0}$
where $+$ denotes rational addition.
\end{theorem}
\begin{proof}
Let $a$ and $b$ be expressed in canonical form:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of rational addition:
:$\dfrac {p_1} {q_1} + \dfrac {p_2} {q_2} = \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}$
From Integers form Ordered Integral Domain, it follows that:
{{begin-eqn}}
{{eqn | l = p_1 q_2
| o = >
| r = 0
}}
{{eqn | l = p_2 q_1
| o = >
| r = 0
}}
{{eqn | l = q_1 q_2
| o = >
| r = 0
}}
{{eqn | ll= \leadsto
| l = p_1 q_2 + p_2 q_1
| o = >
| r = 0
}}
{{eqn | ll= \leadsto
| l = \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}
| o = >
| r = 0
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21851
|
\section{Strictly Positive Rational Numbers are Closed under Multiplication}
Tags: Algebraic Closure, Rational Multiplication, Rational Numbers, Multiplication
\begin{theorem}
:$\forall a, b \in \Q_{>0}: a b \in \Q_{>0}$
\end{theorem}
\begin{proof}
Let $a$ and $b$ be expressed in canonical form:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of rational multiplication:
:$\dfrac {p_1} {q_1} \times \dfrac {p_2} {q_2} = \dfrac {p_1 \times p_2} {q_1 \times q_2}$
From Integers form Ordered Integral Domain, it follows that:
{{begin-eqn}}
{{eqn | l = p_1 \times p_2
| o = >
| r = 0
}}
{{eqn | l = q_1 \times q_2
| o = >
| r = 0
}}
{{eqn | ll= \leadsto
| l = \dfrac {p_1} {q_1} \times \dfrac {p_2} {q_2}
| o = >
| r = 0
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21852
|
\section{Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group}
Tags: Abelian Groups, Examples of Abelian Groups, Group Examples, Abelian Groups: Examples, Infinite Groups: Examples, Rational Multiplication, Examples of Infinite Groups, Abelian Group Examples, Rational Numbers
\begin{theorem}
Let $\Q_{> 0}$ be the set of strictly positive rational numbers, i.e. $\Q_{> 0} = \set {x \in \Q: x > 0}$.
The structure $\struct {\Q_{> 0}, \times}$ is a countably infinite abelian group.
\end{theorem}
\begin{proof}
From Strictly Positive Rational Numbers under Multiplication form Subgroup of Non-Zero Rational Numbers we have that $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$, where $\Q_{\ne 0}$ is the set of rational numbers without zero: $\Q_{\ne 0} = \Q \setminus \set 0$.
From Subgroup of Abelian Group is Abelian it follows that $\struct {\Q_{> 0}, \times}$ is an abelian group.
From Positive Rational Numbers are Countably Infinite, it follows that $\struct {\Q_{> 0}, \times}$ is a countably infinite group.
{{qed}}
\end{proof}
|
21853
|
\section{Strictly Positive Rational Numbers under Multiplication form Subgroup of Non-Zero Rational Numbers}
Tags: Rational Multiplication, Rational Numbers, Subgroups
\begin{theorem}
Let $\Q_{> 0}$ be the set of strictly positive rational numbers, that is $\Q_{> 0} = \set { x \in \Q: x > 0}$.
The structure $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$, where $\Q_{\ne 0}$ is the set of rational numbers without zero: $\Q_{\ne 0} = \Q \setminus \set 0$.
\end{theorem}
\begin{proof}
From Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group we have that $\struct {\Q_{\ne 0}, \times}$ is a group.
We know that $\Q_{> 0} \ne \O$, as (for example) $1 \in \Q_{> 0}$.
Let $a, b \in \Q_{> 0}$.
Then:
:$a b \in \Q_{\ne 0}$ and $ab > 0$
Hence:
:$a b \in \Q_{> 0}$
Let $a \in \Q_{> 0}$.
Then:
:$a^{-1} = \dfrac 1 a \in \Q_{> 0}$
So, by the Two-Step Subgroup Test, $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$.
{{qed}}
Category:Rational Multiplication
Category:Subgroups
\end{proof}
|
21854
|
\section{Strictly Positive Real Numbers are Closed under Division}
Tags: Real Numbers, Algebraic Closure, Real Division
\begin{theorem}
The set $\R_{>0}$ of strictly positive real numbers is closed under division:
:$\forall a, b \in \R_{>0}: a \div b \in \R_{>0}$
\end{theorem}
\begin{proof}
From the definition of division:
:$a \div b := a \times \paren {\dfrac 1 b}$
where $\dfrac 1 b$ is the inverse for real number multiplication.
From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, the algebraic structure $\struct {\R_{>0}, \times}$ forms a group.
Thus it follows that:
:$\forall a, b \in \R_{>0}: a \times \paren {\dfrac 1 b} \in \R$
Therefore real number division is closed in $\R_{>0}$.
{{qed}}
\end{proof}
|
21855
|
\section{Strictly Positive Real Numbers are Closed under Multiplication}
Tags: Real Numbers, Strictly Positive Real Numbers are Closed under Multiplication, Real Multiplication
\begin{theorem}
The set $\R_{>0}$ of strictly positive real numbers is closed under multiplication:
:$\forall a, b \in \R_{> 0}: a \times b \in \R_{> 0}$
\end{theorem}
\begin{proof}
Let $a, b \in \R_{> 0}$
We have that the Real Numbers form Ordered Integral Domain.
It follows from Positive Elements of Ordered Ring that:
:$a \times b \in \R_{> 0}$.
{{qed}}
\end{proof}
|
21856
|
\section{Strictly Positive Real Numbers are not Closed under Subtraction}
Tags: Algebraic Closure, Real Subtraction
\begin{theorem}
The set $\R_{>0}$ of strictly positive real numbers is not closed under subtraction.
\end{theorem}
\begin{proof}
;Proof by Counterexample
Let $a = 1$ and $b = 2$.
Then:
:$a - b = -1$
but $-1$ is not a (strictly) positive real number.
{{qed}}
\end{proof}
|
21857
|
\section{Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers}
Tags: Abelian Group Examples, Subgroups, Real Numbers, Real Multiplication
\begin{theorem}
Let $\R_{>0}$ be the set of strictly positive real numbers, that is:
:$\R_{>0} = \set {x \in \R: x > 0}$
The structure $\struct {\R_{>0}, \times}$ forms a subgroup of $\struct {\R_{\ne 0}, \times}$, where $\R_{\ne 0}$ is the set of real numbers without zero, that is:
:$\R_{\ne 0} = \R \setminus \set 0$
\end{theorem}
\begin{proof}
From Non-Zero Real Numbers under Multiplication form Abelian Group we have that $\struct {\R_{\ne 0}, \times}$ is a group.
We know that $\R_{>0} \ne \O$, as (for example) $1 \in \R_{>0}$.
Now, verify that the conditions for Two-Step Subgroup Test are satisfied:
\end{proof}
|
21858
|
\section{Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group}
Tags: Abelian Groups, Examples of Abelian Groups, Group Examples, Abelian Groups: Examples, Real Numbers, Infinite Groups: Examples, Examples of Infinite Groups, Abelian Group Examples, Real Multiplication
\begin{theorem}
Let $\R_{>0}$ be the set of strictly positive real numbers:
:$\R_{>0} = \set {x \in \R: x > 0}$
The structure $\struct {\R_{>0}, \times}$ is an uncountable abelian group.
\end{theorem}
\begin{proof}
From Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers we have that $\struct {\R_{>0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$, where $\R_{\ne 0}$ is the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
From Subgroup of Abelian Group is Abelian it also follows that $\struct {\R_{>0}, \times}$ is an abelian group.
Its infinite nature follows from the nature of real numbers.
{{Handwaving|Strictly state how the positive real numbers are uncountable. Straightforward, but we have no page for it.}}
{{qed}}
\end{proof}
|
21859
|
\section{Strictly Precede and Step Condition and not Precede implies Joins are equal}
Tags: Join and Meet Semilattices
\begin{theorem}
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $p, q, u \in S$ be such that:
:$p \prec q$ and $\paren {\forall s \in S: p \prec s \implies q \preceq s}$ and $u \npreceq p$
Then:
:$p \vee u = q \vee u$
\end{theorem}
\begin{proof}
From the definition of join, it is required to prove that:
:$\forall s \in S: p \preceq s \land u \preceq s \implies q \vee u \preceq s$
Let $s \in S$ be such that:
:$p \preceq s$ and $u \preceq s$
We have:
:$p \ne s$
By definition of strictly precede:
:$p \prec s$
By assumption:
:$q \preceq s$
Thus by definition of the join $q \vee u$:
:$q \vee u \preceq s$
and hence, as desired:
:$q \vee u = p \vee u$
{{qed}}
\end{proof}
|
21860
|
\section{Strictly Precedes is Strict Ordering}
Tags: Order Theory, Strict Orderings
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $\prec$ be the relation on $S$ defined as:
:$a \prec b \iff \paren {a \ne b} \land \paren {a \preceq b}$
That is, $a \prec b$ {{iff}} $a$ strictly precedes $b$.
Then:
:$a \preceq b \iff \paren {a = b} \lor \paren {a \prec b}$
and $\prec$ is a strict ordering on $S$.
\end{theorem}
\begin{proof}
We are given that $\struct {S, \preceq}$ is an ordered set.
\end{proof}
|
21861
|
\section{Strictly Succeed is Dual to Strictly Precede}
Tags: Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $a, b \in S$.
The following are dual statements:
:$a$ strictly succeeds $b$
:$a$ strictly precedes $b$
\end{theorem}
\begin{proof}
By definition, $a$ strictly succeeds $b$ {{iff}}:
:$b \preceq a$ and $b \ne a$
The dual of this statement is:
:$a \succeq b$ and $b \ne a$
by Dual Pairs (Order Theory).
By definition, this means $a$ strictly precedes $b$.
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}}
\end{proof}
|
21862
|
\section{Strictly Well-Founded Relation determines Strictly Minimal Elements}
Tags: Von Neumann Hierarchy, Axiom of Foundation, Class Theory
\begin{theorem}
Let $A$ be a class.
Let $\RR$ be a strictly well-founded relation on $A$.
Let $B$ be a nonempty class such that $B \subseteq A$.
Then $B$ has a strictly minimal element under $\RR$.
\end{theorem}
\begin{proof}
{{NotZFC}}
First a lemma:
\end{proof}
|
21863
|
\section{Strictly Well-Founded Relation determines Strictly Minimal Elements/Lemma}
Tags: Von Neumann Hierarchy, Axiom of Foundation, Class Theory
\begin{theorem}
Let $A$ be a non-empty class.
Let $\RR$ be a strictly well-founded relation on $A$.
Then $A$ has a strictly minimal element under $\RR$.
\end{theorem}
\begin{proof}
Proceed by induction:
$\map R A$ is a set, so $\map F 0$ is a set.
Suppose that $\map R {\map F n}$ is a set.
We know that for each $y \in \map F n$, $\map R {\map {\RR^{-1} } y}$ is a set, so by the Axiom of Unions, $\map F {n + 1}$ is a set.
{{qed|lemma}}
Let $\ds a = \bigcup_{n \mathop \in \omega} \map F n$.
By the Axiom of Unions, $a$ is a set.
Since $\map F n \subseteq A$ for each $n \in \omega$, $a \subseteq A$.
By Non-Empty Class has Element of Least Rank, $\map F 0 \ne \O$, so $a \ne \O$.
Since $\RR$ is strictly well-founded, $a$ has a strictly minimal element $m$ under $\RR$.
{{AimForCont}} that $m$ is not a strictly minimal element under $\RR$ of $A$.
Then, by Characterization of Minimal Element,
:$\map {\RR^{-1} } m \ne \O$
By Non-Empty Class has Element of Least Rank, $\map {\RR^{-1} } m$ has an element $w$ of least rank.
{{begin-eqn}}
{{eqn | q = \exists n \in \N
| l = m
| o = \in
| r = \map F n
| c = Definition of $a$
}}
{{eqn | l = w
| o = \in
| r = \map F {n + 1}
| c = Definition of $F$
}}
{{eqn | l = w
| o = \in
| r = a
| c = Definition of $a$
}}
{{end-eqn}}
Therefore, $a \mathop \cap \map {\RR^{-1} } m \ne \O$, contradicting the fact that $m$ is a strictly minimal element under $\RR$ in $a$.
Thus we conclude that $m$ is a strictly minimal element under $\RR$ of $A$.
{{qed}}
\end{proof}
|
21864
|
\section{Strictly Well-Founded Relation has no Relational Loops}
Tags: Set Theory
\begin{theorem}
Let $\prec$ be a strictly well-founded relation on $A$ and let $x_1, x_2, \ldots, x_n \in A$.
Then:
:$\neg \paren {x_1 \prec x_2 \land x_3 \prec x_4 \cdots \land x_n \prec x_1}$
That is, there are no relational loops within $A$.
\end{theorem}
\begin{proof}
Since $x_1, x_2, \ldots, x_n \in A$, there exists a $y$ such that $y = \set {x_1, x_2, \ldots, x_n}$.
Then $y$ is a non-empty subset of $A$.
So, by the definition of a strictly well-founded relation:
:$\exists w \in y: \forall z \in y: \neg w \prec z$
Now, suppose $x_1 \prec x_2 \land x_2 \prec x_3 \cdots \land x_n \prec x_1$.
But since the elements of $y$ are $x_1, x_2, \ldots, x_n$, then this contradicts the previous statement, since:
:$\forall w \in y: \exists z \in y: w \prec z$
Thus a founded relation has no relational loops.
{{explain|is a "founded relation" the same as a "foundational relation", that is a (strictly) well-founded relation? If so, use the same terminology throughout the entire page; if not, provide a definition for the former.}}
{{qed}}
\end{proof}
|
21865
|
\section{Strictly Well-Founded Relation is Antireflexive}
Tags: Reflexive Relations, Relation Theory, Well-Founded Relations, Antireflexive Relations, Foundational Relations
\begin{theorem}
Let $\RR$ be a strictly well-founded relation on a set or class $A$.
Then $\RR$ is antireflexive.
\end{theorem}
\begin{proof}
Let $p \in A$.
Then $\set p \ne \O$ and $\set p \subseteq A$.
Thus, by the definition of strictly well-founded relation:
:$\exists x \in \set p: \forall y \in \set p: \neg \paren {y \mathrel \RR x}$
Since $x \in \set p$, it must be that $x = p$.
It follows that $p \not \mathrel \RR p$.
Since this holds for all $p \in A$, $\RR$ is antireflexive.
{{qed}}
Category:Well-Founded Relations
Category:Reflexive Relations
\end{proof}
|
21866
|
\section{Strictly Well-Founded Relation is Antireflexive/Corollary}
Tags: Well-Founded Relations, Foundational Relations, Relation Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Suppose that $S$ is non-empty.
Then $\preceq$ is not a strictly well-founded relation.
\end{theorem}
\begin{proof}
Since $S$ is non-empty, it has an element $x$.
By the definition of ordering, $\preceq$ is a reflexive relation.
Thus $x \preceq x$.
By Strictly Well-Founded Relation is Antireflexive, $\preceq$ is not a strictly well-founded relation.
{{qed}}
Category:Well-Founded Relations
\end{proof}
|
21867
|
\section{Strictly Well-Founded Relation is Asymmetric}
Tags: Well-Founded Relations, Foundational Relations, Symmetric Relations, Relation Theory
\begin{theorem}
Let $\struct {S, \RR}$ be a relational structure, where $S$ is a set or a proper class.
Let $\RR$ be a strictly well-founded relation.
Then $\RR$ is asymmetric.
\end{theorem}
\begin{proof}
Let $p, q \in S$ and suppose that $p \mathrel \RR q$.
Then $\set {p, q} \ne \O$ and $\set {p, q} \subseteq S$.
By the definition of strictly well-founded relation, $\set {p, q}$ has a strictly minimal element under $\RR$.
Since $p \mathrel \RR q$, $q$ is not an $\RR$-minimal element of $\set {p, q}$.
Thus $p$ is a strictly minimal element under $\RR$ of $\set {p, q}$.
Thus $q \not \mathrel \RR p$.
Since for all $p, q \in S$, $p \mathrel \RR q \implies q \not \mathrel \RR p$, $\RR$ is asymmetric.
{{qed}}
Category:Well-Founded Relations
Category:Symmetric Relations
\end{proof}
|
21868
|
\section{Strictly Well-Founded Relation is Well-Founded}
Tags: Well-Founded Relations
\begin{theorem}
Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a strictly well-founded relation on $S$.
Then $\RR$ is a well-founded relation on $S$.
\end{theorem}
\begin{proof}
We have that $\RR$ is a strictly well-founded relation on $S$.
By definition:
:$\forall T: \paren {T \subseteq S \land T \ne \O} \implies \exists y \in T: \forall z \in T: \neg \paren {z \mathrel \RR y}$
It immediately follows that:
:$\forall T: \paren {T \subseteq S \land T \ne \O} \implies \exists y \in T: \forall z \in \paren {T \setminus \set y}: \neg \paren {z \mathrel \RR y}$
which is the definition of a well-founded relation on $S$.
{{qed}}
\end{proof}
|
21869
|
\section{Strong Law of Large Numbers}
Tags: Probability Theory
\begin{theorem}
Let $P$ be a population.
Let $P$ have mean $\mu$ and finite variance.
Let $\sequence {X_n}_{n \ge 1}$ be a sequence of random variables forming a random sample from $P$.
Let:
:$\ds {\overline X}_n = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
:$\ds {\overline X}_n \xrightarrow {\text {a.s.} } \mu$
where $\xrightarrow {\text {a.s.} }$ denotes almost sure convergence.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Probability Theory
\end{proof}
|
21870
|
\section{Strongly Locally Compact Space is Weakly Locally Compact}
Tags: Strongly Locally Compact Spaces, Compact Spaces, Weakly Locally Compact Spaces, Local Compactness, Locally Compact Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a strongly locally compact space.
Then $T$ is weakly locally compact.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be strongly locally compact.
Let $x \in S$.
By definition, there exists an open set $U_x$ of $T$ such that:
:$x \in U_x$
:${U_x}^-$ (the closure of $U_x$) is compact.
From Set is Subset of its Topological Closure, $U_x \subseteq {U_x}^-$ and so $x \in {U_x}^-$.
Thus $x$ is contained in a compact neighborhood.
As this holds for all $x$, $T$ is a weakly locally compact.
{{qed}}
\end{proof}
|
21871
|
\section{Structure Induced by Abelian Group Operation is Abelian Group}
Tags: Abelian Groups, Group Theory, Pointwise Operations, Mappings, Mapping Theory
\begin{theorem}
Let $\struct {G, \circ}$ be an abelian group whose identity is $e$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$.
Then $\struct {G^S, \oplus}$ is an abelian group.
\end{theorem}
\begin{proof}
From Structure Induced by Group Operation is Group, $\struct {G^S, \oplus}$ is a group.
From Structure Induced by Commutative Operation is Commutative, so is the pointwise operation it induces on $G^S$.
Hence the result.
{{qed}}
Category:Abelian Groups
Category:Pointwise Operations
\end{proof}
|
21872
|
\section{Structure Induced by Associative Operation is Associative}
Tags: Abstract Algebra, Mappings, Associativity, Pointwise Operations, Mapping Theory
\begin{theorem}
Let $\struct {T, \circ}$ be an algebraic structure.
Let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $\circ$ be associative.
Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also associative.
\end{theorem}
\begin{proof}
Let $f, g, h \in T^S$.
Let $\struct {T, \circ}$ be an associative algebraic structure.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\paren {f \oplus g} \oplus h} } x
| r = \map {\paren {f \oplus g} } x \circ \map h x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \paren {\map f x \circ \map g x} \circ \map h x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \map f x \circ \paren {\map g x \circ \map h x}
| c = $\circ$ is associative
}}
{{eqn | r = \map f x \circ \map {\paren {g \oplus h} } x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \map {\paren {f \oplus \paren {g \oplus h} } } x
| c = {{Defof|Pointwise Operation}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
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