id
stringlengths 1
260
| contents
stringlengths 1
234k
|
|---|---|
21173
|
\section{Set of Finite Subsets of Countable Set is Countable}
Tags: Countable Sets, Subsets, Subset, Set of Finite Subsets of Countable Set is Countable
\begin{theorem}
Let $A$ be a countable set.
Then the set of finite subsets of $A$ is countable.
\end{theorem}
\begin{proof}
Let $A^{(n)}$ be the set of subsets of $A$ with no more than $n$ elements.
Thus:
:$A^{(0)} = \left\{{\varnothing}\right\}$
:$A^{(1)} = A^{(0)} \cup \left\{{\left\{{a}\right\}: a \in A}\right\}$
and $\forall n \ge 0$:
:$A^{(n+1)} = \left\{{a^{(n)} \cup a^{(1)}: a^{(n)} \in A^{(n)} \land a^{(1)} \in A^{(1)}}\right\}$
Let us verify by induction that each $A^{(n)}$ is countable.
Let $A$ be countable.
:$A^{(1)}$ is countable, as its cardinality is $1 + \left|{A}\right|$.
Suppose $A^{(n)}$ is countable.
Then by Union of Countable Sets of Sets, so $A^{(n+1)}$ also countable.
By induction, each $A^{(n)}$ is countable.
Denote with $A^f$ the set of finite subsets of $A$.
It is apparent that every finite subset is in some $A^{(n)}$, and so:
:$A^f = \displaystyle \bigcup_{n \mathop \in \N} A^{(n)}$
The result follows from Countable Union of Countable Sets is Countable.
{{WIP|actually, this theorem is provable in ZF}}
{{qed}}
\end{proof}
|
21174
|
\section{Set of Finite Subsets under Induced Operation is Closed}
Tags: Subset Products, Power Set
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the algebraic structure consisting of the power set of $S$ and the operation induced on $\powerset S$ by $\circ$.
Let $T \subseteq \powerset S$ be the set of all finite subsets of $S$.
Then the algebraic structure $\struct {T, \circ_\PP}$ is closed.
\end{theorem}
\begin{proof}
Let $X, Y \in T$.
Then:
{{begin-eqn}}
{{eqn | q =
| l = X \circ_\PP Y
| r = \set {x \circ y: x \in X, y \in Y}
| c = {{Defof|Operation Induced on Power Set}}
}}
{{eqn | ll= \leadsto
| l = \card {X \circ_\PP Y}
| o = \le
| r = \card X \times \card Y
| c =
}}
{{eqn | ll= \leadsto
| l = X \circ_\PP Y
| o = \in
| r = T
| c =
}}
{{end-eqn}}
{{handwaving|Between step $1$ and step $2$ we really need a result Order of Subset Product is not Greater than Cardinality of Cartesian Product which is straightforward but tedious, and I lack patience.}}
{{qed}}
\end{proof}
|
21175
|
\section{Set of Finite Suprema is Directed}
Tags: Join and Meet Semilattices
\begin{theorem}
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $X$ be a non-empty subset of $S$.
Then
:$\set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$ is directed.
where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $X$.
\end{theorem}
\begin{proof}
By Existence of Non-Empty Finite Suprema in Join Semilattice:
:for every $A \in \map {\operatorname {Fin} } S$ if $A \ne \O$, then $A$ admits a supremum.
By definition of non-empty set:
:$\exists a: a \in X$
By definitions of subset and singleton:
:$\set x \subseteq X$
By Singleton is Finite:
:$\set x$ is finite
By definitions of non-empty set and singleton:
:$\set x \ne \O$
By definition of $\operatorname {Fin}$:
:$\sup {\set x} \in \set {\sup A: A \in \map {\operatorname{Fin} } X \land A \ne \O}$
Thus by definition:
:$\set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$ is a non-empty set.
Let $x, y \in \set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$
Then
:$\exists A \in \map {\operatorname {Fin} } X: x = \sup A$ and $A \ne \O$
and
:$\exists B \in \map {\operatorname {Fin} } X: y = \sup B$ and $B \ne \O$
By Finite Union of Finite Sets is Finite:
:$A \cup B$ is finite
By Union of Subsets is Subset:
:$A \cup B \subseteq X$
By definitions of non-empty set and union:
:$A \cup B \ne \O$
By definition of $\operatorname {Fin}$:
:$\map \sup {A \cup B} \in \set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$
By Set is Subset of Union:
:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
Thus by Supremum of Subset:
:$x \preceq \map \sup {A \cup B}$ and $y \preceq \map \sup {A \cup B}$
Thus by definition:
:$\set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$ is directed.
{{qed}}
\end{proof}
|
21176
|
\section{Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic}
Tags: Mathematical Logic
\begin{theorem}
Let $T$ be the set of theorems of some consistent theory in the language of arithmetic which contains minimal arithmetic.
The set of Gödel numbers of the theorems of $T$ is not definable in $T$.
\end{theorem}
\begin{proof}
{{Questionable|Not only the proof is faulty, this theorem is wrong. If we have a recursively enumerable set A of axioms, then the set of theorems (and hence, the Gödel numbers of those) proven by A is recursively enumerable. Minimal arithmetic and PA are both recursively enumerable, and hence their theorems have a recursively enumerable set of Gödel numbers. Any recursively enumerable set is arithmetical with degree Sigma^0_1 in the arithmetical hierarchy. Moreover, there is already a well known provability predicate Pr(x) in the language of Peano arithmetic. Indeed, if we use the diagonal lemma on ~Pr(x) to get a sentence with T proves G iff ~Pr(G), then we will have an unprovable sentence.|Burak}}
The proof is by contradiction.
Let $\Theta$ be the set of Gödel numbers of the theorems of $T$.
Suppose it is defined in $T$ by the formula $\map \theta y$.
Since $T$ contains $Q$, we may apply the Diagonal Lemma to $\neg \map \theta y$.
This gives us a sentence $G$ such that
:$T \vdash G \leftrightarrow \neg \map \theta {\hat G}$
where $\hat G$ is the Gödel number of $G$ (more accurately, it is the term in the language of arithmetic obtained by applying the function symbol $s$ to $0$ this many times).
{{Questionable|The below argument is not correct. $T$ does not necessarily prove that $\neg \map \theta {\hat G}$ just because it does not prove $G$. PA has a provability predicate $\map \Pr x$ which satisfies the property that if PA proves G, then PA proves $\map \Pr {\hat G}$). The step used here assumes a somewhat similar property in the opposite direction that our provability predicate satisfies if T does not prove G, then T proves $\neg \map \theta {\hat G}$, which we did not have as an assumption on $\map \Theta x$. |Burak}}
Suppose $G$ is not a theorem of $T$.
:Then the Gödel number of $G$ is not in $\Theta$.
:Since $\theta$ defines $\Theta$ in $T$, this means that:
::$T \vdash \neg \map \theta {\hat G}$
:But, by choice of $G$ (specifically, the bi-implication above), this gives us:
::$T\vdash G$
:which contradicts $G$ not being a theorem of $T$
Thus, $G$ is a theorem of $T$.
:But, then the Gödel number of $G$ is in $\Theta$, and
:since $\theta$ defines $\Theta$ in $T$, this means that
::$T \vdash \map \theta {\hat G}$
:But, then this gives us
::$T \vdash \neg G$
:which contradicts $G$ being a theorem of $T$, ''since $T$ is consistent''.
Since assuming $\Theta$ was definable in $T$ necessarily leads to a contradiction, we conclude that it is impossible.
{{qed}}
\end{proof}
|
21177
|
\section{Set of Homomorphisms to Abelian Group is Subgroup of All Mappings}
Tags: Subgroups, Abelian Groups, Group Theory, Morphisms, Homomorphisms
\begin{theorem}
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \oplus}$ be an abelian group.
Let $\struct {T^S, \oplus}$ be the algebraic structure on $T^S$ induced by $\oplus$.
Then the set of all homomorphisms from $\struct {S, \circ}$ into $\struct {T, \oplus}$ is a subgroup of $\struct {T^S, \oplus}$.
\end{theorem}
\begin{proof}
Let $H$ be the set of all homomorphisms from $\struct {S, \circ}$ into $\struct {T, \oplus}$.
\end{proof}
|
21178
|
\section{Set of Ideals forms Complete Lattice}
Tags: Lattice Theory, Ideal Theory
\begin{theorem}
Let $\struct {K, +, \circ}$ be a ring.
Let $\mathbb K$ be the set of all ideals of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice.
\end{theorem}
\begin{proof}
Let $\O \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Ring Ideals is Largest Ideal Contained in all Ideals:
:$\bigcap \mathbb S$ is the largest ideal of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Ring Ideals Containing Subset is Smallest:
:The intersection of the set of all ideals of $K$ containing $\bigcup \mathbb S$ is the smallest ideal of $K$ containing $\bigcup \mathbb S$.
Thus, not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.
The supremum of $\mathbb S$ is the join of the set of all ideals of $\mathbb S$.
From Sum of Ideals is Ideal: General Result, this supremum is:
:$\ds \sum_{S \mathop \in \mathbb S} \struct {S, +, \circ}$
That is:
:$\struct {S_1, +, \circ} + \struct {S_2, +, \circ} + \dotsb$
where addition of ideals is as defined in subset product.
Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice.
{{qed}}
\end{proof}
|
21179
|
\section{Set of Infima for Sequence is Directed}
Tags: Order Theory
\begin{theorem}
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.
Let $\left({A, \precsim}\right)$ be a non-empty directed set.
Let $Z: A \to S$ be a Moore-Smith sequence.
Let $D = \left\{ {\inf \left({Z\left[{\precsim \left({j}\right)}\right]}\right): j \in A}\right\}$ be a subset of $S$.
Then $D$ is directed.
\end{theorem}
\begin{proof}
By definition of non-empty set:
:$\exists j: j \in A$
By definition of $D$:
:$\inf \left({Z\left[{\precsim \left({j}\right)}\right]}\right) \in D$
Hence by definition:
:$D$ is a non-empty set.
Let $x, y \in D$.
By definition of $D$:
:$\exists j_1 \in A: x = \inf \left({Z\left[{\precsim \left({j_1}\right)}\right]}\right)$
and
:$\exists j_2 \in A: y = \inf \left({Z\left[{\precsim \left({j_2}\right)}\right]}\right)$
By definition of directed set:
:$\exists j \in A: j_1 \precsim j \land j_2 \precsim j$
By Preceding implies Image is Subset of Image:
:$\precsim\left({j}\right) \subseteq \mathord\precsim\left({j_1}\right)$ and $\precsim\left({j}\right) \subseteq \mathord\precsim\left({j_2}\right)$
By Image of Subset under Relation is Subset of Image/Corollary 2
:$Z\left[{\precsim\left({j}\right)}\right] \subseteq Z\left[{\precsim\left({j_1}\right)}\right]$ and $Z\left[{\precsim\left({j}\right)}\right] \subseteq Z\left[{\precsim\left({j_2}\right)}\right]$
Thus by definition of $D$:
:$z := \inf \left({Z\left[{\precsim \left({j}\right)}\right]}\right) \in D$
Thus by Infimum of Subset:
:$x \preceq z$ and $y \preceq z$
{{qed}}
\end{proof}
|
21180
|
\section{Set of Infinite Sequences is Uncountable}
Tags: Diagonal Arguments, Countable Sets
\begin{theorem}
Let $S$ be a set which contains more than one element.
Let $S^\infty$ denote the set of all sequences of elements of $S$.
Then $S^\infty$ is uncountable.
\end{theorem}
\begin{proof}
As $S$ has more than one element, it must have at least two.
So, let $a, b \in S$ be those two elements.
Let $Z$ be the set of all sequences from $\set {a, b}$.
Suppose $S^\infty$ were countable.
From Subset of Countably Infinite Set is Countable, $Z$ is likewise countable.
So by definition, it would be possible to set up a bijection $\phi: Z \leftrightarrow \N$ between $Z$ and the set $\N$ of natural numbers:
{{begin-eqn}}
{{eqn | l = 0
| o = \leftrightarrow
| r = \sequence {z_0} = \tuple {z_{00}, z_{01}, z_{02}, \ldots, z_{0n}, \ldots}
}}
{{eqn | l = 1
| o = \leftrightarrow
| r = \sequence {z_1} = \tuple {z_{10}, z_{11}, z_{12}, \ldots, z_{1n}, \ldots}
}}
{{eqn | l =
| o = \vdots
| r =
}}
{{eqn | l = n
| o = \leftrightarrow
| r = \sequence {z_n} = \tuple {z_{n0}, z_{n1}, z_{n2}, \ldots, z_{nn}, \ldots}
}}
{{eqn | l =
| o = \vdots
| r =
}}
{{end-eqn}}
where each of $z_{rs}$ is:
: an element of $\set {a, b}$
: the $s$'th element of $\sequence {z_r}$, the particular sequence which is in correspondence with $r \in \N$.
Now we create the sequence $Q = \sequence {q_{dd} }$ where:
:$q_{dd} = \begin{cases} a & : z_{dd} = b \\ b & : z_{dd} = a \end{cases}$
Thus:
:$\forall n \in \N: q_{nn} \notin \sequence {z_n}$
and so:
:$\forall n \in \N: Q \ne \sequence {z_n}$
So $Q \notin Z$.
That is, we have constructed a sequence which has not been placed into correspondence with an element of $\N$.
Thus by definition $Z$ is uncountable.
Hence, by the Rule of Transposition, $S^\infty$ is likewise uncountable.
{{qed}}
\end{proof}
|
21181
|
\section{Set of Integer Multiples is Integral Ideal}
Tags: Sets of Integer Multiples, Integral Ideals
\begin{theorem}
Let $m \in \Z$ be an integer.
Let $m \Z$ denote the set of integer multiples of $m$.
Then $m \Z$ is an integral ideal.
\end{theorem}
\begin{proof}
First note that $m \times 0 \in m \Z$ whatever $m$ may be.
Thus $m \Z \ne \O$.
Let $a, b \in m \Z$.
Then:
{{begin-eqn}}
{{eqn | l = a + b
| r = m j + m k
| c = for some $j, k \in \Z$ by definition of $m \Z$
}}
{{eqn | r = m \paren {j + k}
| c =
}}
{{eqn | o = \in
| r = m \Z
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = a - b
| r = m j - m k
| c = for some $j, k \in \Z$ by definition of $m \Z$
}}
{{eqn | r = m \paren {j - k}
| c =
}}
{{eqn | o = \in
| r = m \Z
| c =
}}
{{end-eqn}}
Let $r \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = r a
| r = r \paren {m j}
| c = for some $j \in \Z$ by definition of $m \Z$
}}
{{eqn | r = m \paren {r j}
| c =
}}
{{eqn | o = \in
| r = m \Z
| c =
}}
{{end-eqn}}
Thus the conditions for $m \Z$ to be an integral ideal are fulfilled.
Hence the result.
{{qed}}
\end{proof}
|
21182
|
\section{Set of Integer Multiples of GCD}
Tags: Sets of Integer Multiples, Greatest Common Divisor
\begin{theorem}
Let $m, n \in \Z$.
Let $m \Z$ denote the set of integer multiples of $m$
Then:
:$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$
where $\gcd$ denotes greatest common divisor.
\end{theorem}
\begin{proof}
Let $x \in m \Z \cup n \Z$.
Then either:
:$m \divides x$
or:
:$n \divides x$
In both cases:
:$\gcd \set {m, n} \divides x$
and so:
:$x \in \gcd \set {m, n} \Z$
Hence by definition of subset:
:$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$
{{qed}}
\end{proof}
|
21183
|
\section{Set of Integers Bounded Above by Integer has Greatest Element}
Tags: Number Theory, Integers, Set of Integers Bounded Above has Greatest Element
\begin{theorem}
Let $\Z$ be the set of integers.
Let $\le$ be the ordering on the integers.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded above in $\struct {\Z, \le}$.
Then $S$ has a greatest element.
\end{theorem}
\begin{proof}
$S$ is bounded above, so $\exists M \in \Z: \forall s \in S: s \le M$.
Hence $\forall s \in S: 0 \le M - s$.
Thus the set $T = \left\{{M - s: s \in S}\right\} \subseteq \N$.
The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$.
Hence:
: $\left({\forall s \in S: b_T \le M - s}\right) \land \left({\exists g_S \in S: b_T = M - g_S}\right)$
So:
{{begin-eqn}}
{{eqn | o=
| r=\forall s \in S: M - g_S \le M - s
| c=
}}
{{eqn | o=\implies
| r=\forall s \in S: -g_S \le -s
| c=Cancellability of elements of $\Z$
}}
{{eqn | o=\implies
| r=\forall s \in S: g_S \ge s
| c=
}}
{{eqn | o=\implies
| r=g_S \in S \land \left({\forall s \in S: g_S \ge s}\right)
| c=... which is how the greatest element is defined.
}}
{{end-eqn}}
So $g_S$ is the greatest element of $S$.
{{qed}}
\end{proof}
|
21184
|
\section{Set of Integers Bounded Above by Real Number has Greatest Element}
Tags: Number Theory, Set of Integers Bounded Above has Greatest Element
\begin{theorem}
Let $\Z$ be the set of integers.
Let $\le$ be the usual ordering on the real numbers $\R$.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded above in $\struct {\R, \le}$.
Then $S$ has a greatest element.
\end{theorem}
\begin{proof}
Let $S$ be bounded above by $x \in \R$.
By the Archimedean Principle, there exists an integer $n \ge x$.
Then $S$ is bounded above by $n$.
By Set of Integers Bounded Above by Integer has Greatest Element, $S$ has a greatest element.
{{qed}}
\end{proof}
|
21185
|
\section{Set of Integers Bounded Below by Integer has Smallest Element}
Tags: Number Theory, Well-Orderings, Integers, Set of Integers Bounded Below has Smallest Element
\begin{theorem}
Let $\Z$ be the set of integers.
Let $\le$ be the ordering on the integers.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded below in $\struct {\Z, \le}$.
Then $S$ has a smallest element.
\end{theorem}
\begin{proof}
We have that $S$ is bounded below in $\Z$.
So:
: $\exists m \in \Z: \forall s \in S: m \le s$
Hence:
: $\forall s \in S: 0 \le s - m$
Thus:
:$T = \set {s - m: s \in S} \subseteq \N$
The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$.
Hence:
:$\paren {\forall s \in S: b_T \le s - m} \land \paren {\exists b_S \in S: b_T = b_S - m}$
So:
{{begin-eqn}}
{{eqn | o =
| r = s \in S: b_S - m \le s - m
| c =
}}
{{eqn | o = \leadsto
| r = \forall s \in S: b_S \le s
| c = Cancellability of elements of $\Z$
}}
{{eqn | o = \leadsto
| r = b_S \in S \land \paren {\forall s \in S: b_S \le s}
| c = {{Defof|Smallest Element}}
}}
{{end-eqn}}
So $b_S$ is the smallest element of $S$.
{{Qed}}
\end{proof}
|
21186
|
\section{Set of Integers Bounded Below by Real Number has Smallest Element}
Tags: Number Theory, Set of Integers Bounded Below has Smallest Element
\begin{theorem}
Let $\Z$ be the set of integers.
Let $\le$ be the usual ordering on the real numbers $\R$.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded below in $\struct {\R, \le}$.
Then $S$ has a smallest element.
\end{theorem}
\begin{proof}
Let $S$ be bounded below by $x \in \R$.
By the Archimedean Principle, there exists an integer $n \le x$.
Then $S$ is bounded below by $n$.
By Set of Integers Bounded Below by Integer has Smallest Element, $S$ has a smallest element.
{{qed}}
\end{proof}
|
21187
|
\section{Set of Integers can be Well-Ordered}
Tags: Well-Orderings, Integers
\begin{theorem}
The set of integers $\Z$ can be well-ordered with an appropriately chosen ordering.
\end{theorem}
\begin{proof}
Consider the ordering $\preccurlyeq \subseteq \Z \times \Z$ defined as:
:$x \preccurlyeq y \iff \left({\left\vert{x}\right\vert < \left\vert{y}\right\vert}\right) \lor \left({\left\vert{x}\right\vert = \left\vert{y}\right\vert \land x \le y}\right)$
{{finish|It remains to be shown that $\preccurlyeq$ is an ordering, and also that it is a well-ordering.}}
\end{proof}
|
21188
|
\section{Set of Integers is not Bounded}
Tags: Boundedness, Real Analysis, Definitions: Analysis, Analysis
\begin{theorem}
Let $\R$ be the real number line considered as an Euclidean space.
The set $\Z$ of integers is not bounded in $\R$.
\end{theorem}
\begin{proof}
Let $a \in \R$.
Let $K \in \R_{>0}$.
Consider the open $K$-ball $\map {B_K} a$.
By the Archimedean Principle there exists $n \in \N$ such that $n > a + K$.
As $\N \subseteq \Z$:
:$\exists n \in \Z: a + K < n$
and so:
:$n \notin \map {B_K} a$
As this applies whatever $a$ and $K$ are, it follows that there is no $\map {B_K} a$ which contains all the integers.
Hence the result, by definition of bounded space.
{{qed}}
\end{proof}
|
21189
|
\section{Set of Integers is not Compact}
Tags: Compact Spaces
\begin{theorem}
Let $\Z$ be the set of integers.
Then $\Z$ is not compact.
\end{theorem}
\begin{proof}
Let $\R$ be the real number line considered as an Euclidean space.
From Set of Integers is not Bounded, $\Z$ is not bounded in $\R$.
The result follows by definition of compact.
{{qed}}
\end{proof}
|
21190
|
\section{Set of Integers is not Well-Ordered by Usual Ordering}
Tags: Well-Orderings, Integers
\begin{theorem}
The set of integers $\Z$ is not well-ordered under the usual ordering $\le$.
\end{theorem}
\begin{proof}
{{AimForCont}} $\Z$ is a well-ordered set.
Then by definition, all subsets of $\Z$ has a smallest element.
But take $\Z$ itself.
Suppose $x \in \Z$ is a smallest element.
Then $x - 1 \in \Z$.
But $x - 1 < x$, which contradicts the supposition that $x \in \Z$ is a smallest element.
Hence there can be no such smallest element.
So by Proof by Contradiction, $\Z$ is not well-ordered by $\le$.
{{qed}}
\end{proof}
|
21191
|
\section{Set of Integers under Addition is Isomorphic to Set of Even Integers under Addition}
Tags: Additive Group of Integers, Integers, Even Integers
\begin{theorem}
Let $\struct {\Z, +}$ be the algebraic structure formed by the set of integers under the operation of addition.
Let $\struct {2 \Z, +}$ be the algebraic structure formed by the set of even integers under the operation of addition.
Then $\struct {\Z, +}$ and $\struct {2 \Z, +}$ are isomorphic.
\end{theorem}
\begin{proof}
Let $f: \Z \to 2 \Z$ be the mapping:
:$\forall n \in \Z: \map f n = 2 n$
From Bijection between Integers and Even Integers, $f$ is a bijection.
Let $m, n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {m + n}
| r = 2 \paren {m + n}
| c = Definition of $f$
}}
{{eqn | r = 2 m + 2 n
| c = Integer Multiplication Distributes over Addition
}}
{{eqn | r = \map f m + \map f n
| c = Integer Multiplication Distributes over Addition
}}
{{end-eqn}}
Thus $f$ is an isomorphism by definition.
{{qed}}
\end{proof}
|
21192
|
\section{Set of Inverse Positive Integers with Zero is Compact}
Tags: Compact Spaces, Integers
\begin{theorem}
Let $K$ be the set of inverse positive integers with zero:
:$\ds K := \set {1, \frac 1 2, \frac 1 3, \dots} \cup \set 0$
Let $\struct {\R, \size {\, \cdot \,}}$ be the normed vector space of real numbers.
Then $K$ is compact in real numbers.
\end{theorem}
\begin{proof}
We have that $K \subset \closedint 0 1$.
Hence, $K$ is bounded.
Furthermore:
:$\ds \R \setminus K = \openint {-\infty} 0 \cup \paren {\bigcup_{n \mathop = 1}^\infty \openint {\frac 1 {n + 1}} {\frac 1 n}} \cup \openint 1 \infty$
By Union of Open Sets of Normed Vector Space is Open, $\R \setminus K$ is open.
By definition, $K$ is closed.
By Heine-Borel theorem, $K$ is compact.
{{qed}}
\end{proof}
|
21193
|
\section{Set of Invertible Mappings forms Symmetric Group}
Tags: Group of Permutations, Symmetric Groups, Permutation Theory, Group Examples, Permutations
\begin{theorem}
Let $S$ be a set.
Let $\GG$ be the set of all invertible mappings from $S$ to $S$.
Then $\struct {\GG, \circ}$ is the symmetric group on $S$.
\end{theorem}
\begin{proof}
Let $\struct {S^S, \circ}$ be the algebraic structure formed from the set of all mappings from $S$ to itself.
From Set of all Self-Maps under Composition forms Monoid, $\struct {S^S, \circ}$ is a monoid.
By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
By Bijection iff Inverse is Bijection, it follows that all the invertible elements of $S^S$ are exactly the permutations on $S$.
The result follows from Invertible Elements of Monoid form Subgroup of Cancellable Elements.
{{qed}}
\end{proof}
|
21194
|
\section{Set of Isolated Points of Metric Space is Disjoint from Limit Points}
Tags: Isolated Points, Limit Points
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H'$ be the set of limit points of $H$.
Let $H^i$ be the set of isolated points of $H$.
Then:
:$H' \cap H^i = \O$
\end{theorem}
\begin{proof}
Let $a \in H_i$.
Then by definition of isolated point:
:$\exists \epsilon \in \R_{>0}: \set {x \in H: \map d {x, a} < \epsilon} = \set a$
But by {{Metric-space-axiom|1}}:
:$\map d {a, a} = 0$
and so:
:$\set {x \in H: 0 < \map d {x, a} < \epsilon} = \O$
So by definition $a$ is not a limit point of $H$.
That is:
:$a \notin H'$
or:
:$a \in \relcomp A {H'}$
It follows from Intersection with Complement is Empty iff Subset that:
:$H' \cap H^i = \O$
{{qed}}
\end{proof}
|
21195
|
\section{Set of Isometries in Complex Plane under Composition forms Group}
Tags: Complex Numbers, Isometries, Examples of Groups
\begin{theorem}
Let $S$ be the set of all complex functions $f: \C \to \C$ which preserve distance when embedded in the complex plane.
That is:
:$\size {\map f a - \map f b} = \size {a - b}$
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.
\end{theorem}
\begin{proof}
From Complex Plane is Metric Space, $\C$ can be treated as a metric space.
Hence it is seen that a complex function $f: \C \to \C$ which preserves distance is in fact an isometry on $\C$.
Taking the group axioms in turn:
\end{proof}
|
21196
|
\section{Set of Isometries in Euclidean Space under Composition forms Group}
Tags: Euclidean Metric, Isometries, Examples of Groups
\begin{theorem}
Let $\struct {\R^n, d}$ be a real Euclidean space of $n$ dimensions.
Let $S$ be the set of all mappings $f: \R^n \to \R^n$ which preserve distance:
That is:
:$\map d {\map f a, \map f b} = \map d {a, b}$
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.
\end{theorem}
\begin{proof}
From Euclidean Metric on Real Vector Space is Metric, $\R^n$ is a metric space.
Hence it is seen that a complex function $f: \C \to \C$ which preserves distance is in fact an isometry on $\C$.
Taking the group axioms in turn:
\end{proof}
|
21197
|
\section{Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal}
Tags: Principal Ideal Domains
\begin{theorem}
Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $a_1, a_2, \dotsc, a_n$ be non-zero elements of $D$.
Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$
Then for some $x \in D$:
:$J = \ideal x$
where $\ideal x$ denotes the principal ideal generated by $x$.
\end{theorem}
\begin{proof}
Let the unity of $D$ be $1_D$.
By definition of principal ideal:
:$\ds \ideal a = \set {\sum_{i \mathop = 1}^n r_i \circ a \circ s_i: n \in \N, r_i, s_i \in D}$
Let $x, y \in J$.
By definition of linear combination:
{{begin-eqn}}
{{eqn | l = x
| r = \sum_{i \mathop = 1}^n r_i \circ a_i
| c = for some $n \in \N$ and for some $r_i \in D$ where $i \in \set {1, 2, \dotsc, n}$
}}
{{eqn | r = r_1 \circ a_1 + r_2 \circ a_2 + \dotsb + r_n \circ a_n
| c = for some $r_1, r_2, \dotsc, r_n \in D$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = y
| r = \sum_{i \mathop = 1}^n s_i \circ a_i
| c = for some $n \in \N$ and for some $s_i \in D$ where $i \in \set {1, 2, \dotsc, n}$
}}
{{eqn | r = s_1 \circ a_1 + s_2 \circ a_2 + \dotsb + s_n \circ a_n
| c = for some $s_1, s_2, \dotsc, s_n \in R$
}}
{{eqn | ll= \leadsto
| l = -y
| r = -\sum_{i \mathop = 1}^n s_i \circ a_i
| c =
}}
{{eqn | r = -1_D \times \sum_{i \mathop = 1}^n s_i \circ a_i
| c = Product with Ring Negative: Corollary
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {-1_D} \times \paren {s_i \circ a_i}
| c = {{Ring-axiom|D}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {-\paren {s_i \circ a_i} }
| c = Product with Ring Negative: Corollary
}}
{{eqn | r = \sum_{i \mathop = 1}^n s_i \circ \paren {-a_i}
| c = Product with Ring Negative
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = x + \paren {-y}
| r = \sum_{i \mathop = 1}^n r_i \circ a_i + \sum_{i \mathop = 1}^n s_i \circ \paren {-a_i}
| c =
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + s_i \circ \paren {-a_i} }
| c =
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + \paren {-\paren {s_i \circ a_i} } }
| c =
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + \paren {-s_i} \circ a_i}
| c =
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i + \paren {-s_i} } \circ a_i
| c = {{Ring-axiom|D}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n t_i \circ a_i
| c = where $t_i - r_1 + \paren {-s_i}$
}}
{{eqn | o = \in
| r = J
| c = as $t_i \in D$
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | l = x \circ y
| r = \paren {\sum_{i \mathop = 1}^n r_i \circ a_i} \circ \paren {\sum_{i \mathop = 1}^n s_i \circ a_i }
| c =
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {t_i \circ a_i}
| c = where $t_i \in D$ for $i \in \set {1, 2, \dotsc, n}$
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {a_i \circ t_i}
| c = as $\circ$ is commutative in an integral domain
}}
{{end-eqn}}
{{explain|There exists (or ought to) some convolution result which proves the above -- I just haven't found it yet.}}
Thus by the Test for Ideal, $J$ is an ideal of $D$.
As $D$ is a principal ideal domain, it follows that $J$ is a principal ideal.
Thus by definition of principal ideal:
:$J = \ideal x$
for some $x \in D$.
{{qed}}
\end{proof}
|
21198
|
\section{Set of Linear Subspaces is Closed under Intersection}
Tags: Vector Spaces, Vector Subspaces
\begin{theorem}
Let $\struct {V, +, \circ}_K$ be a $K$-vector space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed family of subspaces of $V$.
Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a subspace of $V$.
\end{theorem}
\begin{proof}
It needs to be demonstrated that $M$ is:
:$(1): \quad$ a closed algebraic structure under $+$
:$(2): \quad$ closed for scalar product $\circ$.
So let $a, b \in M$.
By definition of intersection, $a, b \in M_i$ for all $i \in I$.
As the $M_i$ are subspaces of $V$, $a + b \in M_i$ for all $i \in I$.
That is, by definition of intersection, $a + b \in M$.
It follows that $M$ is closed under $+$.
Now let $\lambda \in K, a \in M$.
By definition of intersection, $a \in M_i$ for all $i \in I$.
As the $M_i$ are subspaces of $V$, $\lambda \circ a \in M_i$ for all $i \in I$.
That is, by definition of intersection, $\lambda \circ a \in M$.
It follows that $M$ is closed under $\circ$.
Hence the result, by definition of subspace.
{{qed}}
Category:Vector Subspaces
\end{proof}
|
21199
|
\section{Set of Linear Transformations is Isomorphic to Matrix Space}
Tags: Linear Algebra, Set of Linear Transformations is Isomorphic to Matrix Space, Matrix Algebra, Linear Transformations
\begin{theorem}
Let $R$ be a ring with unity.
Let $F$, $G$ and $H$ be free $R$-modules of finite dimension $p,n,m>0$ respectively.
Let $\sequence {a_p}$, $\sequence {b_n}$ and $\sequence {c_m}$ be ordered bases
Let $\map {\LL_R} {G, H}$ denote the set of all linear transformations from $G$ to $H$.
Let $\map {\MM_R} {m, n}$ be the $m \times n$ matrix space over $R$.
Let $\sqbrk {u; \sequence {c_m}, \sequence {b_n} }$ be the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$.
Let $M: \map {\LL_R} {G, H} \to \map {\MM_R} {m, n}$ be defined as:
:$\forall u \in \map {\LL_R} {G, H}: \map M u = \sqbrk {u; \sequence {c_m}, \sequence {b_n} }$
Then $M$ is a module isomorphism.
\end{theorem}
\begin{proof}
Let $u, v \in \map {\LL_R} {G, H}$ such that:
:$\map M u = \map M v$
We have that the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$ is defined as the $m \times n$ matrix $\sqbrk \alpha_{m n}$ where:
:$\ds \forall \tuple {i, j} \in \closedint 1 m \times \closedint 1 n: \map u {b_j} = \sum_{i \mathop = 1}^m \alpha_{i j} \circ c_i$
and it is seen that $\map M u$ and $\map M v$ are the same object.
That is:
:$\map M u = \map M v \implies u = v$
and $M$ is seen to be injective.
{{finish|Surjectivity needs proving, so does homomorphism}}
\end{proof}
|
21200
|
\section{Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings}
Tags: Linear Transformations
\begin{theorem}
Let $R$ be a commutative ring.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Then $\map {\LL_R} {G, H}$ is a submodule of the $R$-module $H^G$.
\end{theorem}
\begin{proof}
From Group equals Center iff Abelian, the center of a commutative ring $R$ is $R$ itself.
So, we need to show that every $\phi \in \map {\LL_R} {G, H}$ fulfils the conditions to be a linear transformation:
:$(1): \quad \forall x, y \in G: \map {\paren {\lambda \circ \phi} } {x +_G y} = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y$
:$(2): \quad \forall x \in G: \forall \mu \in R: \map {\paren {\lambda \circ \phi} } {\mu \circ x} = \mu \circ \map {\paren {\lambda \circ \phi} } x$
for all $\phi \in \map {\LL_R} {G, H}$.
From Linear Transformation from Center of Scalar Ring, these conditions are indeed fulfulled.
Hence the result.
{{qed}}
\end{proof}
|
21201
|
\section{Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings/Unitary}
Tags: Unitary Modules, Linear Transformations
\begin{theorem}
Let $R$ be a commutative ring with unity whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\struct {H, +_H, \circ}_R$ be a unitary module.
Then $\map {\LL_R} {G, H}$ is also a unitary module.
\end{theorem}
\begin{proof}
From Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings, $\map {\LL_R} {G, H}$ is a module.
It remains to be shown that $\map {\LL_R} {G, H}$ is a unitary module, that is:
:$\forall \phi \in \map {\LL_R} {G, H}: 1_R \circ \phi = \phi$
So, let $\struct {H, +_H, \circ}_R$ be a unitary $R$-module.
Then:
:$\forall x \in H: 1_R \circ x = x$
Thus:
{{begin-eqn}}
{{eqn | l = \map {\paren {1_R \circ \phi} } x
| r = 1_R \circ \paren {\map \phi x}
| c =
}}
{{eqn | r = \map \phi x
| c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
21202
|
\section{Set of Linear Transformations under Pointwise Addition forms Abelian Group}
Tags: Abelian Groups, Linear Transformations
\begin{theorem}
Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be abelian groups.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\oplus_H$ be defined as pointwise addition on $\map {\LL_R} {G, H}$:
:$\forall u, v \in \map {\LL_R} {G, H}: \forall x \in G: \map {\paren {u \oplus_H v} } x := \map u x +_H \map v x$
Then $\struct {\map {\LL_R} {G, H}, \oplus_H}$ is an abelian group.
\end{theorem}
\begin{proof}
From Structure Induced by Group Operation is Group, $\struct {H^G, \oplus_H}$ is a group
Let $\phi, \psi \in \map {\LL_R} {G, H}$.
From Addition of Linear Transformations:
:$\phi \oplus_H \psi \in \map {\LL_R} {G, H}$
From Negative Linear Transformation:
:$-\phi \in \map {\LL_R} {G, H}$
Thus, from the Two-Step Subgroup Test:
:$\struct {\map {\LL_R} {G, H}, \oplus_H}$ is a subgroup of $\struct {H^G, \oplus_H}$.
It remains to be shown that $\struct {\map {\LL_R} {G, H}, \oplus_H}$ is abelian.
Let $u$ and $v$ be arbitrary elements of $\map {\LL_R} {G, H}$.
Indeed, we have that:
{{begin-eqn}}
{{eqn | q =
| l = \map {\paren {u \oplus_H v} } x
| r = \map u x +_H \map v x
| c = {{Defof|Pointwise Addition of Linear Transformations}}
}}
{{eqn | r = \map v x +_H \map u x
| c = as $H$ is abelian
}}
{{eqn | r = \map {\paren {v \oplus_H u} } x
| c = {{Defof|Pointwise Addition of Linear Transformations}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21203
|
\section{Set of Liouville Numbers is Uncountable}
Tags: Transcendental Numbers
\begin{theorem}
The set of Liouville numbers is uncountable.
\end{theorem}
\begin{proof}
By Corollary to Liouville's Constant is Transcendental, all numbers of the form:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop \ge 1} \frac {a_n} {10^{n!} }
| r = \frac {a_1} {10^1} + \frac {a_2} {10^2} + \frac {a_3} {10^6} + \frac {a_4} {10^{24} } + \cdots
| c =
}}
{{end-eqn}}
where
:$a_1, a_2, a_3, \ldots \in \set {1, 2, \ldots, 9}$
are Liouville numbers.
Therefore each sequence in $\set {1, 2, \ldots, 9}$ defines a unique Liouville number.
By Set of Infinite Sequences is Uncountable, there are uncountable sequences in $\set {1, 2, \ldots, 9}$.
As the set of Liouville numbers has an uncountable subset, it is also uncountable by Sufficient Conditions for Uncountability.
{{qed}}
Category:Transcendental Numbers
\end{proof}
|
21204
|
\section{Set of Local Minimum is Countable}
Tags: Countable Sets, Real Analysis
\begin{theorem}
Let $X$ be a subset of $\R$.
The set:
:$\leftset {x \in X: x}$ is local minimum in $\rightset X$
is countable.
\end{theorem}
\begin{proof}
Define:
:$Y := \leftset {x \in X: x}$ is local minimum in $\rightset X$
By definition of $Y$ and definition of local minimum in set:
:$\forall x \in Y: \exists y \in \R: y < x \land \openint y x \cap X = \O$
By the Axiom of Choice, define a mapping $f: Y \to \powerset \R$ as:
:$\forall x \in Y: \exists y \in \R: \map f x = \openint y x \land y < x \land \map f x \cap X = \O$
We will prove that $f$ is an injection by definition:
Let $x_1, x_2 \in Y$ such that
:$\map f {x_1} = \map f {x_2}$
By definition of $f$:
:$\exists y_1 \in \R: \map f {x_1} = \openint {y_1} {x_1} \land y_1 < x_1 \land \map f {x_1} \cap X = \O$
and:
:$\exists y_2 \in \R: \map f {x_2} = \openint {y_2} {x_2} \land y_2 < x_2 \land \map f {x_2} \cap X = \O$
Then:
:$\openint {y_1} {x_1} = \openint {y_2} {x_2}$
Thus $x_1 = x_2$.
This ends the proof of injection.
By Cardinality of Image of Injection:
:$(1): \quad \card Y = \card {\map {f^\to} Y} = \card {\Img f}$
where
:$\card Y$ denotes the cardinality of $Y$,
:$\map {f^\to} Y$ denotes the image of $Y$ under $f$,
:$\Img f$ denotes the image of $f$.
We will prove that $\Img f$ is pairwise disjoint by definition.
Let $A, B \in \Img f$ such that
:$A \ne B$.
Then by definition of image:
:$\exists x_1 \in Y: \map f {x_1} = A$
and
:$\exists x_2 \in Y: \map f {x_2} = B$.
By difference of $A$ and $B$:
:$x_1 \ne x_2$
By definition of $f$:
:$\exists y_1 \in \R: \map f {x_1} = \openint {y_1} {x_1} \land y_1 < x_1 \land \map f {x_1} \cap X = \O$
and:
:$\exists y_2 \in \R: \map f {x_2} = \openint {y_2} {x_2} \land y_2 < x_2 \land \map f {x_2} \cap X = \O$
Aiming at contradiction suppose that
:$A \cap B \ne \O$.
$x_1 < x_2$ or $x_1 > x_2$.
In case when $x_1 < x_2$, $x_1 \in \map f {x_2}$ what contradicts with $\map f {x_2} \cap X = \O$.
In case when $x_1 > x_2$, analogically.
This ends the proof that $\Img f$ is pairwise disjoint.
By Set of Pairwise Disjoint Intervals is Countable:
:$\Img f$ is countable.
Thus by $(1)$ and Set is Countable if Cardinality equals Cardinality of Countable Set the result:
:$Y$ is countable.
{{qed}}
\end{proof}
|
21205
|
\section{Set of Mappings can be Ordered by Inclusion}
Tags: Order Theory
\begin{theorem}
Let $S \times T$ be the product of two sets.
Let $\FF$ be a set of mappings on $S \times T$.
Then $\FF$ can be ordered by inclusion.
\end{theorem}
\begin{proof}
By the definition of mapping, a mapping is a specific type of relation.
The result then follows from Set of Relations can be Ordered by Inclusion.
{{qed}}
Category:Order Theory
\end{proof}
|
21206
|
\section{Set of Mappings which map to Same Element induces Equivalence Relation}
Tags: Mapping Theory, Equivalence Relations
\begin{theorem}
Let $X$ and $Y$ be sets.
Let $E$ be the set of all mappings from $X$ to $Y$.
Let $b \in X$.
Let $\RR \subseteq E \times E$ be the relation on $E$ defined as:
:$\RR := \set {\tuple {f, g} \in \RR: \map f b = \map g b}$
Then $\RR$ is an equivalence relation.
\end{theorem}
\begin{proof}
Checking in turn each of the criteria for equivalence:
\end{proof}
|
21207
|
\section{Set of Meet Irreducible Elements Excluded Top is Order Generating}
Tags: Order Generating, Continuous Lattices, Meet Irreducible
\begin{theorem}
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a continuous complete lattice.
Let $X = \mathit{IRR}\left({L}\right) \setminus \left\{ {\top}\right\}$
where $\mathit{IRR}\left({L}\right)$ denotes the set of all meet irreducible element of $S$,
:$\top$ denotes the top of $L$.
Then $X$ is order generating.
\end{theorem}
\begin{proof}
We will prove that
:$\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p }\right)$
Let $x, y \in S$ such that
:$y \npreceq x$
By Not Preceding implies There Exists Meet Irreducible Element Not Preceding
:$\exists p \in S: p$ is meet irreducible and $x \preceq p$ and $y \npreceq p$
By definition of greatest element:
:$p \ne \top$ and $p \in \mathit{IRR}\left({L}\right)$
By definitions of difference and singleton:
:$p \in X$
Thus
:$\exists p \in X: x \preceq p \land y \npreceq p$
{{qed|lemma}}
Hence by Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding:
:$X$ is order generating.
{{qed}}
\end{proof}
|
21208
|
\section{Set of Monomials is Closed Under Multiplication}
Tags: Polynomial_Theory, Polynomial Theory, Monomials
\begin{theorem}
Let $M$ be the set of all monomials on the set $\set {X_j: j \in J}$, with multiplication $\circ$ defined by:
:$\ds \paren {\prod_{j \mathop \in J} X_j^{k_j} } \circ \paren {\prod_{j \mathop \in J} X_j^{k_j'} } = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$
Then $M$ is closed under $\circ$.
\end{theorem}
\begin{proof}
Let $\ds m_1 = \prod_{j \mathop \in J} X_j^{k_j}, m_2 = \prod_{j \mathop \in J} X_j^{k_j'}$ be two monomials.
Their product is:
:$\ds m_1 \circ m_2 = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$
If $k_j + k_j' \ne 0$ then either $k_j \ne 0$ or $k_j' \ne 0$ (or both are nonzero).
Therefore if $k_j + k_j' \ne 0$ for infinitely many $j$, then either $m_1$ or $m_2$ is not a monomial.
{{qed}}
Category:Monomials
\end{proof}
|
21209
|
\section{Set of Natural Numbers Equals Union of its Successor}
Tags: Natural Numbers
\begin{theorem}
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction on a Zermelo universe $V$.
Then:
:$\bigcup \omega^+ = \omega$
\end{theorem}
\begin{proof}
We have that:
:$\omega^+ = \omega \cup \set \omega$
and so:
:$\omega \subseteq \omega^+$
{{qed|lemma}}
By definition:
:$\bigcup \omega^+ = \set {x: \exists y \in \omega^+: x \in y}$
Thus:
:$x \in \bigcup \omega^+ \implies x \in \omega$
{{qed|lemma}}
So by definition of set equality:
:$\bigcup \omega^+ = \omega$
{{qed}}
\end{proof}
|
21210
|
\section{Set of Natural Numbers Equals its Union}
Tags: Natural Numbers
\begin{theorem}
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction on a Zermelo universe $V$.
Then:
:$\bigcup \omega = \omega$
\end{theorem}
\begin{proof}
By hypothesis, $\omega \in V$, where $V$ is a Zermelo universe.
By the Axiom of Infinity we have that $\omega$ is a set.
By the Axiom of Transitivity, $\omega$ is transitive.
Hence:
:$\bigcup \omega \subseteq \omega$
{{qed|lemma}}
Let $n \in \omega$.
Then by definition of the von Neumann construction:
:$n + 1 = n \cup \set n$
That is:
:$\exists X \in \omega: n \in X$
where in this case $X = n + 1$.
Thus we have that:
:$n \in \set {x: \exists X \in \omega: x \in X}$
and so by definition of union of class:
:$n \in \bigcup \omega$
Thus we have that:
:$\omega \subseteq \bigcup \omega$
which, together with:
:$\bigcup \omega \subseteq \omega$
gives us, by set equality:
:$\bigcup \omega = \omega$
{{qed}}
\end{proof}
|
21211
|
\section{Set of Natural Numbers is Primitive Recursive}
Tags: Natural Numbers, Primitive Recursive Functions
\begin{theorem}
The set of natural numbers $\N$ is primitive recursive.
\end{theorem}
\begin{proof}
The characteristic function $\chi_\N: \N \to \N$ is defined as:
:$\forall n \in \N: \chi_\N \left({n}\right) = 1$.
So:
: $\chi_\N \left({n}\right) = f^1_1 \left({n}\right)$
The constant function $f^1_1$ is primitive recursive.
Hence the result.
{{qed}}
Category:Primitive Recursive Functions
Category:Natural Numbers
\end{proof}
|
21212
|
\section{Set of Natural Numbers is either Finite or Denumerable}
Tags: Natural Numbers
\begin{theorem}
Let $S$ be a subset of the natural numbers $\N$.
Then $S$ is either finite or denumerable.
\end{theorem}
\begin{proof}
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.
By the Well-Ordering Principle, $\omega$ is well-ordered by the $\le$ relation.
Thus from the Well-Ordering Principle, $S$ has a smallest element.
Let this smallest element of $S$ be denoted $s_0$.
Also from the Well-Ordering Principle, every subset of $S$ also has a smallest element.
{{ProofWanted|establish a sequence where $s_k$ is the smallest element of $S_k$ where $S_k {{=}} S_{k - 1} \setminus \set {s_{k - 1} }$, and then build a mapping $f$ such that $\map f {s_k} \to \set {s_0, s_1, \ldots, s_{k - 1} }$ and show it's progressing, or something, but it's tedious.}}
\end{proof}
|
21213
|
\section{Set of Non-Negative Real Numbers is not Well-Ordered by Usual Ordering}
Tags: Real Numbers, Well-Orderings
\begin{theorem}
The set of non-negative real numbers $\R_{\ge 0}$ is not well-ordered under the usual ordering $\le$.
\end{theorem}
\begin{proof}
{{AimForCont}} $\R_{\ge 0}$ is a well-ordered set.
Then by definition, all subsets of $\R_{\ge 0}$ has a smallest element.
Take the subset $\R_{> 0}$:
:$\R_{> 0} = \left\{ {x \in \R_{\ge 0}: x > 0}\right\} = \R_{\ge 0} \setminus \left\{ {0}\right\}$
Suppose $x \in \R_{> 0}$ is a smallest element.
Then $\dfrac x 2 \in \R_{> 0}$.
But $\dfrac x 2 < x$, which contradicts the supposition that $x \in \R_{> 0}$ is a smallest element.
Hence there can be no such smallest element.
So by Proof by Contradiction, $\R_{\ge 0}$ is not well-ordered by $\le$.
{{qed}}
\end{proof}
|
21214
|
\section{Set of Non-Zero Natural Numbers is Primitive Recursive}
Tags: Primitive Recursive Functions
\begin{theorem}
Let $\N^*$ be defined as $\N^* = \N \setminus \set 0$.
The subset $\N^* \subset \N$ is primitive recursive.
\end{theorem}
\begin{proof}
We have that the characteristic function $\chi_{\set 0}$ of $\set 0$ is primitive recursive.
We note that:
:If $n = 0$ then $\map {\chi_{\set 0} } n = 1$ therefore $\map {\chi_{\set 0} } {\map {\chi_{\set 0} } n} = 0$.
:If $n > 0$ then $\map {\chi_{\set 0} } n = 0$ therefore $\map {\chi_{\set 0} } {\map {\chi_{\set 0} } n} = 1$.
Thus $\map {\chi_{\set 0} } {\map {\chi_{\set 0} } n} = \map {\chi_{\N^*} } n$.
So $\chi_{\N^*}$ is obtained by substitution from the primitive recursive function $\chi_{\set 0}$.
Hence the result.
{{qed}}
Category:Primitive Recursive Functions
\end{proof}
|
21215
|
\section{Set of Normal Subgroups of Group is Subsemigroup of Power Set Semigroup}
Tags: Normal Subgroups, Semigroups, Subset Products
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $\circ_\PP$ be the operation induced by $\circ$ on $\powerset G$, the power set of $G$.
Let $\HH$ be the set of all normal subgroups of $\struct {G, \circ}$.
Then the algebraic structure $\struct {\HH, \circ_\PP}$ is a subsemigroup of the algebraic structure $\struct {\powerset G, \circ_\PP}$.
\end{theorem}
\begin{proof}
From Power Set of Group under Induced Operation is Semigroup, we have that $\struct {\powerset G, \circ_\PP}$ is a semigroup.
Note that $\HH \subseteq \powerset G$.
From Subset Product of Normal Subgroups is Normal, $\struct {\HH, \circ_\PP}$ is closed.
By Subsemigroup Closure Test, $\struct {\HH, \circ_\PP}$ is a subsemigroup of $\struct {\powerset G, \circ_\PP}$.
{{qed}}
\end{proof}
|
21216
|
\section{Set of Normal Subgroups of Group is Subsemigroup of Power Set under Intersection}
Tags: Semigroups, Power Set, Power Sets, Set Intersection, Normal Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $\HH$ be the set of all normal subgroups of $\struct {G, \circ}$.
Then the algebraic structure $\struct {\HH, \cap}$ is a subsemigroup of the algebraic structure $\struct {\powerset G, \cap}$.
\end{theorem}
\begin{proof}
From Power Set with Intersection is Commutative Monoid, we have that $\struct {\powerset G, \cap}$ is ''a fortiori'' a semigroup.
Note that $\HH \subseteq \powerset G$.
Let $H_1$ and $H_2$ be normal subgroups of $\struct {G, \circ}$.
From Intersection of Normal Subgroups is Normal, $H_1 \cap H_2$ is also a normal subgroup of $\struct {G, \circ}$.
Hence $\struct {\HH, \cap}$ is closed.
By Subsemigroup Closure Test, $\struct {\HH, \cap}$ is a subsemigroup of $\struct {\powerset G, \cap}$.
{{qed}}
\end{proof}
|
21217
|
\section{Set of Numbers of form n - 1 over n is Bounded Above}
Tags: Suprema
\begin{theorem}
Let $S$ be the subset of the set of real numbers $\R$ defined as:
:$S = \set {\dfrac {n - 1} n: n \in \Z_{>0} }$
$S$ is bounded above with supremum $1$.
$S$ has no greatest element.
\end{theorem}
\begin{proof}
We have that:
:$\dfrac {n - 1} n = 1 - \dfrac 1 n$
As $n > 0$ it follows from Reciprocal of Strictly Positive Real Number is Strictly Positive that $\dfrac 1 n > 0$.
Thus $1 - \dfrac 1 n < 1$ and so $S$ is bounded above by $1$.
Next it is to be shown that $1$ is the supremum of $S$.
Suppose $x$ is the supremum of $S$ such that $x < 1$.
Then:
:$1 - x = \epsilon$
where $\epsilon \in \R_{>0}$.
By the Archimedean Principle we have that:
:$\exists m \in \Z_{>0}: n > \dfrac 1 \epsilon$
and so from Reciprocal Function is Strictly Decreasing:
:$\exists m \in \Z_{>0}: \dfrac 1 n < \epsilon$
Thus:
:$1 - \dfrac 1 m > 1 - \epsilon = x$
and so $x$ is not the supremum of $S$ after all.
Thus $\sup S$ cannot be less than $1$.
It follows that:
:$\sup S = 1$
Next it is noted that:
:$\forall n \in \Z_{>0}: \dfrac 1 n > 0$
and so:
:$\forall x \in S: x < 1$
Thus as $\sup S \notin S$ it follows from Greatest Element is Supremum that $S$ has no greatest element.
{{qed}}
\end{proof}
|
21218
|
\section{Set of Odd Integers is Countably Infinite}
Tags: Countable Sets, Odd Integers
\begin{theorem}
Let $\Bbb O$ be the set of odd integers.
Then $\Bbb O$ is countably infinite.
\end{theorem}
\begin{proof}
Let $f: \Bbb O \to \Z$ be the mapping defined as:
:$\forall x \in \Bbb O: \map f x = \dfrac {x + 1} 2$
$f$ is well-defined as $x + 1$ is even and so $\dfrac {x + 1} 2 \in \Z$.
Let $x, y \in \Bbb O$ such that $\map f x = \map f y$.
Then:
{{begin-eqn}}
{{eqn | l = \map f x
| r = \map f y
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac {x + 1} 2
| r = \dfrac {y + 1} 2
| c = Definition of $f$
}}
{{eqn | ll= \leadsto
| l = x + 1
| r = y + 1
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = y
| c =
}}
{{end-eqn}}
Thus $f$ is injective by definition.
Consider the inverse $f^{-1}$.
By inspection:
:$\forall x \in \Z: \map {f^{-1} } x = 2 x - 1$
$f^{-1}$ is well-defined, and $2 x - 1$ is odd.
Thus $f^{-1}$ is a mapping from $\Z$ to $\Bbb O$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f^{-1} } x
| r = \map {f^{-1} } y
| c =
}}
{{eqn | ll= \leadsto
| l = 2 x - 1
| r = 2 y - 1
| c = Definition of $f^{-1}$
}}
{{eqn | ll= \leadsto
| l = 2 x
| r = 2 y
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = y
| c =
}}
{{end-eqn}}
Thus $f^{-1}$ is injective by definition.
It follows by the Cantor-Bernstein-Schröder Theorem that there exists a bijection between $\Z$ and $\Bbb O$.
{{qed}}
\end{proof}
|
21219
|
\section{Set of Orbits forms Partition}
Tags: Group Actions
\begin{theorem}
Let $G$ be a group.
Let $X$ be a set.
Let $G$ act on $X$.
Then the set of orbits of the group action forms a partition of $X$.
\end{theorem}
\begin{proof}
Follows from the Fundamental Theorem on Equivalence Relations.
{{qed}}
\end{proof}
|
21220
|
\section{Set of Order 3 Vectors under Cross Product does not form Ring}
Tags: Examples of Rings, Vectors
\begin{theorem}
Let $S$ be the set of all vectors in a vector space of dimension $3$.
Let $\times$ denote the cross product operation.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.
\end{theorem}
\begin{proof}
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\times$ is associative on $S$.
However, from Vector Cross Product is not Associative, this is not the case here.
The result follows.
{{qed}}
\end{proof}
|
21221
|
\section{Set of Order m times n Matrices does not form Ring}
Tags: Matrices, Examples of Rings
\begin{theorem}
Let $m, n \in \N_{>0}$ be non-zero natural numbers such that $m > n$.
Let $S$ be the set of all matrices of order $m \times n$.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.
Note that $\times$ denotes conventional matrix multiplication.
\end{theorem}
\begin{proof}
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed.
That is:
:$\forall x, y \in S: x \times y \in S$
Let $\mathbf A$ and $\mathbf B$ be elements of $S$.
The matrix multiplication operation is defined on $\mathbf A$ and $\mathbf B$ {{iff}} $\mathbf A$ is of order $m \times p$ and $\mathbf A$ is of order $p \times n$, for some $m, n, p \in \N_{>0}$.
That is, the second dimension of $\mathbf A$ must be equal to the first dimension of $\mathbf B$.
But in this case, the second dimension of $\mathbf A$ is $n$, and the first dimension of $\mathbf B$ is $m$.
As we have been given that $m \ne n$, it follows that matrix multiplication operation is not defined on $S$.
Hence the result.
{{qed}}
\end{proof}
|
21222
|
\section{Set of Pairwise Disjoint Intervals is Countable}
Tags: Countable Sets
\begin{theorem}
Let $X$ be a subset of $\powerset \R$ such that:
:$(1): \quad X$ is pairwise disjoint:
::::$\forall A, B \in X: A \ne B \implies A \cap B = \O$.
:$(2): \quad$ every element of $X$ contains an open interval:
::::$\forall A \in X: \exists x, y \in \R: x < y \land \openint x y \subseteq A$.
Then $X$ is countable.
\end{theorem}
\begin{proof}
By Between two Real Numbers exists Rational Number:
:$\forall A \in X: \exists x, y \in \R, q \in \Q: x < y \land q \in \openint x y \subseteq A$
By the Axiom of Choice define a mapping $f: X \to \Q$:
:$\forall A \in X: \map f A \in A$
First it needs to be shown that $f$ is an injection by definition.
Let $A, B \in X$ such that:
:$\map f A = \map f B$
By definition of $f$:
:$\map f A \in A$ and $\map f B \in B$
By definition of intersection:
:$\map f A \in A \cap B$
Then by definition of empty set:
:$A \cap B \ne \O$
Thus by definition of pairwise disjoint:
:$A = B$
Hence $f$ is an injection.
By Set is Subset of Itself, $X$ is a subset of $X$.
Thus by Cardinality of Image of Injection:
:$\card X = \card {f^\to \sqbrk X}$
By definition of image:
:$f^\to \sqbrk X \subseteq \Q$
By Rational Numbers are Countably Infinite:
:$\Q$ is countable.
Hence by Subset of Countable Set is Countable:
:$f^\to \sqbrk X$ is countable.
Thus by Set is Countable if Cardinality equals Cardinality of Countable Set the result:
:$X$ is countable.
{{qed}}
\end{proof}
|
21223
|
\section{Set of Points at which Sequence of Measurable Functions does not Converge to Given Measurable Function is Measurable}
Tags: Measurable Functions, Convergence
\begin{theorem}
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \R$ be a $\Sigma$-measurable function.
For each $n \in \N$, let $f_n : X \to \R$ be a $\Sigma$-measurable function.
Then:
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x}$ is $\Sigma$-measurable.
\end{theorem}
\begin{proof}
From Expression for Set of Points at which Sequence of Functions does not Converge to Given Function, we have:
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x} = \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$
From Pointwise Difference of Measurable Functions is Measurable, we have:
:$f_n - f$ is $\Sigma$-measurable for each $n \in \N$.
From Absolute Value of Measurable Function is Measurable, we have:
:$\size {f_n - f}$ is $\Sigma$-measurable for each $n \in \N$.
From Characterization of Measurable Functions, we have that:
:$\ds \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k, n \in \N$.
Since $\sigma$-Algebras are closed under countable union, we have:
:$\ds \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k, N \in \N$.
Then, from Sigma-Algebra Closed under Countable Intersection:
:$\ds \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k \in \N$.
Again, since Since $\sigma$-Algebras are closed under countable union, we have:
:$\ds \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable.
So:
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x}$ is $\Sigma$-measurable.
{{qed}}
Category:Measurable Functions
Category:Convergence
\end{proof}
|
21224
|
\section{Set of Points for which Measurable Function is Real-Valued is Measurable}
Tags: Measurable Sets, Measurable Functions, Set of Points for which Measurable Function is Real-Valued is Measurable
\begin{theorem}
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable.
Then:
:$\set {x \in X : \map f x \in \R}$ is $\Sigma$-measurable.
\end{theorem}
\begin{proof}
Since $f$ is $\Sigma$-measurable, we have that:
:for all $n \in \N$ the set $\set {x \in X : \map f x \le n}$ is $\Sigma$-measurable
and:
:for all $n \in \N$ the set $\set {x \in X : -n \le \map f x}$ is $\Sigma$-measurable.
From $\sigma$-Algebra Closed under Countable Intersection, we have:
:$\set {x \in X : -n \le \map f x \le n} = \set {x \in X : \map f x \le n} \cap \set {x \in X : -n \le \map f x}$ is $\Sigma$-measurable.
Since $\sigma$-algebras are closed under countable union, we also have:
:$\ds \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$ is $\Sigma$-measurable.
We will finally show that:
:$\ds \set {x \in X : \map f x \in \R} = \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$
which gives the claim.
If $x \in X$ has $\map f x \in \R$, then we have:
:$-\paren {\floor {\size {\map f x} } + 1} \le \map f x \le \floor {\size {\map f x} } + 1$
with:
:$\floor {\size {\map f x} } + 1 \in \N$
So:
:$\ds x \in \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$
Now, let:
:$\ds x \in \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$
Then:
:$-n \le \map f x \le n$
for some $n \in \N$, so certainly:
:$\map f x \in \R$
So we have:
:$\ds \set {x \in X : \map f x \in \R} = \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$
and hence the claim.
{{qed}}
Category:Measurable Sets
Category:Measurable Functions
Category:Set of Points for which Measurable Function is Real-Valued is Measurable
\end{proof}
|
21225
|
\section{Set of Points for which Measurable Function is Real-Valued is Measurable/Corollary}
Tags: Set of Points for which Measurable Function is Real-Valued is Measurable
\begin{theorem}
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable.
Then:
:$\set {x \in X : \size {\map f x} = +\infty}$ is $\Sigma$-measurable.
\end{theorem}
\begin{proof}
We have:
:$\set {x \in X : \size {\map f x} = +\infty} = X \setminus \set {x \in X : \map f x \in \R}$
From Set of Points for which Measurable Function is Real-Valued is Measurable, we have:
:$\set {x \in X : \map f x \in \R}$ is $\Sigma$-measurable.
Since $\sigma$-algebras are closed under complementation, we have that:
:$\set {x \in X : \size {\map f x} = +\infty}$ is $\Sigma$-measurable.
{{qed}}
Category:Set of Points for which Measurable Function is Real-Valued is Measurable
\end{proof}
|
21226
|
\section{Set of Points on Line Segment is Infinite}
Tags: Infinite Sets, Lines
\begin{theorem}
The set of points on a line segment is infinite.
\end{theorem}
\begin{proof}
Let $S$ denote the set of points on a line segment.
{{AimForCont}} $S$ is finite.
Then there exists $n \in \N$ such that $S$ has $n$ elements.
Let $s_1$ and $s_2$ be two arbitrary adjacent points in $S$.
That is, such that there are no points in $S$ between $s_1$ and $s_2$.
But there exists (at least) one point on the line segment between $s_1$ and $s_2$ which is not in $S$.
Hence there must be more than $n$ elements of $S$.
From that contradiction it follows by Proof by Contradiction that $S$ is not finite.
{{Qed}}
\end{proof}
|
21227
|
\section{Set of Polynomials over Infinite Set has Same Cardinality}
Tags: Polynomial Theory
\begin{theorem}
Let $S$ be a set of infinite cardinality $\kappa$.
Let $S \sqbrk x$ be the set of polynomial forms over $S$ in the indeterminate $x$.
Then $S \sqbrk x$ has cardinality $\kappa$.
\end{theorem}
\begin{proof}
Since $S \sqbrk x$ contains a copy of $S$ as constant polynomials, we have an injection $S \to S \sqbrk x$.
We define an injection from $S \sqbrk x$ to the set $\FF$ of finite sequences over $S$ as follows:
Each polynomial in $f \in S \sqbrk x$ is of the form:
:$f = a_0 + a_1 x + a_2 x^2 + \dotsb + a_n x^n$
where $a_n$ is non-zero and each $a_i$ is in $S$.
We send each polynomial $f$ to the sequence of its coefficients $\sequence {a_0, \dotsc, a_n}$.
By the definition of equality of polynomials, this is injective.
Now the set of finite sequences over $S$ is a countable union of sets of cardinality $\kappa$.
From Cardinality of Infinite Union of Infinite Sets, $\FF$ has cardinality $\kappa$.
Therefore there is a bijection $\FF \leftrightarrow S$.
Composing this with the injection $S \sqbrk x \to \FF$, we have an injection $S \sqbrk x \to S$.
So by the Cantor-Bernstein-Schröder Theorem there is a bijection $S \sqbrk x \leftrightarrow S$.
Hence, we have:
:$\card {S \sqbrk x} = \card S = \kappa$
{{qed}}
Category:Polynomial Theory
\end{proof}
|
21228
|
\section{Set of Polynomials over Integral Domain is Subring}
Tags: Subrings, Polynomials, Polynomial Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is a subring of $R$.
\end{theorem}
\begin{proof}
By application of the Subring Test:
As $D$ is an integral domain, it has a unity $1_D$ and so $x = 1_D x$.
Hence $x \in D \sqbrk x$ and so $D \sqbrk x \ne \O$.
Let $p, q \in D \sqbrk x$.
Then let:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k, q = \sum_{k \mathop = 0}^n b_k \circ x^k$
Thus:
:$\ds -q = -\sum_{k \mathop = 0}^n b_k \circ x^k = \sum_{k \mathop = 0}^n \paren {-b_k} \circ x^k$
and so:
:$q \in D \sqbrk x$
Thus as Polynomials Closed under Addition, it follows that:
:$p + \paren {-q} \in D \sqbrk x$
Finally, from Polynomials Closed under Ring Product, we have that $p \circ q \in D \sqbrk x$.
All the criteria of the Subring Test are satisfied.
Hence the result.
{{qed}}
\end{proof}
|
21229
|
\section{Set of Positive Integers does not form Ring}
Tags: Integers, Examples of Rings
\begin{theorem}
Let $\Z_{\ge 0}$ denote the set of positive integers.
Then the algebraic structure $\struct {\Z_{\ge 0}, +, \times}$ does not form a ring.
\end{theorem}
\begin{proof}
For $\struct {\Z_{\ge 0}, +, \times}$ to be a ring, it is necessary for the algebraic structure $\struct {\Z_{\ge 0}, +}$ to form a group.
But from the corollary to Natural Numbers under Addition do not form Group:
:$\struct {\Z_{\ge 0}, +}$ is not a group.
{{qed}}
\end{proof}
|
21230
|
\section{Set of Prime Numbers is Primitive Recursive}
Tags: Primitive Recursive Functions
\begin{theorem}
The set $\Bbb P$ of prime numbers is primitive recursive.
\end{theorem}
\begin{proof}
A prime number is defined as an element of $\N$ with '''exactly two''' positive divisors.
So, we have that $n > 0$ is prime {{iff}} $\map \tau n = 2$, where $\tau: \N \to \N$ is the divisor counting function.
Thus we can define the characteristic function of the set of prime numbers $\Bbb P$ as:
:$\forall n > 0: \map {\chi_\Bbb P} n := \map {\chi_{\operatorname {eq} } } {\map \tau n, 2}$
Now we let $g: \N^2 \to \N$ be the function given by:
:$\map g {n, z} = \begin{cases}
0 & : z = 0 \\
\ds \sum_{y \mathop = 1}^z \map {\operatorname {div} } {n, y} & : z > 0
\end{cases}$
As:
:$\operatorname {div}$ is primitive recursive
:$\ds \sum_{y \mathop = 1}^z$ is primitive recursive
it follows that $g$ is primitive recursive.
Then for $n > 0$:
:$\ds \map g {n, n} = \sum_{y \mathop = 1}^n \map {\operatorname {div} } {n, y} = \map \tau n$
and from Divisor Counting Function is Primitive Recursive we have that $g$ is primitive recursive.
Then let $h: \N \to \N$ be the function defined as:
:$\map h n = \map g {n, n}$
which is also primitive recursive.
So we have, for all $n \in \N$:
:$\map {\chi_\Bbb P} n = \map {\chi_{\operatorname {eq} } } {\map h n, 2}$
Hence the result.
{{qed}}
Category:Primitive Recursive Functions
\end{proof}
|
21231
|
\section{Set of Rational Cuts forms Ordered Field}
Tags: Totally Ordered Fields, Cuts
\begin{theorem}
Let $\RR$ denote the set of rational cuts.
Let $\struct {\RR, +, \times, \le}$ denote the ordered structure formed from $\RR$ and:
:the operation $+$ of addition of cuts
:the operation $\times$ of multiplication of cuts
:the ordering $\le$ of cuts.
Then $\struct {\RR, + \times, \le}$ is an ordered field.
\end{theorem}
\begin{proof}
We demonstrate that $\struct {\RR, +, \times}$ is a field by showing it is a subfield of the structure $\struct {\CC, +, \times}$, where $\CC$ denotes the set of all cuts.
We do this by establishing that all $4$ criteria of the Subfield Test are satisfied.
We note that $0^* \in \RR$, where $0^*$ is the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
So $\RR \ne \O$.
Thus criterion $(1)$ is satisfied.
{{qed|lemma}}
{{finish}}
{{qed}}
\end{proof}
|
21232
|
\section{Set of Rational Numbers Strictly between Zero and One has no Greatest or Least Element}
Tags: Rational Numbers, Infima, Suprema
\begin{theorem}
Let $S \subseteq \Q$ be the subset of the set of rational numbers defined as:
:$S = \set {r \in \Q: 0 < r < 1}$
Then $S$ has no greatest or smallest element.
However, $S$ has a supremum $1$ and an infimum $0$.
\end{theorem}
\begin{proof}
We have that:
:$\forall r \in S: 0 < r$
and:
:$\forall r \in S: r < 1$
Hence $0$ and $1$ are lower and upper bounds of $S$ respectively.
Let $s \in S$.
Then $s \in \Q: 0 < s < 1$.
{{AimForCont}} $s$ is the greatest element of $S$.
But then we have:
:$0 < s < \dfrac {s + 1} 2 < 1$
and so $\dfrac {s + 1} 2 \in S$ but $s < \dfrac {s + 1} 2$.
This contradicts our assertion that $s$ is the greatest element of $S$.
{{AimForCont}} $s$ is the smallest element of $S$.
But then we have:
:$0 < \dfrac s 2 < s < 1$
and so $\dfrac s 2 \in S$ but $\dfrac s 2 < s$.
This contradicts our assertion that $s$ is the smallest element of $S$.
So $S$ cannot have either a greatest or smallest element.
Let $H$ be such that $0 < H < 1$.
Suppose $H$ is a smaller upper bound of $S$ than $1$.
Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $H < r < 1$.
But then $r$ is a rational number between $0$ and $1$ which is greater than $H$.
Hence $H$ cannot be an upper bound of $S$.
Thus, by definition, $1$ is the supremum of $S$
Suppose $H$ is a greater lower bound of $S$ than $0$.
Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $0 < r < H$.
But then $r$ is a rational number between $0$ and $1$ which is smaller than $H$.
Hence $H$ cannot be a lower bound of $S$.
Thus, by definition, $0$ is the infimum of $S$.
{{Qed}}
\end{proof}
|
21233
|
\section{Set of Rational Numbers is not Closed in Reals}
Tags: Real Number Line with Euclidean Topology, Rational Numbers, Real Number Space
\begin{theorem}
Let $\Q$ be the set of rational numbers.
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then $\Q$ is not closed in $\R$.
\end{theorem}
\begin{proof}
Let $\alpha \in \R \setminus \Q$.
Let $I := \openint a b$ be an open interval in $\R$ such that $\alpha \in I$.
By Between two Real Numbers exists Rational Number:
:$\exists \beta \in \Q: \beta \in I$.
Thus $I$ contains elements of $\Q$ and so $\R \setminus \Q$ is not open in $\R$.
Thus by definition, $\Q$ is not closed in $\R$.
{{qed}}
\end{proof}
|
21234
|
\section{Set of Rational Numbers is not G-Delta Set in Reals}
Tags: G-Delta Sets, Real Number Line with Euclidean Topology, Rational Numbers, Real Number Space
\begin{theorem}
Let $\Q$ be the set of rational numbers.
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then $\Q$ is not a $G_\delta$ set in $\R$.
\end{theorem}
\begin{proof}
{{AimForCont}} $\Q$ is a $G_\delta$ set in $\R$.
Let $\Q = \ds \bigcap_{i \mathop \in \N} V_i$.
Since Rational Numbers are Countably Infinite, there exists an enumeration of $\Q$.
Write $\Q = \set {q_i}_{i \mathop \in \N}$.
Define $U_i = V_i \setminus \set {q_i}$.
We show that $U_i$ is dense in $\R$.
:Let $A \subseteq \R$ be an open set of $\struct {\R, \tau}$.
:From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau}$.
:So there exists $\openint a b \subseteq A$ for some $a < b$.
:One of $q_i \le a$, $q_i > a$ must hold.
:Suppose $q_i \le a$.
:By Between two Real Numbers exists Rational Number we have:
::$\exists r \in Q: a < r < b$.
:Since $r > a \ge q_i$:
::$r \in \Q \setminus \set {q_i} \subseteq V_i \setminus \set {q_i} = U_i$.
:Suppose $q_i > a$.
:By Between two Real Numbers exists Rational Number we have:
::$\exists r \in Q: a < r < \min \set {q_i, b}$.
:Since $r < q_i$:
::$r \in \Q \setminus \set {q_i} \subseteq V_i \setminus \set {q_i} = U_i$.
:In both cases we see that $r \in \openint a b \cap U_i \subseteq A \cap U_i$.
:Therefore $A \cap U_i \ne \O$.
:Thus $U_i$ is dense in $\R$.
Also we have:
{{begin-eqn}}
{{eqn | l = \bigcap_{i \mathop \in \N} U_i
| r = \bigcap_{i \mathop \in \N} \paren {V_i \setminus \set {q_i} }
}}
{{eqn | r = \paren {\bigcap_{i \mathop \in \N} V_i} \setminus \paren {\bigcup_{i \mathop \in \N} \set {q_i} }
| c = Difference with Union
}}
{{eqn | r = \Q \setminus \Q
}}
{{eqn | r = \O
| c = Set Difference with Self is Empty Set
}}
{{end-eqn}}
By Real Number Line is Complete Metric Space and Baire Category Theorem, $\struct {\R, \tau}$ is a Baire space.
By definition of Baire Space, $\ds \bigcap_{i \mathop \in \N} U_i$ is dense in $\R$.
But Empty Set is Nowhere Dense.
This is a contradiction.
Hence $\Q$ is not a $G_\delta$ set in $\R$.
{{qed}}
\end{proof}
|
21235
|
\section{Set of Rational Numbers whose Numerator Divisible by p is Closed under Addition}
Tags: Prime Numbers, Algebraic Closure, Rational Addition, Rational Numbers
\begin{theorem}
Let $p$ be a prime number.
Let $A_p$ be the set of all rational numbers which, when expressed in canonical form has a numerator which is divisible by $p$.
Then $A_p$ is closed under rational addition.
\end{theorem}
\begin{proof}
Let $a, b \in A_p$.
Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
:$n_1, n_2 \in \Z$
:$d_1, d_2 \in \Z_{>0}$
:$p n_1 \perp d_1, p n_2 \perp d_2$
Then:
{{begin-eqn}}
{{eqn | l = a + b
| r = \frac {p n_1} {d_1} + \frac {p n_2} {d_2}
| c =
}}
{{eqn | r = \frac {p n_1 d_2 + p n_2 d_1} {d_1 d_2}
| c = {{Defof|Rational Addition}}
}}
{{eqn | r = \frac {p \paren {n_1 d_2 + n_2 d_1} } {d_1 d_2}
| c =
}}
{{end-eqn}}
From Euclid's Lemma for Prime Divisors, if $p \divides d_1 d_2$ then either $p \divides d_1$ or $p \divides d_2$.
But neither of these is the case, so $p \nmid d_1 d_2$.
Hence by Prime not Divisor implies Coprime:
:$p \perp d_1 d_2$
where $\perp$ denotes coprimeness.
So when $\dfrac {p \paren {n_1 d_2 + n_2 d_1} } {d_1 d_2}$ is expressed in canonical form, $p$ will still be a divisor of the numerator.
Hence the result.
{{qed}}
\end{proof}
|
21236
|
\section{Set of Rational Numbers whose Numerator Divisible by p is Closed under Multiplication}
Tags: Prime Numbers, Algebraic Closure, Rational Multiplication, Rational Numbers
\begin{theorem}
Let $p$ be a prime number.
Let $A_p$ be the set of all rational numbers which, when expressed in canonical form has a numerator which is divisible by $p$.
Then $A_p$ is closed under rational multiplication.
\end{theorem}
\begin{proof}
Let $a, b \in A_p$.
Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
:$n_1, n_2 \in \Z$
:$d_1, d_2 \in \Z_{>0}$
:$p n_1 \perp d_1, p n_2 \perp d_2$
Then:
{{begin-eqn}}
{{eqn | l = a \times b
| r = \frac {p n_1} {d_1} \times \frac {p n_2} {d_2}
| c =
}}
{{eqn | r = \frac {p n_1 p n_2} {d_1 d_2}
| c = {{Defof|Rational Multiplication}}
}}
{{eqn | r = \frac {p^2 \paren {n_1 n_2} } {d_1 d_2}
| c =
}}
{{end-eqn}}
From Euclid's Lemma for Prime Divisors, if $p \divides d_1 d_2$ then either $p \divides d_1$ or $p \divides d_2$.
But neither of these is the case, so $p \nmid d_1 d_2$.
Hence by Prime not Divisor implies Coprime:
:$p \perp d_1 d_2$
where $\perp$ denotes coprimeness.
So when $\dfrac {p^2 \paren {n_1 n_2} } {d_1 d_2}$ is expressed in canonical form, $p$ will still be a divisor of the numerator.
Hence the result.
{{qed}}
\end{proof}
|
21237
|
\section{Set of Rationals Greater than Root 2 has no Smallest Element}
Tags: Rational Numbers
\begin{theorem}
Let $B$ be the set of all positive rational numbers $p$ such that $p^2 > 2$.
Then $B$ has no smallest element.
\end{theorem}
\begin{proof}
{{AimForCont}} $p \in B$ is the smallest element of $B$.
Then by definition of $B$:
:$p^2 > 2$
Let $q = p - \dfrac {p_2 - 2} {2 p}$.
Then $q > p$, and:
{{begin-eqn}}
{{eqn | l = q
| r = p - \dfrac {p_2 - 2} {2 p}
| c =
}}
{{eqn | r = \dfrac p 2 + \dfrac 1 p
| c =
}}
{{end-eqn}}
Hence:
:$0 < q < p$
and so:
{{begin-eqn}}
{{eqn | l = q^2
| r = p^2 - \dfrac {p^2 - 2} + \paren {\dfrac {p^2 - 2} {2 p} }^2
| c =
}}
{{eqn | o = >
| r = p^2 - \paren {p^2 - 2}
| c =
}}
{{eqn | r = 2
| c =
}}
{{end-eqn}}
That means $q \in B$.
This contradicts our assertion that $p$ is the smallest element of $B$.
It follows that $B$ can have no smallest element.
{{qed}}
\end{proof}
|
21238
|
\section{Set of Rationals Less than Root 2 has no Greatest Element}
Tags: Rational Numbers
\begin{theorem}
Let $A$ be the set of all positive rational numbers $p$ such that $p^2 < 2$.
Then $A$ has no greatest element.
\end{theorem}
\begin{proof}
{{AimForCont}} $p \in A$ is the greatest element of $A$.
Then by definition of $A$:
:$p^2 < 2$
Let $h \in \Q$ be a rational number such that $0 < h < 1$ such that:
:$h < \dfrac {2 - p^2} {2 p + 1}$
This is always possible, because by definition $2 - p^2 > 0$ and $2 p + 1 > 0$.
Let $q = p + h$.
Then $q > p$, and:
{{begin-eqn}}
{{eqn | l = q^2
| r = p^2 + \paren {2 p + h} h
| c =
}}
{{eqn | o = <
| r = p^2 + \paren {2 p + 1} h
| c =
}}
{{eqn | o = <
| r = p^2 + \paren {2 - p^2}
| c =
}}
{{eqn | r = 2
| c =
}}
{{end-eqn}}
That means $q \in A$.
This contradicts our assertion that $p$ is the greatest element of $A$.
It follows that $A$ can have no greatest element.
{{qed}}
\end{proof}
|
21239
|
\section{Set of Reciprocals of Positive Integers is Nowhere Dense in Reals}
Tags: Analysis, Real Number Line with Euclidean Metric, Integer Reciprocal Space, Real Number Space, Denseness
\begin{theorem}
Let $N$ be the set defined as:
:$N := \set {\dfrac 1 n: n \in \Z_{>0} }$
where $\Z_{>0}$ is the set of (strictly) positive integers.
Let $\R$ denote the real number line with the usual (Euclidean) metric.
Then $N$ is nowhere dense in $\R$.
\end{theorem}
\begin{proof}
From Zero is Limit Point of Integer Reciprocal Space, the only limit point of $N$ is $0$.
Hence:
:$\map \cl N = \set {\dfrac 1 n: n \in \Z_{>0} } \cup \set 0$
where $\map \cl N$ denotes the closure of $N$ in $\R$.
Trivially, $\map \cl N$ contains no open real intervals.
Hence no subset of $\map \cl N$ is open in $\R$.
Hence the union of all the subset of $\map \cl N$ which are open in $\R$ is empty.
That is, by definition, the interior of $N$ is empty.
That is:
:$\paren {\map \cl N}^\circ = \O$
and the result follows by definition of nowhere dense.
{{qed}}
\end{proof}
|
21240
|
\section{Set of Relations can be Ordered by Inclusion}
Tags: Subsets, Order Theory
\begin{theorem}
Let $S \times T$ be the product of two sets.
Let $\RR$ be a set of relations on $S \times T$.
Then $\RR$ can be ordered by inclusion.
\end{theorem}
\begin{proof}
Let $R$ be a relation on $S \times T$.
By the definition of relation, $R$ is associated with a subset $R \subseteq S \times T$.
Thus $\RR$ is a subset of the power set $\powerset {S \times T}$.
The result follows from Subset Relation is Ordering.
{{qed}}
\end{proof}
|
21241
|
\section{Set of Ring Elements forming Zero Product with given Element is Ideal}
Tags: Ideal Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $a \in R$ be an arbitrary element of $R$.
Let $A$ be the subset of $R$ defined as:
:$A = \set {x \in R: x \circ a = 0_R}$
Then $A$ is an ideal of $A$.
\end{theorem}
\begin{proof}
By definition of ring zero:
:$\forall x \in R: x \circ 0_R = 0_R$
Hence $0_R \in A$ and so $A \ne \O$.
Let $a, b \in A$.
{{begin-eqn}}
{{eqn | q = \forall x \in R
| l = x \circ b
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = -\paren {x \circ b}
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = x \circ \paren {-b}
| r = 0_R
| c =
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | q = \forall x \in R
| l = x \circ a
| r = 0_R
| c =
}}
{{eqn | lo= \land
| l = x \circ \paren {-b}
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {x \circ a} + \paren {x \circ \paren {-b} }
| r = 0_R
| c = as $0_R$ is identity of $\struct {R, +}$
}}
{{eqn | ll= \leadsto
| l = x \circ \paren {a + \paren {-b} }
| r = 0_R
| c = {{Ring-axiom|D}}
}}
{{eqn | ll= \leadsto
| l = a + \paren {-b}
| o = \in
| r = A
| c = Definition of $A$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in R
| l = x \circ a
| r = 0_R
| c =
}}
{{eqn | lo= \land
| l = x \circ b
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {x \circ a} \circ b
| r = 0_R
| c = {{Defof|Ring Zero}}
}}
{{eqn | ll= \leadsto
| l = x \circ \paren {a \circ b}
| r = 0_R
| c = {{Ring-axiom|M1}}
}}
{{eqn | ll= \leadsto
| l = a \circ b
| o = \in
| r = A
| c = Definition of $A$
}}
{{end-eqn}}
Hence the result, from Test for Ideal:
{{qed}}
\end{proof}
|
21242
|
\section{Set of Sequence Codes is Primitive Recursive}
Tags: Primitive Recursive Functions
\begin{theorem}
Let $\operatorname{Seq}$ be the set of all code numbers of finite sequences in $\N$.
Then $\operatorname{Seq}$ is primitive recursive.
\end{theorem}
\begin{proof}
By the definition of a primitive recursive set, it is sufficient to show that the characteristic function $\chi_{\operatorname{Seq}}$ of $\operatorname{Seq}$ is primitive recursive.
Let $p: \N \to \N$ be the prime enumeration function.
Let $\map {\operatorname{len} } n$ be the length of $n$.
We note that $\map {\chi_{\operatorname{Seq}} } n = 1$ {{iff}} $\map p y$ divides $n$ for $1 \le y \le \map {\operatorname{len} } n$.
That is, {{iff}} $\map {\operatorname{div} } {n, \map p y} = 1$ for $1 \le y \le \map {\operatorname{len} } n$, where $\operatorname{div}$ is the divisor relation.
We then see that $\map {\operatorname{div} } {n, \map p y} = 1$ for $1 \le y \le \map {\operatorname{len} } n$ {{iff}} their product equals $1$.
So we can define $\chi_{\operatorname{Seq}}$ by:
:$\map {\chi_{\operatorname{Seq}} } n = \begin{cases}
\ds \prod_{y \mathop = 1}^{\map {\operatorname{len} } n} \map {\operatorname{div} } {n, \map p y} & : n > 1 \\
0 & : \text{otherwise}
\end{cases}$
Then we define $g: \N^2 \to \N$ as:
:$\map g {n, z} = \begin{cases}
1 & : z = 0 \\
\ds \prod_{y \mathop = 1}^z \map {\operatorname{div} } {n, \map p y} & : z \ne 0
\end{cases}$
We then apply Bounded Product is Primitive Recursive to the primitive recursive function $\map {\operatorname{div} } {n, \map p y}$, and see that $g$ is primitive recursive.
Finally, we have that:
:$\map {\chi_{\operatorname{Seq}} } n = \begin{cases}
\map g {n, \map {\operatorname{len} } n} & : n > 1 \\
0 & : \text{otherwise}
\end{cases}$
is obtained by substitution from:
* the primitive recursive function $\operatorname{len}$
* the primitive recursive function $g$
* the primitive recursive relation $>$
* the constant $1$
So $\chi_{\operatorname{Seq}}$ is primitive recursive.
Hence the result.
{{qed}}
Category:Primitive Recursive Functions
\end{proof}
|
21243
|
\section{Set of Sets can be Defined as Family}
Tags: Indexed Families, Set Systems
\begin{theorem}
Let $\Bbb S$ be a set of sets.
Then $\Bbb S$ can be defined as an indexed family of sets.
\end{theorem}
\begin{proof}
Let $S: \Bbb S \to \Bbb S$ denote the identity mapping on $\Bbb S$:
:$\forall i \in \Bbb S: S_i = i$
where we use $S_i$ to mean the image of $i$ under $S$:
:$S_i := \map S i$
Then we can consider $S$ as an indexing function from $\Bbb S$ to $\Bbb S$.
Hence in this case $\Bbb S$ is at the same time both:
:an indexing set
and:
:the set indexed by itself.
It follows that each of the sets $i \in \Bbb S$ is both:
:an index
and:
:a term $S_i$ of the family of elements of $\Bbb S$ indexed by $\Bbb S$.
Thus we would write $\Bbb S$ as:
:$\family {S_i}_{i \mathop \in \Bbb S}$
{{qed}}
\end{proof}
|
21244
|
\section{Set of Singletons is Smallest Basis of Discrete Space}
Tags: Discrete Spaces, Discrete Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $\BB = \set {\set x : x \in S}$.
Then $\BB$ is the smallest basis of $T$.
That is:
:$\BB$ is a basis of $T$
and:
:for every basis $\CC$ of $T$, $\BB \subseteq \CC$.
\end{theorem}
\begin{proof}
By Basis for Discrete Topology $\BB$ is a basis of $T$.
It remains to be shown that $\BB$ is the smallest basis of $T$.
Let $\CC$ be a basis of $T$.
Let $A \in \BB$.
By definition of the set $\BB$:
:$\exists x \in S: A = \set x$
By definition of basis:
:$\exists B \in \CC: x \in B \subseteq A$
Then by Singleton of Element is Subset:
:$\set x \subseteq B$
Hence $B = A$ by definition of set equality.
Thus $A \in \CC$.
{{qed}}
\end{proof}
|
21245
|
\section{Set of Strictly Positive Real Numbers has no Smallest Element}
Tags: Real Analysis
\begin{theorem}
Let $\R_{>0}$ denote the set of strictly positive real numbers.
Then $\R_{>0}$ has no smallest element.
\end{theorem}
\begin{proof}
{{AimForCont}} $\R_{>0}$ has a smallest element.
Let $m$ be that smallest element.
Then we have that:
:$0 < \dfrac m 2 < m$
But as $0 < \dfrac m 2$ it follows that $\dfrac m 2 \in \R_{>0}$.
This contradicts our assertion that $m$ is the smallest element of $\R_{>0}$.
Hence the result by Proof by Contradiction.
{{qed}}
\end{proof}
|
21246
|
\section{Set of Subfields forms Complete Lattice}
Tags: Lattice Theory, Rings, Fields, Subfields
\begin{theorem}
Let $\struct {F, +, \circ}$ be a field.
Let $\mathbb F$ be the set of all subfields of $F$.
Then $\struct {\mathbb F, \subseteq}$ is a complete lattice.
\end{theorem}
\begin{proof}
Let $\O \subset \mathbb S \subseteq \mathbb F$.
By Intersection of Subfields is Largest Subfield Contained in all Subfields:
:$\bigcap \mathbb S$ is the largest subfield of $F$ contained in each of the elements of $\mathbb S$.
By Intersection of Subfields Containing Subset is Smallest:
:The intersection of the set of all subfields of $F$ containing $\bigcup \mathbb S$ is the smallest subfield of $F$ containing $\bigcup \mathbb S$.
Thus:
:Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.
:The supremum of $\mathbb S$ is the intersection of the set of all subfields of $R$.
Therefore $\struct {\mathbb F, \subseteq}$ is a complete lattice.
{{qed}}
\end{proof}
|
21247
|
\section{Set of Subgroups forms Complete Lattice}
Tags: Lattice Theory, Examples of Complete Lattices, Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group, and let $\mathbb G$ be the set of all subgroups of $G$.
Then $\struct {\mathbb G, \subseteq}$ is a complete lattice.
\end{theorem}
\begin{proof}
Let $\O \subset \mathbb H \subseteq \mathbb G$.
By Intersection of Subgroups: General Result, $\bigcap \mathbb H$ is the largest subgroup of $G$ contained in each of the elements of $\mathbb H$.
Thus, not only is $\bigcap \mathbb H$ a lower bound of $\mathbb H$, but also the largest, and therefore an infimum.
The supremum of $\mathbb H$ is the smallest subgroup of $G$ containing $\bigcup \mathbb H$.
Therefore $\struct {\mathbb G, \subseteq}$ is a complete lattice.
{{qed}}
\end{proof}
|
21248
|
\section{Set of Subgroups of Abelian Group form Subsemigroup of Structure Induced on Power Set}
Tags: Abelian Groups, Subset Products
\begin{theorem}
Let $\struct {G, \circ}$ be an abelian group.
Let $\circ_\PP$ be the operation induced on $\powerset G$, the power set of $G$.
Let $\SS$ be the set of all subgroups of $G$.
Then $\struct {\SS, \circ_\PP}$ is a subsemigroup of $\struct {\powerset G, \circ_\PP}$.
\end{theorem}
\begin{proof}
A fortiori, an abelian group is a commutative semigroup.
Also a fortiori, a subgroup of $G$ is also a subsemigroup of $G$.
The result then follows directly from Set of Subsemigroups of Commutative Semigroup form Subsemigroup of Structure Induced on Power Set.
{{qed}}
\end{proof}
|
21249
|
\section{Set of Subrings forms Complete Lattice}
Tags: Lattice Theory, Rings, Subrings
\begin{theorem}
Let $\struct {K, +, \circ}$ be a ring.
Let $\mathbb K$ be the set of all subrings of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice.
\end{theorem}
\begin{proof}
Let $\P \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Subrings is Largest Subring Contained in all Subrings:
:$\bigcap \mathbb S$ is the largest subring of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Subrings Containing Subset is Smallest:
:The intersection of the set of all subrings of $K$ containing $\bigcup \mathbb S$ is the smallest subring of $K$ containing $\bigcup \mathbb S$.
Thus:
:Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.
:The supremum of $\mathbb S$ is the intersection of the set of all subrings of $K$.
Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice.
{{qed}}
\end{proof}
|
21250
|
\section{Set of Subsemigroups of Commutative Semigroup form Subsemigroup of Structure Induced on Power Set}
Tags: Commutative Semigroups, Subset Products
\begin{theorem}
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\circ_\PP$ be the operation induced on $\powerset S$, the power set of $S$.
Let $\TT$ be the set of all subsemigroups of $S$.
Then $\struct {\TT, \circ_\PP}$ is a subsemigroup of $\struct {\powerset S, \circ_\PP}$.
\end{theorem}
\begin{proof}
First we establish that from Power Set of Semigroup under Induced Operation is Semigroup:
:$\struct {\powerset S, \circ_\PP}$ is a semigroup.
From Subset Product within Commutative Structure is Commutative:
:$\struct {\powerset S, \circ_\PP}$ is a commutative semigroup.
Let $A$ and $B$ be arbitrary subsemigroups of $S$.
As $A$ and $B$ are subsemigroups of $S$, they themselves are closed for $\circ$.
That is:
:$\forall x, y \in A: x \circ y \in A$
and:
:$\forall x, y \in B: x \circ y \in B$
By definition of operation induced on $\powerset S$:
:$A \circ_\PP B = \set {a \circ b: a \in A, b \in B}$
We are to show that:
:$\forall x, y \in A \circ_\PP B: x \circ y \in A \circ_\PP B$
Let $x, y \in A \circ_\PP B$ such that $x = a_x \circ b_x$, $y = a_y \circ b_y$.
We have:
{{begin-eqn}}
{{eqn | l = x \circ y
| r = \paren {a_x \circ b_x} \circ \paren {a_y \circ b_y}
| c =
}}
{{eqn | r = a_x \circ \paren {b_x \circ a_y} \circ b_y
| c = {{SemigroupAxiom|1}}
}}
{{eqn | r = a_x \circ \paren {a_y \circ b_x} \circ b_y
| c = $\circ$ is commutative
}}
{{eqn | r = \paren {a_x \circ a_y} \circ \paren {b_x \circ b_y}
| c = {{SemigroupAxiom|1}}
}}
{{eqn | o = \in
| r = A \circ_\PP B
| c = {{SemigroupAxiom|0}}: both $A$ and $B$ are closed for $\circ$
}}
{{end-eqn}}
That is:
:$x, y \in A \circ_\PP B \implies x \circ y \in A \circ_\PP B$
and $A \circ_\PP B$ is seen to be $\struct {\TT, \circ_\PP}$.
Hence the result.
{{qed}}
\end{proof}
|
21251
|
\section{Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure}
Tags: Infinite Sets
\begin{theorem}
Let $\BB$ be a set of subsets of $\R$, the set of all real numbers.
Let:
:$\card \BB < \mathfrak c$
where
:$\card \BB$ denotes the cardinality of $\BB$
:$\mathfrak c = \card \R$ denotes continuum.
Let $\FF = \set {\bigcup \GG: \GG \subseteq \BB}$.
Then:
:$\exists x, y \in \R: x < y \land \hointr x y \notin \FF$
\end{theorem}
\begin{proof}
Define:
:$ Z = \leftset {x \in \R: \exists U \in \FF: x}$ is local minimum in $\rightset U$
By Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum:
:$\card Z < \mathfrak c$
Then by Cardinalities form Inequality implies Difference is Nonempty:
:$\R \setminus Z \ne \O$
Hence by definition of empty set:
:$\exists z: z \in \R \setminus Z$
By definition of difference:
:$z \in \R \land z \notin Z$
Thus $z < z + 1$.
We will show that $z$ is local minimum in $\hointr z {z + 1}$.
Thus:
:$z \in \hointr z {z + 1}$
Hence:
:$z - 1 < z$
Thus:
:$\openint {z - 1} z \cap \hointr z {z + 1} = \O$
Then by definition $z$ is a local minimum in $\hointr z {z + 1}$.
Because $z \notin Z$:
:$\hointr z {z + 1} \notin \FF$
{{qed}}
\end{proof}
|
21252
|
\section{Set of Subsets is Cover iff Set of Complements is Free}
Tags: Set Theory, Definition Equivalences
\begin{theorem}
Let $S$ be a set.
Let $\CC$ be a set of sets.
Then $\CC$ is a cover for $S$ {{iff}} $\set {\relcomp S X: X \in \CC}$ is free.
\end{theorem}
\begin{proof}
Let $S$ be a set.
\end{proof}
|
21253
|
\section{Set of Subsets of Element of Minimally Inductive Class is Finite}
Tags: Minimally Inductive Classes, Finite Sets
\begin{theorem}
Let $M$ be a class which is minimally inductive under a progressing mapping $g$.
Let $x \in M$.
Let $S$ be the set of all $y \in M$ such that $y \subseteq x$.
Then $S$ is finite.
\end{theorem}
\begin{proof}
The proof proceeds by general induction.
For all $x \in M$, let $\map P x$ be the proposition:
:$\set {y \in M: y \subseteq x}$ is finite.
\end{proof}
|
21254
|
\section{Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum}
Tags: Infinite Sets
\begin{theorem}
Let $\BB$ be a set of subsets of $\R$.
Let:
:$\size \BB < \mathfrak c$
where
:$\size \BB$ denotes the cardinality of $\BB$
:$\mathfrak c = \size \R$ denotes continuum.
Let
:$X = \leftset {x \in \R: \exists U \in \set {\bigcup \GG: \GG \subseteq \BB}: x}$ is local minimum in $\rightset U$
Then:
:$\size X < \mathfrak c$
\end{theorem}
\begin{proof}
We will prove that:
:$(1): \quad \size \BB \aleph_0 < \mathfrak c$
where $\aleph_0 = \size \N$ by Aleph Zero equals Cardinality of Naturals.
In the case when $\size \BB = \mathbf 0$ we have by Zero of Cardinal Product is Zero:
:$\size \BB \aleph_0 = \mathbf 0 < \mathfrak c$
In the case when $\mathbf 0 < \size \BB < \aleph_0$:
{{begin-eqn}}
{{eqn | l = \size \BB \aleph_0
| r = \aleph_0 \size \BB
| c = Product of Cardinals is Commutative
}}
{{eqn | r = \size {\N \times \BB}
| c = {{Defof|Product of Cardinals}}
}}
{{eqn | r = \map \max {\size \N, \size \BB}
| c = Cardinal Product Equal to Maximum
}}
{{eqn | r = \aleph_0
| c = because $\size \BB < \aleph_0$
}}
{{eqn | o = <
| r = \mathfrak c
| c = Aleph Zero is less than Continuum
}}
{{end-eqn}}
In the case when $\size \BB \ge \aleph_0$ we have:
{{begin-eqn}}
{{eqn | l = \size \BB \aleph_0
| r = \size {\BB \times \N}
| c = {{Defof|Product of Cardinals}}
}}
{{eqn | r = \map \max {\size \BB, \size \N}
| c = Cardinal Product Equal to Maximum
}}
{{eqn | r = \size \BB
| c = because $\size \BB \ge \aleph_0$
}}
{{eqn | o = <
| r = \mathfrak c
| c = {{hypothesis}}
}}
{{end-eqn}}
Define:
:$Y = \leftset {x \in \R: \exists U \in \BB: x}$ is local minimum in $\rightset U$
We will show that $X \subseteq Y$ by definition of subset.
Let $x \in X$.
By definition of $X$:
:$\exists U \in \leftset {\bigcup \GG: \GG \subseteq \BB}: x$ is local minimum in $\rightset U$
:$\exists \GG \subseteq \BB: U = \bigcup \GG$
By definition of local minimum:
:$x \in U$
By definition of union:
:$\exists V \in \GG: x \in V$
By definition of subset
:$V \in \BB$
By definition of local minimum
:$\exists y \in \R: y < x \land \openint y x \cap U = \O$
By Set is Subset of Union:
:$V \subseteq U$
Then:
:$\exists y \in \R: y < x \land \openint y x \cap V = \O$
By definition:
:$x$ is local minimum in $V$
Thus by definition of $Y$
:$x \in Y$
So
:$(2): \quad X \subseteq Y$
Define $\family {Z_A}_{A \mathop \in \BB}$ as:
:$Z_A = \leftset {x \in \R: x}$ is local minimum in $\rightset A$
We will prove that:
:$(3): \quad Y \subseteq \ds \bigcup_{A \mathop \in \BB} Z_A$
Let $x \in Y$.
By definition of $Y$:
:$\exists U \in \BB: x$ is local minimum in $U$
By definition of $Z_U$:
:$x \in Z_U$
Thus by definition of union:
:$x \in \ds \bigcup_{A \mathop \in \BB} Z_A$
This ends the proof of inclusion.
{{qed|lemma}}
By Set of Local Minimum is Countable:
:$\forall A \in \BB: Z_A$ is countable
By Countable iff Cardinality not greater Aleph Zero:
:$\forall A \in \BB: \size {Z_A} \le \aleph_0$
By Cardinality of Union not greater than Product:
:$(4): \quad \ds \size {\bigcup_{A \mathop \in \BB} Z_A} \le \size \BB \aleph_0$
Thus:
{{begin-eqn}}
{{eqn | l = \size X
| o = \le
| r = \size Y
| c = $(2)$ and Subset implies Cardinal Inequality
}}
{{eqn | o = \le
| r = \size {\bigcup_{A \mathop \in \BB} Z_A}
| c = $(3)$ and Subset implies Cardinal Inequality
}}
{{eqn | o = \le
| r = \size \BB \aleph_0
| c = $(4)$
}}
{{eqn | o = <
| r = \mathfrak c
| c = $(1)$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21255
|
\section{Set of Successive Numbers contains Unique Multiple}
Tags: Number Theory
\begin{theorem}
Let $m \in \Z_{\ge 1}$.
Then $\set {m, m + 1, \ldots, m + n - 1}$ contains a unique integer that is a multiple of $n$.
That is, in any set containing $n$ successive integers, $n$ divides exactly one of those integers.
\end{theorem}
\begin{proof}
Let $S_m = \set {m, m + 1, \ldots, m + n - 1}$ be a set containing $n$ successive integers.
The proof proceeds by induction on $m$, the smallest number in $S_m$.
\end{proof}
|
21256
|
\section{Set of Transpositions is not Subgroup of Symmetric Group}
Tags: Symmetric Groups
\begin{theorem}
Let $S$ be a finite set with $n$ elements such that $n > 2$.
Let $G = \struct {\map \Gamma S, \circ}$ denote the symmetric group on $S$.
Let $H \subseteq G$ denote the set of all transpositions of $S$ along with the identity mapping which moves no elements of $S$.
Then $H$ does not form a subgroup of $G$.
\end{theorem}
\begin{proof}
First it is noted that $H \subseteq G$, and that the identity mapping is an element of $H$.
Hence to demonstrate that $H$ is a subgroup of $G$, one may use the Two-Step Subgroup Test.
Let $\phi \in H$ be a transposition.
Then from Transposition is Self-Inverse, $\phi^{-1} \in H$.
So $H$ is closed under inversions.
Let $a, b, c \in S$.
Let $\pi, \phi \in H$ such that:
:$\pi = \tuple {a \ b}$
:$\phi = \tuple {b \ c}$
Then by inspection:
:$\phi \circ \pi = \tuple {a \ c \ b}$
Thus $\phi \circ \pi$ is a $3$-cycle and so not a transposition.
Hence $H$ is not closed under composition.
So, by the Two-Step Subgroup Test, $H$ is not a subgroup of $G$.
{{Qed}}
\end{proof}
|
21257
|
\section{Set of Upper Closures of Compact Elements is Basis implies Complete Scott Topological Lattice is Algebraic}
Tags: Topological Order Theory, Continuous Lattices
\begin{theorem}
Let $L = \struct {S, \preceq, \tau}$ be a complete Scott Definition:Topological Lattice.
Let $\BB = \left\{ {x^\succeq: x \in K\left({L}\right)}\right\}$ be a basis of $L$
where
:$x^\succeq$ denotes the upper closure of $x$,
:$K\left({L}\right)$ denotes the compact subset of $L$.
Then $L$ is algebraic.
\end{theorem}
\begin{proof}
Thus by Compact Closure is Directed:
:$\forall x \in S:x^{\mathrm{compact} }$ is directed
where $x^{\mathrm{compact} }$ denotes the compact closure of $x$.
Thus by definition of complete lattice:
:$L$ is up-complete.
Let $x \in S$.
By definition of lower closure of element:
:$x$ is upper closure for $x^\preceq$
By definition of supremum:
:$\sup \left({x^\preceq}\right) \preceq x$
By Compact Closure is Intersection of Lower Closure and Compact Subset:
:$x^{\mathrm{compact} } = x^\preceq \cap K\left({L}\right)$
By Intersection is Subset:
:$x^{\mathrm{compact} } \subseteq x^\preceq$
By Supremum of Subset:
:$\sup \left({x^{\mathrm{compact} } }\right) \preceq \sup \left({x^\preceq}\right)$
By definition of transitivity:
:$\sup \left({x^{\mathrm{compact} } }\right) \preceq x$
{{AimForCont}}
:$x \ne \sup \left({x^{\mathrm{compact} } }\right)$
We will prove that
:$x \notin \left({\left({x^{\mathrm{compact} } }\right) }\right)^\preceq$
{{AimForCont}}
:$x \in \left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq$
By definition of lower closure of element:
:$x \preceq \sup \left({x^{\mathrm{compact} } }\right)$
Thus by definition of antisymmetry:
:this contrasicts $x \ne \sup \left({x^{\mathrm{compact} } }\right)$
{{qed|lemma}}
By definition of relative complement:
:$x \in \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)$
By Complement of Lower Closure of Element is Open in Scott Topological Ordered Set:
:$\complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)$ is open.
By definition of basis:
:$\complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right) = \bigcup\left\{ {G \in \BB: G \subseteq \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)}\right\}$
By definition of union
:$\exists X \in \left\{ {G \in \BB: G \subseteq \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)}\right\}: x \in X$
Then
:$X \in \BB \land X \subseteq \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)$
By definition of $\BB$:
:$\exists k \in K\left({L}\right): X = k^\succeq$
By definition of compact subset:
:$k$ is compact.
By definition of upper closure of element:
:$k \preceq x$
By definition of compact closure:
:$k \in x^{\mathrm{compact} }$
By definitions of supremum and upper bound:
:$k \preceq \sup \left({x^{\mathrm{compact} } }\right)$
By definition of upper closure of element:
:$\sup \left({x^{\mathrm{compact} } }\right) \in X$
By gefinition of subset:
:$\sup \left({x^{\mathrm{compact} } }\right) \in \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)$
By definition of difference:
:$\sup \left({x^{\mathrm{compact} } }\right) \notin \left({\sup \left({x^{\mathrm{compact} } }\right)}\right)^\preceq$
Thus by definition of lower closure of element:
:this contradicts $\sup \left({x^{\mathrm{compact} } }\right) \in \left({\sup \left({x^{\mathrm{compact} } }\right)}\right)^\preceq$
{{qed}}
\end{proof}
|
21258
|
\section{Set of Words Generates Group}
Tags: Set of Words Generates Group, Generated Subgroups, Group Theory
\begin{theorem}
Let $S \subseteq G$ where $G$ is a group.
Let $\hat S$ be defined as $S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.
Then $\gen S = \map W {\hat S}$, where $\map W {\hat S}$ is the set of words of $\hat S$.
\end{theorem}
\begin{proof}
Let $H = \gen S$ where $S \subseteq G$.
$H$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$.
Thus $\hat S \subseteq H$.
By {{GroupAxiom|0}}, $H$ must also contain all products of a finite number of elements of $\hat S$.
Thus $\map W {\hat S} \subseteq H$.
Now we prove that $\map W {\hat S} \le G$.
By the Two-Step Subgroup Test:
Let $x, y \in \map W {\hat S}$.
As $x$ and $y$ are both products of a finite number of elements of $\hat S$, it follows that so is their product $x y$.
Thus $x y \in \map W {\hat S}$ and {{GroupAxiom|0}} is satisfied.
Let $x = s_1 s_2 \ldots s_n \in \map W {\hat S}$.
Then $x^{-1} = s_n^{-1} \ldots s_2^{-1} s_1^{-1} \in \map W {\hat S}$.
Thus the conditions of the Two-Step Subgroup Test are fulfilled, and $\map W {\hat S} \le G$.
Thus $\map W {\hat S}$ is the subgroup of $G$ generated by $S$.
{{qed}}
\end{proof}
|
21259
|
\section{Set of all Self-Maps under Composition forms Monoid}
Tags: Examples of Monoids, Abstract Algebra, Monoids, Mappings, Composite Mappings, Mapping Theory
\begin{theorem}
Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let the operation $\circ$ represent composition of mappings.
Then the algebraic structure $\struct {S^S, \circ}$ is a monoid whose identity element is the identity mapping on $S$.
\end{theorem}
\begin{proof}
By Set of all Self-Maps under Composition forms Semigroup, $\struct {S^S, \circ}$ is a semigroup.
By Identity Mapping is Left Identity and Identity Mapping is Right Identity the identity mapping on $S$ is the identity element of $\struct {S^S, \circ}$.
Since $\struct {S^S, \circ}$ is a semigroup with an identity element, it is a monoid by definition.
{{qed}}
\end{proof}
|
21260
|
\section{Set of all Self-Maps under Composition forms Semigroup}
Tags: Semigroups, Mapping Theory, Examples of Semigroups, Composite Mappings
\begin{theorem}
Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let the operation $\circ$ represent composition of mappings.
Then the algebraic structure $\struct {S^S, \circ}$ is a semigroup.
\end{theorem}
\begin{proof}
Let $f, g \in S^S$.
As the domain of $g$ and codomain of $f$ are the same, the composition $f \circ g$ is defined.
By the definition of composition, $f \circ g$ is a mapping from the domain of $g$ to the codomain of $f$.
Thus $f \circ g: S \to S$, so $f \circ g \in S^S$.
Since this holds for all $f, g \in S^S$, $\struct {S^S, \circ}$ is closed.
By Composition of Mappings is Associative, $\circ$ is associative.
Since $\struct {S^S, \circ}$ is closed and $\circ$ is associative:
:$\struct {S^S, \circ}$ is a semigroup.
{{qed}}
\end{proof}
|
21261
|
\section{Set together with Condensation Points is not necessarily Closed}
Tags: Condensation Points, Set Closures, Omega-Accumulation Points
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\CC$ denote the set of condensation points of $H$.
Then it is not necessarily the case that $H \cup \CC$ is a closed set of $T$.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.
Let $H \subseteq \R$ be a finite subset of $\R$.
Let $\CC$ denote the set of condensation points of $H$.
From Finite Set of Right Order Topology with Condensation Points is not Closed, $H \cup \CC$ is not a closed set of $T$.
{{qed}}
\end{proof}
|
21262
|
\section{Set together with Omega-Accumulation Points is not necessarily Closed}
Tags: Set Closures, Omega-Accumulation Points
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.
Then it is not necessarily the case that $H \cup \Omega$ is a closed set of $T$.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.
Let $H \subseteq \R$ be a finite subset of $\R$.
Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.
From Finite Set of Right Order Topology with Omega-Accumulation Points is not Closed, $H \cup \Omega$ is not a closed set of $T$.
{{qed}}
\end{proof}
|
21263
|
\section{Set with Complement forms Partition}
Tags: Set Partitions, Set Complement, Universe
\begin{theorem}
Let $\O \subset S \subset \mathbb U$.
Then $S$ and its complement $\map \complement S$ form a partition of the universal set $\mathbb U$.
\end{theorem}
\begin{proof}
Follows directly from Set with Relative Complement forms Partition:
If $\O \subset T \subset S$, then $\set {T, \relcomp S T}$ is a partition of $S$.
{{Qed}}
Category:Set Complement
Category:Universe
Category:Set Partitions
\end{proof}
|
21264
|
\section{Set with Dispersion Point is Biconnected}
Tags: Dispersion Points, Biconnectedness, Connectedness, Biconnected Sets
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a connected set in $T$.
Let $p \in H$ be a dispersion point of $H$.
Then $H$ is biconnected.
\end{theorem}
\begin{proof}
{{AimForCont}} $H$ is not biconnected.
Then by definition there exist disjoint non-degenerate connected sets $U, V$ such that $H = U \cup V$.
{{WLOG}}, let $p \in U$.
Then $V \subset H \setminus \set p$.
As $p$ is a dispersion point of $H$, $H \setminus \set p$ is totally disconnected.
Thus, by definition, $H \setminus \set p$ contains no non-degenerate connected sets.
But this contradicts our definition of $V$, as $V \subset H \setminus \set p$.
It follows by Proof by Contradiction that $H$ is biconnected.
{{qed}}
\end{proof}
|
21265
|
\section{Set with Two Parallel Elements is Dependent}
Tags: Matroid Theory
\begin{theorem}
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x, y \in S$.
Let $x, y$ be parallel elements.
If $x, y \in A$ then $A$ is dependent.
\end{theorem}
\begin{proof}
Let $x, y \in A$.
From Doubleton of Elements is Subset:
:$\set{x, y} \subseteq A$
By the definition of parallel elements:
:$\set {x, y}$ is dependent
From Superset of Dependent Set is Dependent:
:$A$ is dependent
{{qed}}
\end{proof}
|
21266
|
\section{Sets in Modified Fort Space are Disconnected}
Tags: Modified Fort Space, Connectedness, Disconnected Spaces
\begin{theorem}
Let $T = \struct {S, \tau_{a, b}}$ be a modified Fort space.
Let $H$ be a subset of $S$ with more than one point.
Then $H$ is disconnected.
\end{theorem}
\begin{proof}
By Isolated Points in Subsets of Modified Fort Space:
:$\exists x \in H: x$ is isolated
By Point in Topological Space is Open iff Isolated, $\set x$ is open in $T$.
By Modified Fort Space is $T_1$ and definition of $T_1$ space, $\set x$ is closed in $T$.
Therefore $\relcomp S {\set x}$ is open in $T$.
Then we have:
{{begin-eqn}}
{{eqn | l = H
| o = \subseteq
| r = S
}}
{{eqn | r = \set x \cup \relcomp S {\set x}
| c = Union with Relative Complement
}}
{{eqn | l = H \cap \set x \cap \relcomp S {\set x}
| o = \subseteq
| r = \set x \cap \relcomp S {\set x}
| c = Intersection is Subset
}}
{{eqn | r = \O
| c = Intersection with Relative Complement is Empty
}}
{{eqn | ll = \leadsto
| l = H \cap \set x \cap \relcomp S {\set x}
| r = \O
| c = Subset of Empty Set
}}
{{eqn | l = H \cap \set x
| r = \set x
}}
{{eqn | o = \ne
| r = \O
}}
{{eqn | l = H \cap \relcomp S {\set x}
| r = H \setminus \set x
| c = Set Difference as Intersection with Relative Complement
}}
{{eqn | o = \ne
| r = \O
| c = $H$ has more than one point
}}
{{end-eqn}}
This shows that $H$ is disconnected.
{{qed}}
\end{proof}
|
21267
|
\section{Sets of Operations on Set of 3 Elements/Automorphism Group of A}
Tags: Sets of Operations on Set of 3 Elements
\begin{theorem}
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
Then:
:$\AA$ has $3$ elements.
\end{theorem}
\begin{proof}
Recall the definition of (group) automorphism:
:$\phi$ is an automorphism on $\struct {S, \circ}$ {{iff}}:
::$\phi$ is a permutation of $S$
::$\phi$ is a homomorphism on $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$
Hence $\AA$ can be defined as the set of operations $\circ$ on $S$ such that every permutation on $S$ is an automorphism of $\struct {S, \circ}$.
The set $\map \Gamma S$ of all permutations on $S = \set {a, b, c}$ has $6$ elements.
From Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.
Hence it is not necessary to analyse the effect of $I_S$ on the various elements of $S$.
Let us denote each of the remaining permutations on $S$ as follows:
{{begin-eqn}}
{{eqn | l = p
| o = :
| r = \map p a = b, \map p b = c, \map p c = a
}}
{{eqn | l = q
| o = :
| r = \map q a = c, \map q b = a, \map q c = b
}}
{{eqn | l = r
| o = :
| r = \map r a = a, \map r b = c, \map r c = b
}}
{{eqn | l = s
| o = :
| r = \map s a = c, \map s b = b, \map s c = a
}}
{{eqn | l = t
| o = :
| r = \map t a = b, \map t b = a, \map t c = c
}}
{{end-eqn}}
So, let $\circ \in \AA$.
From Permutation of Set is Automorphism of Set under Left Operation and Permutation of Set is Automorphism of Set under Right Operation:
:${\to} \in \AA$
:${\gets} \in \AA$
where $\to$ and $\gets$ are the right operation and left operation respectively.
Next we note that from Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent:
:$\forall x \in S: x \circ x = x$
Let us explore various options for $a \circ b$.
Let $a \circ b = a$.
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
{{begin-eqn}}
{{eqn | l = \map p a \circ \map p b
| r = \map p a
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ c
| r = b
| c =
}}
{{eqn | l = \map q a \circ \map q b
| r = \map q a
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ a
| r = c
| c =
}}
{{eqn | l = \map r a \circ \map r b
| r = \map r a
| c =
}}
{{eqn | ll= \leadsto
| l = a \circ c
| r = a
| c =
}}
{{eqn | l = \map s a \circ \map s b
| r = \map s a
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ b
| r = c
| c =
}}
{{eqn | l = \map t a \circ \map t b
| r = \map t a
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ a
| r = b
| c =
}}
{{end-eqn}}
This is none other than the left operation, which has already been counted.
{{qed|lemma}}
Let $a \circ b = b$.
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
{{begin-eqn}}
{{eqn | l = \map p a \circ \map p b
| r = \map p b
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ c
| r = c
| c =
}}
{{eqn | l = \map q a \circ \map q b
| r = \map q b
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ a
| r = a
| c =
}}
{{eqn | l = \map r a \circ \map r b
| r = \map r b
| c =
}}
{{eqn | ll= \leadsto
| l = a \circ c
| r = c
| c =
}}
{{eqn | l = \map s a \circ \map s b
| r = \map s b
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ b
| r = b
| c =
}}
{{eqn | l = \map t a \circ \map t b
| r = \map t b
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ a
| r = a
| c =
}}
{{end-eqn}}
This is none other than the right operation, which has already been counted.
{{qed|lemma}}
Let $a \circ b = c$.
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
{{begin-eqn}}
{{eqn | l = \map p a \circ \map p b
| r = \map p c
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ c
| r = a
| c =
}}
{{eqn | l = \map q a \circ \map q b
| r = \map q c
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ a
| r = b
| c =
}}
{{eqn | l = \map r a \circ \map r b
| r = \map r c
| c =
}}
{{eqn | ll= \leadsto
| l = a \circ c
| r = b
| c =
}}
{{eqn | l = \map s a \circ \map s b
| r = \map s c
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ b
| r = a
| c =
}}
{{eqn | l = \map t a \circ \map t b
| r = \map t c
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ a
| r = c
| c =
}}
{{end-eqn}}
Thus we have identified another operation on $S$ for which all permutations on $S$ are automorphisms of $\struct {S, \circ}$.
{{qed|lemma}}
We now note that once the product element $a \circ b$ has been selected to be either $a$, $b$ or $c$, the full structure of $\struct {S, \circ}$ is forced.
Hence there are no other operations $\circ$ on $S$ but these ones we have counted.
That is, there is a total of $3$ such operations.
{{qed}}
\end{proof}
|
21268
|
\section{Sets of Operations on Set of 3 Elements/Automorphism Group of A/Commutative Operations}
Tags: Sets of Operations on Set of 3 Elements
\begin{theorem}
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
Then:
:Exactly $1$ of the operations of $\AA$ is commutative.
\end{theorem}
\begin{proof}
Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:
:$\begin {array} {c|ccc}
\to & a & b & c \\
\hline
a & a & b & c \\
b & a & b & c \\
c & a & b & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\gets & a & b & c \\
\hline
a & a & a & a \\
b & b & b & b \\
c & c & c & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
From Cayley Table for Commutative Operation is Symmetrical about Main Diagonal, it is apparent by inspection that of the above:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
is the only Cayley table of a commutative operation.
{{qed}}
\end{proof}
|
21269
|
\section{Sets of Operations on Set of 3 Elements/Automorphism Group of A/Isomorphism Classes}
Tags: Sets of Operations on Set of 3 Elements
\begin{theorem}
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
Then:
:The elements of $\AA$ are each in its own isomorphism class.
\end{theorem}
\begin{proof}
Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:
:$\begin {array} {c|ccc}
\to & a & b & c \\
\hline
a & a & b & c \\
b & a & b & c \\
c & a & b & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\gets & a & b & c \\
\hline
a & a & a & a \\
b & b & b & b \\
c & c & c & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
We have from Algebraic Structures formed by Left and Right Operations are not Isomorphic for Cardinality Greater than 1 that $\struct {S, \to}$ and $\struct {S, \gets}$ are not isomorphic.
Observe from the Cayley table $\circ$ that:
:$(1): \quad \forall x, y \in S: x \ne y \implies x \circ y \ne x \land x \circ y \ne y$
{{AimForCont}} there exists an isomorphism $\phi$ from $\struct {S, \to}$ to $\struct {S, \circ}$.
We have:
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {a \to b}
| r = \map \phi a \circ \map \phi b
| c = {{Defof|Isomorphism (Abstract Algebra)|Isomorphism}}
}}
{{eqn | r = \map \phi c
| c = from $(1)$
}}
{{eqn | r = \map \phi b
| c = {{Defof|Right Operation}}
}}
{{end-eqn}}
Hence:
:$\map \phi c = \map \phi b$
which means $\phi$ is not an injection.
That is, $\phi$ is not a bijection and hence not an isomorphism.
Hence by Proof by Contradiction there exists no isomorphism from $\struct {S, \to}$ to $\struct {S, \circ}$.
In a similar way:
{{AimForCont}} there exists an isomorphism $\phi$ from $\struct {S, \gets}$ to $\struct {S, \circ}$.
We have:
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {a \gets b}
| r = \map \phi a \circ \map \phi b
| c = {{Defof|Isomorphism (Abstract Algebra)|Isomorphism}}
}}
{{eqn | r = \map \phi c
| c = from $(1)$
}}
{{eqn | r = \map \phi a
| c = {{Defof|Left Operation}}
}}
{{end-eqn}}
Hence:
:$\map \phi c = \map \phi a$
which means $\phi$ is not an injection.
That is, $\phi$ is not a bijection and hence not an isomorphism.
Hence by Proof by Contradiction there exists no isomorphism from $\struct {S, \gets}$ to $\struct {S, \circ}$.
It has been shown that none of the elements of $\AA$ is isomorphism to any of the others.
The result follows by definition of isomorphism class.
{{qed}}
\end{proof}
|
21270
|
\section{Sets of Operations on Set of 3 Elements/Automorphism Group of A/Operations with Identity}
Tags: Sets of Operations on Set of 3 Elements
\begin{theorem}
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
Then:
:None of the operations of $\AA$ has an identity element.
\end{theorem}
\begin{proof}
Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:
:$\begin{array}{c|ccc}
\to & a & b & c \\
\hline
a & a & b & c \\
b & a & b & c \\
c & a & b & c \\
\end{array}
\qquad
\begin{array}{c|ccc}
\gets & a & b & c \\
\hline
a & a & a & a \\
b & b & b & b \\
c & c & c & c \\
\end{array}
\qquad
\begin{array}{c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end{array}$
The result is apparent by inspection.
{{qed}}
\end{proof}
|
21271
|
\section{Sets of Operations on Set of 3 Elements/Automorphism Group of B}
Tags: Sets of Operations on Set of 3 Elements
\begin{theorem}
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\BB$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }$, where $I_S$ is the identity mapping on $S$.
Then:
:$\BB$ has $3^3 - 3$ elements.
\end{theorem}
\begin{proof}
Recall the definition of (group) automorphism:
:$\phi$ is an automorphism on $\struct {S, \circ}$ {{iff}}:
::$\phi$ is a permutation of $S$
::$\phi$ is a homomorphism on $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$
From Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.
Hence it is not necessary to analyse the effect of $I_S$ on $S$.
Let us denote each of the remaining elements of $\set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }$ as follows:
{{begin-eqn}}
{{eqn | l = p
| o = :
| r = \map p a = b, \map p b = c, \map p c = a
}}
{{eqn | l = q
| o = :
| r = \map q a = c, \map q b = a, \map q c = b
}}
{{end-eqn}}
We select various product elements $x \circ y \in S$ and determine how $p$ and $q$ constrain other product elements as follows:
Then by definition of the above mappings $p$ and $q$, and the definition of a homomorphism, we obtain as follows:
;$(1): \quad a \circ a$
{{Begin-table}}|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ a
| r = a
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p a
| r = \map p a
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ b
| r = b
| c =
}}
{{eqn | l = \map q a \circ \map q a
| r = \map q a
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ c
| r = c
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & & \\
b & & b & \\
c & & & c \\
\end {array}$
|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ a
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p a
| r = \map p b
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ b
| r = c
| c =
}}
{{eqn | l = \map q a \circ \map q a
| r = \map q b
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ c
| r = a
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & b & & \\
b & & c & \\
c & & & a \\
\end {array}$
|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ a
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p a
| r = \map p c
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ b
| r = a
| c =
}}
{{eqn | l = \map q a \circ \map q a
| r = \map q c
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ c
| r = b
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & c & & \\
b & & a & \\
c & & & b \\
\end {array}$
{{End-table}}
So selecting $a \circ a$ fixes $b \circ b$ and $c \circ c$.
;$(2): \quad a \circ b$
{{Begin-table}}|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ b
| r = a
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p b
| r = \map p a
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ c
| r = b
| c =
}}
{{eqn | l = \map q a \circ \map q b
| r = \map q a
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ a
| r = c
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & a & \\
b & & & b \\
c & c & & \\
\end {array}$
|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ b
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p b
| r = \map p b
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ c
| r = c
| c =
}}
{{eqn | l = \map q a \circ \map q b
| r = \map q b
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ a
| r = a
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & b & \\
b & & & c \\
c & a & & \\
\end {array}$
|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ b
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p b
| r = \map p c
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ c
| r = a
| c =
}}
{{eqn | l = \map q a \circ \map q b
| r = \map q c
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ a
| r = b
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & c & \\
b & & & a \\
c & b & & \\
\end {array}$
{{End-table}}
So selecting $a \circ b$ fixes $b \circ c$ and $c \circ a$.
;$(3): \quad a \circ c$
{{Begin-table}}|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ c
| r = a
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p c
| r = \map p a
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ a
| r = b
| c =
}}
{{eqn | l = \map q a \circ \map q c
| r = \map q a
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ b
| r = c
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & a \\
b & b & & \\
c & & c & \\
\end {array}$
|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ c
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p c
| r = \map p b
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ a
| r = c
| c =
}}
{{eqn | l = \map q a \circ \map q c
| r = \map q b
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ b
| r = a
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & b \\
b & c & & \\
c & & a & \\
\end {array}$
|-
| align="left" |
{{begin-eqn}}
{{eqn | l = a \circ c
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = \map p a \circ \map p c
| r = \map p c
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ a
| r = a
| c =
}}
{{eqn | l = \map q a \circ \map q c
| r = \map q c
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ b
| r = b
| c =
}}
{{end-eqn}}
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & c \\
b & a & & \\
c & & b & \\
\end {array}$
{{End-table}}
So selecting $a \circ c$ fixes $b \circ a$ and $c \circ b$.
There are $3$ elements $x$ of $S$ with which $a \circ x$ can be made.
For each of these $3$, there are $3$ different elements $y$ such that $a \circ x = y$.
These collectively fix all the possible values of $b \circ x$ and $c \circ x$.
Hence they exhaust all possible operations $\circ$ on $S$ for which $p$ and $q$ are automorphisms.
Thus there are independently:
:$3$ different options for $a \circ a$
:$3$ different options for $a \circ b$
:$3$ different options for $a \circ c$
and therefore $3 \times 3 \times 3 = 3^3$ operations $\circ$ on $S$ for which $p$ and $q$ are automorphisms.
Note that from Automorphism Group of $\AA$, $3$ of these are also such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$.
So these are excluded from our count.
The result follows.
{{qed}}
\end{proof}
|
21272
|
\section{Sets of Operations on Set of 3 Elements/Automorphism Group of B/Commutative Operations}
Tags: Sets of Operations on Set of 3 Elements
\begin{theorem}
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\BB$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }$, where $I_S$ is the identity mapping on $S$.
Then:
:Exactly $8$ of the operations of $\BB$ is commutative.
\end{theorem}
\begin{proof}
Recall Automorphism Group of $\BB$.
Consider each of the categories of $\BB$ induced by each of $a \circ a$, $a \circ b$ and $a \circ c$, illustrated by the partially-filled Cayley tables to which they give rise:
;$(1): \quad a \circ a$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & & \\
b & & b & \\
c & & & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & b & & \\
b & & c & \\
c & & & a \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & c & & \\
b & & a & \\
c & & & b \\
\end {array}$
;$(2): \quad a \circ b$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & a & \\
b & & & b \\
c & c & & \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & b & \\
b & & & c \\
c & a & & \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & c & \\
b & & & a \\
c & b & & \\
\end {array}$
;$(3): \quad a \circ c$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & c \\
b & a & & \\
c & & b & \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & b \\
b & c & & \\
c & & a & \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & c \\
b & a & & \\
c & & b & \\
\end {array}$
With a view to Cayley Table for Commutative Operation is Symmetrical about Main Diagonal, we inspect the various combinations of these partial Cayley tables.
It is apparent that the result of $x \circ x$ makes no difference to whether an operation is commutative.
However, note that each of the partial operations in $(2)$ can be conjoined with exactly $1$ of the partial operations in $(3)$ to make a commutative operation:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & a & c \\
b & a & & b \\
c & c & b & \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & b & a \\
b & b & & c \\
c & a & c & \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & c & b \\
b & c & & a \\
c & b & a & \\
\end {array}$
Hence by the Product Rule for Counting there are $3 \times 3 = 9$ commutative operations which can be constructed thus.
However, note that one of these:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
has already been accounted for in Automorphism Group of $\AA$: Commutative Operations.
This commutative operations is such that the group of automorphisms of $\struct {S, \circ}$ forms the complete symmetric group on $S$, not just the given permutations.
Hence the result.
{{qed}}
\end{proof}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.