nyuuzyou/wikis huggingface dataset:
Indiscrete Space is Hereditarily Compact

== Theorem ==
Let $\struct {S, \tau}$ be an indiscrete topological space.
Then $\struct {S, \tau}$ is hereditarily compact.
== Proof ==
Let $\struct {H, \tau_H}$ be a subspace of $T$.
From Subset of Indiscrete Space is Compact, $\struct {H, \tau_H}$ is compact.
The result follows by definition of hereditarily compact.
== Also see ==
* Hausdorff Space is Hereditarily Compact iff Finite
Category:Indiscrete Topology
Category:Hereditarily Compact Spaces
Indiscrete Space is Irreducible

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is irreducible.
== Proof ==
There is only one non-empty open set in $T$.
So there can be no two open sets in $T$ which are disjoint.
Hence (trivially) $T$ is irreducible.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $9$
Category:Irreducible Spaces
Category:Indiscrete Topology
Indiscrete Space is Non-Meager

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is non-meager.
== Proof ==
Let $H \subseteq S$ such that $H \ne \O$.
From Limit Points of Indiscrete Space, every point of $S$ is a limit point of $H$.
So, by definition:
:$H^- = S$
where $H^-$ is the closure of $H$ in $T$.
Now the interior of $S$ is $S$ itself (trivially, by definition).
So:
:$\paren {H^-}^\circ = S \ne \O$
where $\paren {H^-}^\circ$ is the interior of the closure of $H$.
So, by definition, $H$ is not nowhere dense.
(Note that if $H = \O$, then Empty Set is Nowhere Dense applies, which is why the stipulation above that $H \ne \O$.)
It follows that $H$ can not be the countable union of nowhere dense subsets of $S$.
So $H$ is, by definition, not meager, so must be non-meager.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $5$
Category:Indiscrete Topology
Category:Non-Meager Spaces
Indiscrete Space is Path-Connected

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is path-connected.
== Proof ==
Let $a, b \in S$.
Consider any mapping $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$.
From Mapping to Indiscrete Space is Continuous, we have that $f$ is ‎continuous.
The result follows by definition of path-connectedness.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $9$
Category:Path-Connected Spaces
Category:Indiscrete Topology
Indiscrete Space is Pseudometrizable

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is pseudometrizable.
== Proof ==
Let $d: S \times S \to \R$ be the mapping defined as:
:$\forall x, y \in S: \map d {x, y} = 0$
Then clearly $d$ is a pseudometric.
Let $\struct {S, \tau_{\struct {S, d} } }$ be the topological space induced by $d$.
Since $\struct {S, \tau_{\struct {S, d} } }$ is a topological space, by Empty Set is Element of Topology we have that $\O$ is open.
For any $\epsilon \in \R_{>0}$ we have:
:$\map {B_\epsilon} a := \set {x \in S: \map d {x, a} < \epsilon} = S$
where $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$.
Let $U\subset S$ be a non-empty open set.
By Open Sets in Pseudometric Space we see that for every $x \in U$, there must exist an $\epsilon > 0$ such that $\map {B_\epsilon} x \subset U$.
However, $S \subset \map {B_\epsilon} x$.
Thus $U = S$.
Hence it is seen that $\O$ and $S$ are the only open sets of this resulting pseudometric space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $11$
Category:Pseudometrizable Topologies
Category:Indiscrete Topology
Indiscrete Space is Second-Countable

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is a second-countable space.
== Proof ==
The only basis for $T$ is $\set S$ which is trivially countable.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $7$
Category:Second-Countable Spaces
Category:Indiscrete Topology
Indiscrete Space is Separable

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space such that $S$ has more than one element.
Then $T$ is separable.
== Proof 1 ==
== Proof 2 ==
== Sources ==
Category:Indiscrete Topology
Category:Separable Spaces
Category:Indiscrete Space is Separable
Indiscrete Space is Separable/Proof 1

== Theorem ==
== Proof ==
By definition, $T$ is separable there exists a countable subset of $S$ which is everywhere dense in $T$.
Let $x \in T$.
Then $\set x \subseteq T$ and $\set x$ is (trivially) countable.
From Subset of Indiscrete Space is Everywhere Dense we have that $\set x$ is everywhere dense.
Hence the result by definition of separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $7$
Category:Indiscrete Space is Separable
Indiscrete Space is T3

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space .
Then $T$ is a $T_3$ space.
== Proof ==
Let $F \subseteq S$ be a closed set in $S$.
Let $y \in S$ such that $y \notin F$.
The only way this can happen is if $F = \O$.
So there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
That is, $U = \O$ and $V = S$.
Hence (trivially) the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $10$
Category:T3 Spaces
Category:Indiscrete Topology
Indiscrete Space is T4

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is a $T_4$ space.
== Proof ==
We have that an indiscrete space is a $T_5$ space.
Then we have that a $T_5$ Space is a $T_4$ Space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $10$
Category:T4 Spaces
Category:Indiscrete Topology
Indiscrete Space is T5

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is a $T_5$ space.
== Proof ==
By definition, the only two separated sets in $T$ are $S$ and $\O$.
Then there exist two disjoint open sets $S$ and $\O$ containing $S$ and $\O$ respectively.
Hence (trivially) $T$ is a $T_5$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $10$
Category:T5 Spaces
Category:Indiscrete Topology
Indiscrete Space is Ultraconnected

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is ultraconnected.
== Proof ==
There is only one non-empty closed set in $T$.
So there can be no two closed sets in $T$ which are disjoint.
Hence (trivially) $T$ is ultraconnected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $9$
Category:Ultraconnected Spaces
Category:Indiscrete Topology
Indiscrete Topology is Coarsest Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be an indiscrete topological space.
:$\tau$ is the coarsest topology on $S$.
Hence it is comparable with all other topologies on $S$.
== Proof ==
Let $\phi$ be any topology on $S$.
Then by definition of topology, $\O \in \phi$ and $S \in \phi$
Hence by definition of subset, $\tau \subseteq \phi$. 
Hence by definition of coarser topology, $\tau$ is coarser than $\phi$.
== Also see ==
* Discrete Topology is Finest Topology
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.6$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $1$
* 
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Strong and weak operator topologies on $\map {CL} {X, Y}$
Category:Indiscrete Topology
Indiscrete Topology is Topology

== Theorem ==
Let $S$ be a set.
Let $\tau$ be the indiscrete topology on $S$.
:$\tau$ is a topology on $S$. 
== Proof ==
Let $T = \struct {S, \set {\O, S} }$ be the indiscrete space on $S$.
Confirming the criteria for $T$ to be a topology:
:$(1): \quad$ Trivially, by definition, $\O \in \tau$ and $S \in \tau$.
:$(2): \quad \O \cup \O = \O \in \tau$, $\O \cup S = S \in \tau$ and $S \cup S = S \in \tau$ from Union with Empty Set and Set Union is Idempotent.
:$(3): \quad \O \cap \O = \O \in \tau$, $\O \cap S = \O \in \tau$ and $S \cap S = S \in \tau$ from Intersection with Empty Set and Set Intersection is Idempotent.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.6$
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology
* : $\S 2$: Topological Spaces: Topologies
Category:Indiscrete Topology
Indiscrete Topology is not Metrizable

== Theorem ==
Let $S$ be a set with more than one element.
The indiscrete topology on $S$ is not metrizable.
== Proof ==
In order to be metrizable, there needs to be a metric $d$ on $S$, so that $\struct {S, d}$ is a metric space.
As $S$ has more than one element, $\exists x, y \in S: x \ne y$.
Then:
:$\epsilon = \map d {x, y} > 0$
So the open $\epsilon$-ball $\map {B_{\epsilon / 2} } x$ is $d$-open.
Hence $\map {B_{\epsilon / 2} } x$ is in the topology which $d$ induces.
But $x \in \map {B_{\epsilon / 2} } x$ while $y \notin \map {B_{\epsilon / 2} } x$.
Thus $\map {B_{\epsilon / 2} } x \ne \O$ and $\map {B_{\epsilon / 2} } x \ne S$.
So the topology induced by $d$ is not the indiscrete topology.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 12$
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.6$
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $11$
Category:Metrizable Topologies
Category:Indiscrete Topology
Induced Algebra Homomorphism on Unitization

== Theorem ==
Let $A, B$ be algebras over $\C$.
Let $\phi : A \to B$ be an algebra homomorphism.
Let $A_+$ and $B_+$ be the unitizations of $A$ and $B$ respectively.
Define $\psi : A_+ \to B_+$ by:
:$\map \psi {a, \lambda} = \tuple {\map \phi a, \lambda}$ 
for each $\tuple {a, \lambda} \in A_+$.
Then $\psi$ is a unital algebra homomorphism.
== Proof ==
Let $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$ and $t \in \C$.
We then have:
and:
We finally have:
:$\map \psi { {\mathbf 1}_{A_+} } = \map \psi { {\mathbf 0}_A, 1} = \tuple {\map \phi { {\mathbf 0}_A}, 1} = \tuple { {\mathbf 0}_B, 1} = {\mathbf 1}_{B_+}$
Hence $\psi$ is unital.
Category:Unitizations of Algebras over Fields
Induced Group Product is Homomorphism iff Commutative

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H_1, H_2$ be subgroups of $G$.
Let $\phi: H_1 \times H_2 \to G$ be defined such that:
:$\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$
Then $\phi$ is a homomorphism every element of $H_1$ commutes with every element of $H_2$.
=== Corollary ===
== Proof ==
We have $\tuple {h_1, h_2} \circ \tuple {k_1, k_2} = \tuple {h_1 \circ k_1, h_2 \circ k_2}$ by definition of group direct product.
=== Necessary Condition ===
Let $\phi$ be a homomorphism.
Then:
This follows whatever $k_1$ and $h_2$ are.
So in order for $\phi$ to be a homomorphism, every element of $H_1$ must commute with every element of $H_2$.
=== Sufficient Condition ===
Let every element of $H_1$ commute with every element of $H_2$.
Let $\tuple {h_1, h_2}, \tuple {k_1, k_2} \in H_1 \times H_2$.
Then:
Thus $\phi$ is shown to be a homomorphism.
Category:Group Homomorphisms
Induced Group Product is Homomorphism iff Commutative/Corollary

== Corollary to Induced Group Product is Homomorphism iff Commutative ==
Let $\struct {G, \circ}$ be a group.
Let $\phi: G \times G \to G$ be defined such that:
:$\forall a, b \in G: \map \phi {a, b} = a \circ b$
Then $\phi$ is a homomorphism $G$ is abelian.
== Proof ==
We have that $G$ is a subgroup of itself.
The result then follows from Induced Group Product is Homomorphism iff Commutative by putting $H_1 = H_2 = G$.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \epsilon$
Category:Abelian Groups
Category:Group Homomorphisms
Induced Homomorphism of Polynomial Forms

== Theorem ==
Let $R$ and $S$ be commutative rings with unity.
Let $\phi: R \to S$ be a ring homomorphism.
Let $R \sqbrk X$ and $S \sqbrk X$ be the rings of polynomial forms over $R$ and $S$ respectively in the indeterminate $X$.
Then the map $\overline \phi: R \sqbrk X \to S \sqbrk X$ given by:
:$\map {\overline \phi} {a_0 + a_1 X + \cdots + a_n X^n} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n$
is a ring homomorphism.
== Proof using evaluation homomorphism ==
== Proof by direct verification ==
== Also see ==
* Definition:Polynomial Evaluation Homomorphism
Category:Induced Homomorphism of Polynomial Forms
Category:Polynomial Theory
Induced Homomorphism of Polynomial Forms/Proof 2

== Theorem ==
== Proof ==
Let:
:$f = a_0 + \cdots + a_n X^n$
:$g = b_0 + \cdots + b_m X^m \in R \sqbrk X$
We have:
and
Thus $\overline \phi$ is a ring homomorphism.
Category:Induced Homomorphism of Polynomial Forms
Induced Metric on Surface of Revolution

== Theorem ==
Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
Let the smooth local parametrization of $C$ be:
:$\map \gamma t = \tuple {\map x t, \map y t}$
Then the induced metric on $S_C$ is:
:$g = \paren {\map {x'} t^2 + \map {y'} t^2} d t^2 + \map y t^2 d \theta^2$
=== Corollary ===
== Proof ==
By Smooth Local Parametrization of Surface of Revolution, the smooth local parametrization of $S_C$ can be written as:
:$\map X {t, \theta} = \tuple {\map y t \cos \theta, \map y t \sin \theta, \map x t}$
By definition, the induced metric on $S_C$ is:
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Surfaces of Revolution
Category:Solid Geometry
Category:Induced Metrics
Induced Metric on Surface of Revolution/Corollary

== Theorem ==
Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
Let the smooth local parametrization of $C$ be:
:$\map \gamma t = \tuple {\map x t, \map y t}$
Let $\gamma$ be a unit-speed curve.
Then the induced metric on $S_C$ is:
:$g = d t^2 + \map y t^2 d \theta^2$
== Proof ==
By definition of the unit-speed curve:
:$\size {\map {\gamma'} t}_g = 1$
In our case we are working with the Euclidean space.
Hence:
:$\sqrt {\map {x'^2} t + \map {y'^2} t} = 1$
or:
:$ \map {x'^2} t + \map {y'^2} t = 1$
Substitution of this into the induced metric of $S_C$ yields the desired result.
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Surfaces of Revolution
Category:Solid Geometry
Category:Induced Metrics
Induced Neighborhood Space is Neighborhood Space

== Theorem ==
Let $S$ be a set.
Let $\tau$ be a topology on $S$, thus forming the topological space $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\struct {S, \tau}$.
Then $\struct {S, \NN}$ is a neighborhood space.
== Proof ==
Let $x \in S$.
Let $\NN_x$ be the neighborhood filter of $x$.
From Basic Properties of Neighborhood in Topological Space, $\NN_x$ fulfils the neighborhood space axioms.
Hence the result.
Category:Neighborhood Spaces
Induced Outer Measure Restricted to Semiring is Pre-Measure

== Theorem ==
Let $\SS$ be a semiring over a set $X$.
Let $\mu: \SS \to \overline \R_{\ge 0}$ be a pre-measure on $\SS$, where $\overline \R_{\ge 0}$ denotes the set of positive extended real numbers.
Let $\mu^*: \powerset X \to \overline \R_{\ge 0}$ be the outer measure induced by $\mu$.
Then:
:$\ds \mu^*\restriction_\SS \, = \mu$
where $\restriction$ denotes restriction.
== Proof ==
Let $S \in \SS$.
It follows immediately from the definition of the induced outer measure that $\map {\mu^*} S \le \map \mu S$.
Therefore, it suffices to show that if $\ds \sequence {A_n}_{n \mathop = 0}^\infty$ is a countable cover for $S$, then:
:$\ds \map \mu S \le \sum_{n \mathop = 0}^\infty \map \mu {A_n}$
If the above statement is true, then it follows directly from the definition of infimum that $\map \mu S \le \map {\mu^*} S$, thus proving the theorem.
Define, for all natural numbers $n \in \N$:
:$\ds B_n = A_n \setminus A_{n - 1} \setminus \cdots \setminus A_0$
where $\setminus$ denotes set difference.
We take $B_0 = A_0$.
Using the mathematical induction, we will prove that for all natural numbers $m < n$, $B_{n, m} = A_n \setminus A_{n-1} \setminus \cdots \setminus A_{n-m}$ is the finite union of pairwise disjoint elements of $\SS$.
We take $B_{n, 0} = A_n$.
The base case $m = 0$ is trivial.
Now assume the induction hypothesis that the above statement is true for some natural number $m < n - 1$, and let $D_1, D_2, \ldots, D_N$ be pairwise disjoint elements of $\SS$ such that:
:$\ds B_{n, m} = \bigcup_{k \mathop = 1}^N D_k$
Then:
By the definition of a semiring, for all natural numbers $k \le N$, $D_k \setminus A_{n - m - 1}$ is the finite union of pairwise disjoint elements of $\SS$.
Hence $B_{n, m+1}$ is the finite union of pairwise disjoint elements of $\SS$, completing the induction step.
Therefore, $B_{n, n-1} = B_n$ is the finite union of pairwise disjoint elements of $\SS$, as desired.
Using the above result and applying the axiom of countable choice, we can, for all $n \in \N$, choose a finite set $\FF_n$ of pairwise disjoint elements of $\SS$ for which:
:$\ds B_n = \bigcup \FF_n$
Now, $x \in S$ there exists an $n \in \N$ such that $x \in S \cap A_n$.
Taking the smallest such $n$, which exists because $\N$ is well-ordered, it follows that $x \notin A_0, A_1, \ldots, A_{n - 1}$, and so $x \in S \cap B_n$.
Therefore:
:$\ds S = \bigcup_{n \mathop = 0}^\infty \paren {S \cap B_n}$
Hence:
Category:Outer Measures
Category:Pre-Measures
Induced Relation Generates Order Isomorphism

== Theorem ==
Let $\struct {A_1, \preceq_1}$ be an ordered set.
Let $\phi: A_1 \to A_2$ be a bijection.
Let:
:$\preceq_2 \mathop{:=} \set {\tuple {\map \phi x, \map \phi y}: x \in A_1 \land y \in A_1 \land x \mathop{\preceq_1} y}$
Then $\phi: \struct {A_1, \preceq_1} \to \struct {A_2, \preceq_2}$ is an order isomorphism.
== Proof ==
Take any $x, y \in A_1$ such that $x \preceq_1 y$.
Since $x, y \in A_1$, it follows by the definition of a mapping that:
:$\map \phi x, \map \phi y \in A_2$
So $x \in A_1$ and $y \in A_1$ and $x \preceq_1 y$.
It follows from the definition of $\preceq_2$ that:
:$\map \phi x \preceq_2 \map \phi y$
Conversely, suppose that:
:$\map \phi x \preceq_2 \map \phi y$
By the definition of $\preceq_2$, it follows that:
:$x \preceq_1 y$
Therefore, the biconditional holds:
:$x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
By definition, it follows that:
:$\phi: \struct {A_1, \preceq_1} \to \struct {A_2, \preceq_2}$
is an order isomorphism.
== Sources ==
* : $\S 6.33$
Category:Order Isomorphisms
Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution

== Theorem ==
Let $\map {y_1} x$ be a particular solution to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
such that $y_1$ is not the trivial solution.
Let $\map {y_2} x$ be the real function defined as:
:$\map {y_2} x = \map v x \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
Then $y_2$ and $y_1$ are linearly independent.
== Proof ==
This will be demonstrated by calculating the Wronskian of $y_1$ and $y_2$ and demonstrating that it is non-zero everywhere.
First we take the derivative of $v$ $x$:
:$v' = \dfrac 1 { {y_1}^2} e^{- \int P \rd x}$
As $\ds -\int P \rd x$ is a real function, $e^{-\int P \rd x}$ is non-zero wherever it is defined.
Hence from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent, $y_1$ and $y_2$ are linearly independent.
== Sources ==
* : $\S 3.16$: Problem $1$
Category:Homogeneous LSOODEs
Induced Structure Identity

== Theorem ==
Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $e$ be an identity for $\circ$.
Then the constant mapping $f_e: S \to T$ defined as:
:$\forall x \in S: \map {f_e} x = e$
is the identity for the pointwise operation $\oplus$ induced on $T^S$ by $\circ$.
== Proof ==
Let $\struct {T, \circ}$ be an algebraic structure with an identity $e$.
Let $f \in T^S$.
Then:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.6: \ 2^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.4$
Category:Identity Elements
Category:Pointwise Operations
Induced Structure from Doubleton is Isomorphic to External Direct Product with Self

== Theorem ==
Let $A$ be a set.
Let $\struct {S, \odot}$ be an algebraic structure.
Let $S^A$ denote the set of mappings from $A$ to $S$.
Let $\struct {S^A, \odot}$ denote the algebraic structure on $S^A$ induced by $\odot$.
Then $\struct {S^A, \odot}$ is isomorphic to the external direct product of $\struct {S, \odot}$ with itself.
== Proof ==
Let $A = \set {a, b}$ where $a$ and $b$ are arbitrary.
Let $\phi: S^A \to S^2$ be defined as:
:$\forall f \in S^A: \map \phi f = \tuple {\map f a, \map f b}$
We are to show that $\phi$ is an isomorphism.
First we demonstrate that $\phi$ is a homomorphism.
So, let $f, g \in S^A$.
Recall that:
:by definition of pointwise operation:
::$\forall f, g \in S^A: \map {\paren {f \odot g} } x = \map f x \odot \map g x$
:by definition of external direct product:
::$\forall \tuple {x_1, x_2}, \tuple {y_1, y_2} \in S \times S: \tuple {x_1, x_2} \odot \tuple {y_1, y_2} = \tuple {x_1 \odot y_1, x_2 \odot y_2}$
We have:
and so $\phi$ has been shown to be a homomorphism.
It remains to be shown that $\phi$ is a bijection.
Let $\map \phi f = \map \phi g$.
We have:
Thus by definition $\phi$ is an injection.
Let $\tuple {s_1, s_2} \in S$ be arbitrary.
By the nature of $S^A$, there exists a mapping $h: A \to S$ such that:
:$\map h a = s_1$
:$\map h b = s_2$
Hence:
:$\map \phi h = \tuple {s_1, s_2}$
As $\tuple {s_1, s_2}$ is arbitrary, it follows that $\phi$ is a surjection.
Thus $\phi$ is both an injection and a surjection, hence by definition a bijection.
We have that $\phi$ is a homomorphism which is a bijection.
That is, $\phi$ is an isomorphism.
The result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.11 \ \text{(d)}$
Category:External Direct Products
Category:Isomorphisms (Abstract Algebra)
Category:Pointwise Operations
Category:Doubletons
Induced Topology on Subspace of Subspace Coincides with Induced Topology on Subspace

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space
Let $S_2 \subseteq S_1$ be a subset of $S_1$.
Let $S_3 \subseteq S_2$ be a subset of $S_2$.
Let $T_2 = \struct {S_2, \tau_2}$ be the topological subspace of $T_1$ such that $\tau_2$ is the subspace topology on $T_2$ induced by $\tau_1$.
Let $T_3 = \struct {S_3, \tau_3}$ be the topological subspace of $T_1$ such that $\tau_3$ is the subspace topology on $T_3$ induced by $\tau_1$.
Then $\tau_3$ is the same set of subsets of $S_1$ as is the subspace topology on $T_3$ induced by $\tau_2$.
== Proof ==
Let $\tau_P$ denote the subspace topology on $T_3$ induced by $\tau_1$.
Let $\tau_Q$ denote the subspace topology on $T_3$ induced by $\tau_2$.
The object of this exercise is to demonstrate that $\tau_P = \tau_Q$.
This will be done by showing that an arbitrary set $V$ is in $\tau_P$ $V$ is in $\tau_Q$.
We have that:
:$\tau_P = \set {U \cap S_3: U \in \tau_1}$
:$\tau_Q = \set {U \cap S_3: U \in \tau_2}$
Then:
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 11$
Category:Topological Subspaces