nyuuzyou/wikis huggingface dataset:
Finite Product of Finite Sets is Finite

== Theorem ==
Let $\sequence {S_n}$ be a sequence of finite sets.
Let $\ds \prod_{k \mathop = 1}^n S_k$ be their Cartesian product.
Then $\ds \prod_{k \mathop = 1}^n S_k$ is also a finite set.
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.8$
Category:Cartesian Product
Finite Product of Sigma-Compact Spaces is Sigma-Compact

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$ be a finite set of topological spaces.
Let $\ds \struct {S, \tau} = \prod_{i \mathop = 1}^n \struct {S_i, \tau_i}$ be the product space of $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$.
Let each of $\struct {S_i, \tau_i}$ be $\sigma$-compact.
Then $\struct {S, \tau}$ is also $\sigma$-compact.
== Proof ==
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be $\sigma$-compact.
Then $\ds S_1 = \bigcup_{n \mathop = 1}^\infty C_n$ and $\ds S_2 = \bigcup_{n \mathop = 1}^\infty D_n$ where all the $C_n$ and $D_n$ are compact sets.
From Tychonoff's Theorem, the set $\KK = \set {C_n \times D_m: n, m \in \N}$ is a countable set of compact sets of $S_1 \times S_2$ with the product topology.
Consider $\tuple {a, b} \in S_1 \times S_2$.
Then $a \in S_1$ and $b \in S_2$.
Using the $\sigma$-compactness of $S_1$ and $S_2$, $\exists n, m$ such that $a \in C_n$ and $b \in D_m$.
So $\tuple {a, b} \in C_n \times D_m \in \KK$.
Thus, $\bigcup \KK = S_1 \times S_2$ and then $S_1 \times S_2$ is $\sigma$-compact.
Using induction on the number of factors of the product we get the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Sigma-Compact Spaces
Category:Product Spaces
Finite Product of Weakly Locally Compact Spaces is Weakly Locally Compact

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a (strictly) positive integer.
Let $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$ be a finite set of topological spaces.
Let $\ds \struct {S, \tau} = \prod_{i \mathop = 1}^n \struct {S_i, \tau_i}$ be the product space of $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$.
Let each of $\struct {S_i, \tau_i}$ be weakly locally compact.
Then $\struct {S, \tau}$ is also weakly locally compact.
== Proof ==
Pick any $\tuple {x_1, \dots, x_n} \in S$.
Since each of $\struct {S_i, \tau_i}$ is weakly locally compact, $x_i$ has a compact neighborhood.
That is, there is a compact set $N_i$ such that:
:$\exists U_i \in \tau_i: x_i \in U_i \subseteq N_i$
From Cartesian Product of Family of Subsets:
:$\ds \tuple {x_1, \dots, x_n} \in \prod_{i \mathop = 1}^n U_i \subseteq \prod_{i \mathop = 1}^n N_i$
By definition of product space:
:$\ds \prod_{i \mathop = 1}^n U_i \in \tau$
From Finite Topological Product of Compact Spaces:
:$\ds \prod_{i \mathop = 1}^n N_i$ is compact
Hence:
:$\ds \prod_{i \mathop = 1}^n N_i$ is a compact neighborhood of $\tuple {x_1, \dots, x_n}$.
As $\tuple {x_1, \dots, x_n}$ is arbitrary, $\struct {S, \tau}$ is weakly locally compact.
== Also see ==
* Compactness Properties Preserved under Projection Mapping
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Weakly Locally Compact Spaces
Category:Product Spaces
Finite Rank Operator is Compact

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $T \in \map {B_{00} } {H, K}$ be a bounded finite rank operator.
Then $T \in \map {B_0} {H, K}$, that is, $T$ is compact.
== Proof ==
Let $A$ be a bounded set in $H$.
Since $T$ is bounded, the image $T \sqbrk A$ is a bounded set by definition.
Since $T$ is finite rank operator, its range $\Rng T$ is finite dimensional.
The closure of a bounded set in $\Rng T$ is compact, by Heine-Borel Theorem.
So the closure of $T \sqbrk A$ in $\Rng T$ is compact.
Hence, $T$ is a compact linear operator.
== Sources ==
* $\text {II}.4$ Exercise $2$
Category:Linear Transformations on Hilbert Spaces
Category:Compact Linear Transformations
Finite Rank Operators Dense in Compact Linear Transformations

== Theorem ==
Let $H, K$ be Hilbert spaces.
Then:
:$\map {B_{00} } {H, K}$ is everywhere dense in $\map {B_0} {H, K}$
where:
:$\map {B_{00} } {H, K}$ is the space of continuous finite rank operators from $H$ to $K$
:$\map {B_0} {H, K}$ is the space of compact linear transformations from $H$ to $K$.
That is, for every $T \in \map {B_0} {H, K}$, there is a sequence $\sequence {T_n}_{n \mathop \in \N}$ in $\map {B_{00} } {H, K}$ such that:
:$\ds \lim_{n \mathop \to \infty} \norm {T_n - T} = 0$
where $\norm {\, \cdot \,}$ denotes the norm on bounded linear transformations.
== Proof ==
== Sources ==
* $\text {II}.4.4$
Category:Compact Linear Transformations
Finite Ring with Multiplicative Norm is Field

== Theorem ==
Let $R$ be a finite ring with a multiplicative norm.
Then $R$ is a field.
== Proof ==
From Ring with Multiplicative Norm has No Proper Zero Divisors, $R$ has no proper zero divisors.
From Finite Ring with No Proper Zero Divisors is Field, $R$ is a field.
Category:Ring Theory
Category:Field Theory
Category:Norm Theory
Finite Ring with No Proper Zero Divisors is Field

== Theorem ==
Let $\struct {R, +, \circ}$ be a finite non-null ring with no proper zero divisors.
Then $R$ is a field.
== Proof ==
As $R$ is non-null, there is at least one nonzero element in $R$. 
Consider the two maps from $R$ to itself, for each nonzero $a \in R$:
:$\varphi_R: x \mapsto a \circ x$
:$\varphi_L: x \mapsto x \circ a$
By Ring Element is Zero Divisor iff not Cancellable, all nonzero elements in $R$ are cancellable. Thus:
:$a \circ x = a \circ y \implies x=y$
:$x \circ a = y \circ a \implies x=y$
Therefore, both maps are by definition injective.
By Equivalence of Mappings between Finite Sets of Same Cardinality, the maps are then also surjective.
First, we show that $R$ has a unity. 
Since $\varphi_R$ is surjective, any element $b \in R$ can be written in the form:
:$b = a \circ x_b$ 
for some $x_b \in R$.
As $\varphi_L$ is surjective, there is some $x_L \in R$ such that:
:$a = x_L \circ a$
We now have:
Thus, $x_L$ is a multiplicative left identity. 
By similar arguments, $R$ also has a multiplicative right identity, which can be denoted $x_R$.
By Left and Right Identity are the Same, it follows that $x_L = x_R$, so they are the unity of $R$. 
Now we show that $R$ is a division ring, i.e. that each nonzero element of $R$ has an inverse.
Let $1_R$ denote the unity of $R$.
Since $\varphi_R$ is surjective, it follows that:
:$a \circ y_R = 1_R$
for some $y_R \in R$. 
Since $\varphi_L$ is surjective, it follows that:
:$y_L \circ a = 1_R$ 
for some $y_L \in R$.
Recalling that the maps were defined for each nonzero $a \in R$, this means that every nonzero element of $R$ has both a left inverse and a right inverse. 
By Left Inverse and Right Inverse is Inverse, each nonzero element of $R$ has an inverse, so $R$ is by definition a division ring. 
It now follows by Wedderburn's Theorem that $R$ is a field. 
Category:Ring Theory
Category:Field Theory
Category:Galois Fields
Finite Semigroup Equal Elements for Different Powers

== Theorem ==
Let $\left({S, \circ}\right)$ be a finite semigroup.
Then:
: $\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$
== Proof ==
List the positive powers $x, x^2, x^3, \ldots$ of any element $x$ of a finite semigroup $\left({S, \circ}\right)$.
Since all are elements of $S$, and the semigroup has a finite number of elements, it follows from the Pigeonhole Principle this list must contain repetitions.
So there must be at least one instance where $x^m = x^n$ for some $m, n \in \N$.
Category:Semigroups
Finite Sequences in Set Form Acyclic Graph

== Theorem ==
Let $S$ be a set.
Let $V$ be the set of finite sequences in $S$.
Let $E$ be the set of unordered pairs $\set {p, q}$ of elements of $V$ such that either:
:$q$ is formed by extending $p$ by one element or
:$p$ is formed by extending $q$ by one element.
That is:
:$\card {\Dom p \symdif \Dom q} = 1$, where $\symdif$ is symmetric difference
:$p \restriction D = q \restriction D$, where $D = \Dom p \cap \Dom q$
Then $T = \struct{V, E}$ is an acyclic graph.
== Proof ==
Category:Graph Theory
Finite Set Contains Subset of Smaller Cardinality

== Theorem ==
Let $S$ be a finite sets.
Let
:$\size S = n$
where $\size {\, \cdot \,}$ denotes cardinality.
Let $0 \le m \le n$.
Then there exists a subset $X \subseteq S$ such that:
:$\size X = m$
== Proof ==
=== Case 1 ===
Let $m = n$. 
Then $X = S$ is a subset $X \subseteq S$ such that:
:$\size X = m$
=== Case 2 ===
Let $m = 0$. 
Then $X = \O$ is a subset $X \subseteq S$ such that:
:$\size X = m$
=== Case 3 ===
Let $0 < m < n$.
By definition of the cardinality of a finite set:
:$S \sim \N_{< n}$
where:
:$\sim$ denotes set equivalence
:$\N_{<n}$ is the set of all natural numbers less than $n$
Consider the set of all natural numbers less than $m$, $\N_{< m}$.
By the definition of a subset:
:$\N_{< m} \subseteq \N_{< n}$
Let $i : \N_{< m} \to \N_{< n}$ be the inclusion mapping.
Let $f : \N_{< n} \to S$ be a bijection between $\N_{< n}$ and $S$.
From Composite of Injections is Injection, the composite $f \circ i : \N_{< m} \to S$ is an injection.
Let $X = \Img {f \circ i}$ be the image of $f \circ i$.
By definition of the image of a mapping:
:$X \subseteq S$
From Injection to Image is Bijection:
:$\N_{< m} \sim X$
By definition of the cardinality of $X$:
:$\size X = m$
== Also see ==
* Cardinality of Subset of Finite Set
Category:Set Theory
Category:Subsets
Category:Cardinality
Finite Set Formed by Substitution has Larger Intersection

== Theorem ==
Let $S$ and $T$ be finite sets.
Let $x \in S \setminus T$.
Let $y \in T \setminus S$.
Let $R$ be the set formed by substituting $x$ for $y$ in $T$, that is:
:$R = \paren{T \setminus \set y} \cup \set x$
Then:
:$\card{R \cap S} = \card{T \cap S} + 1$
== Proof ==
We have:
Category:Finite Sets
Category:Set Intersection
Finite Set Formed by Substitution has Same Cardinality

== Theorem ==
Let $S$ and $T$ be finite sets.
Let $x \in S \setminus T$.
Let $y \in T \setminus S$.
Let $R$ be the set formed by substituting $x$ for $y$ in $T$, that is:
:$R = \paren{T \setminus \set y} \cup \set x$
Then:
:$\card R = \card T$
== Proof ==
We have:
Category:Finite Sets
Category:Cardinality
Finite Set of Continuous Functions between Metric Spaces is Pointwise Equicontinuous

== Theorem ==
Let $\struct {X, d}$ and $\struct {Y, d'}$ be metric spaces. 
Let $\map \CC {X, Y}$ be the set of continuous functions $X \to Y$.
Let $\FF = \set {f_1, f_2, \ldots, f_n}$ be a finite subset of $\map \CC {X, Y}$.
Then $\FF$ is pointwise equicontinuous.
== Proof ==
Let $x \in X$.
Let $\epsilon \in \R_{>0}$.
Let $i$ be a natural number with $1 \le i \le n$. 
Since $f_i$ is continuous at $x$, there exists $\delta_i > 0$ such that whenever: 
:$\map d {x, y} < \delta_i$
we have:
:$\map {d'} {\map {f_i} x, \map {f_i} y} < \epsilon$
Let:
:$\ds \delta = \min_i \set {\delta_i}$
so that: 
:$\delta \le \delta_i$
for each $i$. 
Then, whenever:
:$\map d {x, y} < \delta$
we have: 
:$\map {d'} {\map {f_i} x, \map {f_i} y} < \epsilon$
for each $i$.
So $\FF$ is pointwise equicontinuous.
Category:Pointwise Equicontinuity
Finite Set of Elements in Principal Ideal Domain has GCD

== Theorem ==
Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $a_1, a_2, \dotsc, a_n$ be non-zero elements of $D$.
Then $a_1, a_2, \dotsc, a_n$ all have a greatest common divisor.
== Proof ==
Let $0_D$ and $1_D$ be the zero and unity respectively of $D$.
Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$.
From Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal:
:$J = \ideal x$
for some $x \in D$, where $\ideal x$ denotes the principal ideal generated by $x$.
We have that each $a_i$ can be expressed as a linear combination of $\set {a_1, a_2, \dotsc, a_n}$:
:$a_i = 0_D a_1 + 0_D a_2 + \dotsb + 1_D a_i + \dotsb + 0_D a_n$
Thus:
:$\forall i \in \set {0, 1, \dotsc, n}: a_i \in J$
and so by definition of $J$:
:$\forall i \in \set {0, 1, \dotsc, n}: a_i = t_i x$
for some $t_i \in D$.
Thus $x$ is a common divisor of $a_1, a_2, \dotsc, a_n$.
As $x \in \ideal x = J$, we have:
:$x = c_1 a_1 + c_2 a_2 + \dotsb + c_n a_n$
for some $c_1, c_2, \dotsc, c_n \in D$.
Thus every common divisor of $a_1, a_2, \dotsc, a_n$ also is a divisor of $x$.
Thus $x$ is a greatest common divisor of $a_1, a_2, \dotsc, a_n$.
== Sources ==
* : $\S 62.4$ Factorization in an integral domain: $\text{(i)}$
Category:Principal Ideal Domains
Category:GCD Domains
Category:Greatest Common Divisor
Finite Sets are Comparable

== Theorem ==
Let $S$ and $T$ be finite sets.
Then $S$ and $T$ are comparable by size.
== Proof ==
By definition of finite set, there exist $m, n \in \N$ such that:
:$S \sim \N_{<n}$
:$T \sim \N_{<m}$
That is, there are bijections $f, g$:
:$f: S \to \N_{<n}$
:$g: T \to \N_{<m}$
, suppose that $m \le n$.
Then $\N_{<m} \subseteq \N_{<n}$, and so we can define:
:$h: T \to S, \map h t = \map {f^{-1}} { \map g t }$
where $f^{-1}: \N_{<n} \to S$ is the inverse of $f$.
By Inverse of Bijection is Bijection and Composite of Injections is Injection, it follows that $h$ is injective.
By Injection to Image is Bijection, $h$ is a bijection to $\Img h$.
Since $\Img h \subseteq S$, it follows that $T$ is comparable to $S$.
== Sources ==
* : Chapter $1$: General Background: $\S 4$ Larger and smaller
Category:Comparable Sets
Finite Signed Measure is Complex Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a finite signed measure on $\struct {X, \Sigma}$. 
Then $\mu$ is a complex measure on $\struct {X, \Sigma}$. 
== Proof ==
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then: 
:$\mu = \mu^+ - \mu^-$
for measures $\mu^+$ and $\mu^-$. 
From Jordan Decomposition of Finite Signed Measure, $\mu^+$ and $\mu^-$ are finite measures.
Then, for each $A \in \Sigma$, we have:
So $\mu$ takes only real values.
That is, $\mu$ takes values in $\C$. 
Since $\mu$ is a signed measure, we have: 
:$\map \mu \O = 0$
and: 
:for each sequence $\sequence {S_n}_{n \mathop \in \N}$ of pairwise disjoint $\Sigma$-measurable sets we have: 
::$\ds \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = \sum_{n \mathop = 1}^\infty \map \mu {S_n}$
So $\mu$ is a complex measure.
Category:Finite Signed Measures
Category:Complex Measures
Finite Space Satisfies All Compactness Properties

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.
Then $T$ satisfies the following compactness properties:
: $T$ is compact.
: $T$ is sequentially compact.
: $T$ is countably compact.
: $T$ is weakly countably compact.
: $T$ is a Lindelöf space.
: $T$ is pseudocompact.
: $T$ is $\sigma$-compact.
: $T$ is strongly locally compact.
: $T$ is $\sigma$-locally compact.
: $T$ is weakly $\sigma$-locally compact.
: $T$ is locally compact.
: $T$ is weakly locally compact.
: $T$ is paracompact.
: $T$ is countably paracompact.
: $T$ is metacompact.
: $T$ is countably metacompact.
== Proof ==
We have that:
: A Finite Topological Space is Compact.
: A Finite Space is Sequentially Compact.
The remaining properties are demonstrated in:
: Sequence of Implications of Global Compactness Properties
: Sequence of Implications of Local Compactness Properties
: Sequence of Implications of Paracompactness Properties
Category:Finite Topological Spaces
Category:Compact Spaces
Finite Space is Second-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.
Then $T$ is a second-countable space.
== Proof ==
Let $T = \struct {S, \tau}$ be a topological space where $S$ is finite.
As $S$ is finite, then so is its power set $\powerset S$.
So is $\tau \subseteq \powerset S$.
Now $\tau$ is a basis for itself, as every set in $\tau$ is a union of sets from $\tau$ (just one).
But $\tau$ is finite, and so countable.
The result follows from definition of a second-countable space.
Category:Second-Countable Spaces
Finite Space is Sequentially Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.
Then $T$ is sequentially compact.
== Proof ==
We have:
:Finite Topological Space is Compact
:Compact Space is Countably Compact
:Finite Space is Second-Countable
:Second-Countable Space is First-Countable
:First-Countable Space is Sequentially Compact iff Countably Compact.
Hence the result.
Category:Sequentially Compact Spaces
Category:Finite Topological Spaces
Finite State Machine is Turing Computable

== Theorem ==
Let $F = \tuple {S, A, I, \Sigma, T}$ be a finite state machine.
Then there exists a Turing machine $T$ that:
* Has input language $\Sigma$.
* Accepts exactly the same language as $F$.
* Halts on every input.
== Proof ==
Define the Turing machine:
:$T = \tuple {S \cup \set {H}, \Sigma, \Sigma \cup \set B, \delta, I, B, \set H}$
where:
:$\map \delta {s, \sigma} = \tuple {\map T {s, \sigma}, \sigma, R}$ if $q \in S$ and $\sigma \ne B$
:$\map \delta {s, B} = \tuple {H, B, R}$ if $s \in A$
The machine behaves identically to $F$ while the input is being read.
When there is no more input, if $\sigma \in A$, then $T$ transitions to the $H$ state, which is the accepting state.
If $\sigma \notin A$, then there is no transition defined, so $T$ halts without accepting.
As $T$ only moves $R$, it will eventually reach the end of the input word, and one of these two conditions must hold.
Category:Turing Machines
Finite Subgroup Test

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a non-empty finite subset of $G$.
Then:
:$H$ is a subgroup of $G$
:
:$\forall a, b \in H: a \circ b \in H$
That is, a non-empty finite subset of $G$ is a subgroup it is closed.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $5$: Subgroups: Exercise $15$
Category:Subgroups
Category:Finite Subgroup Test
Finite Subgroup Test/Proof 1

== Theorem ==
== Proof ==
Let $H$ be a finite subset of $G$ such that $a, b \in H \implies a \circ b \in H$.
From the Two-Step Subgroup Test, it follows that we only need to show that $a \in H \implies a^{-1} \in H$.
So, let $a \in H$.
First it is straightforward to show by induction that $\set {x \in G: x = a^n: n \in \N} \subseteq H$.
That is, $a \in H \implies \forall n \in \N: a^n \in H$.
Now, since $H$ is finite, we have that the order of $a$ is finite.
Let the order of $a$ be $m$.
From Inverse Element is Power of Order Less 1 we have that $a^{m-1} = a^{-1}$.
As $a^{m-1} \in H$ (from above) the result follows.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35$
* : Chapter $4$: Subgroups: Proposition $4.2$: Remark $4$
Category:Finite Subgroup Test
Finite Subgroup Test/Proof 2

== Theorem ==
== Proof ==
=== Sufficient Condition ===
Let $H$ be a subgroup of $G$.
Then:
:$\forall a, b \in H: a \circ b \in H$
by definition of subgroup.
=== Necessary Condition ===
Let $H$ be a non-empty finite subset of $G$ such that:
:$\forall a, b \in H: a \circ b \in H$
Let $x \in H$.
We have by hypothesis that $H$ is closed under $\circ$.
Thus all elements of $\set {x, x^2, x^3, \ldots}$ are in $H$.
But $H$ is finite.
Therefore it must be the case that:
:$\exists r, s \in \N: x^r = x^s$
for $r < s$.
So we can write:
Then we have:
But we have that:
So from $(2)$:
:$x^{-1} = x^{s - r - 1}$
it follows that:
: $x^{-1} \in H$
and from the Two-Step Subgroup Test it follows that $H$ is a subgroup of $G$.
== Sources ==
* : $\S 36.5$: Subgroups
* : $\S 1.9$: Subgroups: Lemma $7$
Category:Finite Subgroup Test
Finite Submodule of Function Space

== Theorem ==
Let $\struct {G, +}$ be a group whose identity is $e$.
Let $R$ be a ring.
Let $\struct {G, +, \circ}_R$ be an $R$-module.
Let $S$ be a set.
Let $G^S$ the set of all mappings $f: S \to G$.
Let $G^{\paren S}$ be the set of all mappings $f: S \to G$ such that $\map f x = e$ for all but finitely many elements $x$ of $S$.
Then:
:$\struct {G^{\paren S}, +', \circ}_R$ is a submodule of $\struct {G^S, +, \circ}_R$
where $+'$ is the operation induced on $G^{\paren S}$ by $+$.
== Proof ==
Let $\struct {G, +, \circ}_R$ be an $R$-module and $S$ be a set.
We need to show that $\struct {G^{\paren S}, +'}$ is a group.
Let $f, g \in G^{\paren S}$.
Let:
:$F = \set {x \in S: \map f x \ne e}$
:$G = \set {x \in S: \map g x \ne e}$
From the definition of $f$ and $g$, both $F$ and $G$ are finite.
Since $e + e = e$ by definition of identity element, it follows that if:
:$\map f x + \map g x \ne e$
then necessarily $\map f x \ne e$ or $\map g x \ne e$.
That is:
:$\map {\paren {f +' g} } x = \map f x + \map g x \ne e \implies x \in F \cup G$
But as $F$ and $G$ are both finite, it follows that $F \cup G$ is also finite.
Hence $f +' g \in G^{\paren S}$ and $\struct {G^{\paren S}, +'}$ is closed.
Now let $f \in G^{\paren S}$.
Let $f^*$ be the pointwise inverse of $f$.
Thus:
:$\map {f^*} x = -\paren {\map f x}$
Again, let $F = \set {x \in S: \map f x \ne e}$.
It follows directly that:
:$x \in S \setminus F \implies \map {f^*} x = e$
Hence:
:$\map {f^*} x \ne e \implies x \in F$
and hence:
:$f^* \in G^{\paren S}$
So by the Two-Step Subgroup Test, it follows that $\struct {G^{\paren S}, +'}$ is a subgroup of $\struct {G^S, +}$.
Hence $\struct {G^{\paren S}, +', \circ}_R$ is an $R$-module.
The result follows.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.3$
Category:Submodules
Category:Function Spaces
Finite Subset Bounds Element of Finite Infima Set and Upper Closure

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be meet semilattice.
Let $F$ be filter in $L$.
Let $X$ be non empty finite subset of $S$.
Let $x \in S$ such that
:$x \in \paren {\map {\operatorname{fininfs} } {F \cup X} }^\succeq$
where
:$\operatorname{fininfs}$ denotes the finite infima set
:$X^\succeq$ denotes the upper closure of $X$.
Then there exists $a \in S$: $a \in F \land x \succeq a \wedge \inf X$
== Proof ==
By definition of upper closure of subset:
:$\exists u \in \map {\operatorname{fininfs} } {F \cup X}: u \preceq x$
By definition of finite infima set:
:there exists finite subset $Y$ of $F \cup X$:
:$Y$ admits an infimum and $u = \inf Y$
We will prove that
:$Y \setminus X \subseteq F$
Let $a \in Y \setminus X$.
By definition of difference:
:$a \in Y$ and $a \notin X$
By definition of subset:
:$a \in F \cup X$
Thus by definition of union:
:$a \in F$
Define $Z := Y \setminus X$.
;Case $Z = \O$:
By $F$ is non-empty:
:$\exists b: b \in F$
By Meet Precedes Operands:
:$b \wedge \inf X \preceq \inf X$
By Set Difference with Superset is Empty Set:
:$Y \subseteq X$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$X$ admits an infimum.
By Infimum of Subset:
:$\inf X \preceq u$
Thus by definition of trantisivity:
:$b \wedge \inf X \preceq x$
;Case $Z \ne \O$:
By Set Difference is Subset:
:$Z \subseteq Y$
By Subset of Finite Set is Finite:
:$Z$ is finite.
By Union of Finite Sets is Finite:
:$Z \cup X$ is finite.
By definitions of non-empty set and union:
:$Z \cup X \ne \O$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$X$ admits an infimum
and
:$Z \cup X$ admits an infimum.
By Set is Subset of Union:
:$Y \subseteq Y \cup X$
By Set Difference Union Second Set is Union:
:$Y \subseteq Z \cup X$
By Infimum of Subset:
:$\map \inf {Z \cup X} \preceq u$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$Z$ admits an infimum
By Infimum of Infima:
:$\map \inf {Z \cup X} = \paren {\inf Z} \wedge \inf X$
By Filtered in Meet Semilattice with Finite Infima:
:$\inf Z \in F$
By definition of transitivity:
:$\paren {\inf Z} \wedge \inf X \preceq x$
Thus:
:$\exists a \in S: a \in F \land x \succeq a \wedge \inf X$
== Sources ==
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Category:Join and Meet Semilattices
Category:Upper Closures
Finite Subset Bounds Element of Finite Suprema Set and Lower Closure

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be join semilattice.
Let $I$ be ideal in $L$.
Let $X$ be non empty finite subset of $S$.
Let $x \in S$ such that
:$x \in \paren {\map {\operatorname {finsups} } {F \cup X} }^\preceq$
where
:$\operatorname {finsups}$ denotes the finite suprema set
:$X^\preceq$ denotes the lower closure of $X$.
Then there exists $a \in S$: $a \in I \land x \preceq a \vee \sup X$
== Proof ==
This follows by of the proof of Finite Subset Bounds Element of Finite Infima Set and Upper Closure.
== Sources ==
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* 
Category:Join and Meet Semilattices