nyuuzyou/wikis huggingface dataset:
Mapping Defines Additive Function of Subalgebra of Power Set

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $B$ be an algebra of sets over $S$.
Define $\Sigma: B \to \mathbb A$ using summation as:
:$\ds \map \Sigma T = \sum_{t \mathop \in T} \map f t$
for $T\subseteq S$.
Then $\Sigma$ is an additive function on $B$.
== Proof ==
Note that by Subset of Finite Set is Finite, $B$ consists of finite sets.
The result now follows from Sum over Disjoint Union of Finite Sets.
== Also see ==
* Inclusion-Exclusion Principle
* Sum over Union of Finite Sets
Category:Summations
Mapping Images are Disjoint only if Domains are Disjoint

== Theorem ==
Let $S$ and $T$ be sets.
Let:
:$f \sqbrk S \cap f \sqbrk T = \O$
where $f \sqbrk S$ denotes the image set of $S$.
Then:
:$S \cap T = \O$
== Proof ==
From Image of Intersection under Mapping:
:$f \sqbrk {S \cap T} \subseteq f \sqbrk S \cap f \sqbrk T$
From Empty Set is Subset of All Sets:
:$f \sqbrk {S \cap T} = \O$
From Image of Subset under Mapping is Subset of Image:
:$S \cap T = \O$
== Sources ==
* : $\S 1.10$: Functions: Exercise $5 \ \text{(c)}$
Category:Images
Category:Disjoint Sets
Mapping Induces Partition on Domain

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $F$ be defined as:
:$F = \set {\map {f^{-1} } x: x \in T}$
where $\map {f^{-1} } x$ is the preimage of $x$.
Then $F$ is a partition of $S$.
== Proof ==
Let $\RR_f \subseteq S \times S$ be the relation induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Then from Relation Induced by Mapping is Equivalence Relation, $\RR_f$ is an equivalence relation.
The result follows from Relation Partitions Set iff Equivalence.
Category:Mapping Theory
Mapping Measurable iff Measurable on Generator

== Theorem ==
Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Suppose that $\Sigma'$ is generated by $\GG'$.
Then a mapping $f: X \to X'$ is $\Sigma \, / \, \Sigma'$-measurable :
:$\forall G' \in \GG': \map {f^{-1} } {G'} \in \Sigma$
That is, the preimage of every generator under $f$ is a measurable set.
== Proof ==
=== Necessary Condition ===
Let $f$ be $\Sigma \, / \, \Sigma'$-measurable.
By definition of generated $\sigma$-algebra $\GG' \subseteq \Sigma'$.
Hence, in particular, $f$ satisfies:
:$\forall G' \in \GG': \map {f^{-1} } {G'} \in \Sigma$
=== Sufficient Condition ===
Suppose that:
:$\forall G' \in \GG': \map {f^{-1} } {G'} \in \Sigma$
Consider the pre-image $\sigma$-algebra $\Sigma' '$ on $X'$.
The supposition precisely states that $\GG' \subseteq \Sigma' '$.
By definition of generated $\sigma$-algebra, $\map \sigma {\GG'} \subseteq \Sigma' '$.
By definition of $\Sigma' '$, this precisely means:
:$\forall E' \in \Sigma': \map {f^{-1} } {E'} \in \Sigma$
That is, $f$ is $\Sigma \, / \, \Sigma'$-measurable.
== Sources ==
* : $7.2$
Category:Measure Theory
Mapping Preserves Directed Suprema implies Mapping is Continuous

== Theorem ==
Let $\struct {S, \preceq_1, \tau_1}$ and $\struct {T, \preceq_2, \tau_2}$ be up-complete ordered sets with Scott topologies.
Let $f: S \to T$ be a directed suprema preserving mapping.
Then $f$ is continuous.
== Proof ==
Let $P$ be a closed subset of $T$.
By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
:$P$ is lower and closed under directed suprema.
We will prove that
:for all directed subset $D$ of $S$: $D \subseteq f^{-1} \sqbrk P \implies \sup D \in f^{-1} \sqbrk P$
Let $D$ be a directed subset of $S$ such that
:$D \subseteq f^{-1} \sqbrk P$
By definition of mapping preserves directed suprema:
:$f$ preserves the supremum of $D$.
By definition of up-complete:
:$D$ admits a supremum.
By definition of mapping preserves the supremum:
:$\map \sup {f \sqbrk D} = \map f {\sup D}$
By Directed Suprema Preserving Mapping is Increasing:
:$f$ is increasing.
By Image of Directed Subset under Increasing Mapping is Directed:
:$f \sqbrk D$ is directed.
By Preimage is Subset of Set implies Image of Set is Subset:
:$f \sqbrk D \subseteq P$
By definition of closed under directed suprema:
:$\map \sup {f \sqbrk D} \in P$
Thus by definition of preimage of set:
:$\sup D \in f^{-1} \sqbrk P$
By definition:
:$f^{-1} \sqbrk P$ is closed under directed suprema.
By Preimage of Lower Section under Increasing Mapping is Lower:
:$f^{-1} \sqbrk P$ is lower.
Thus by Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
:$f^{-1} \sqbrk P$ is closed.
== Sources ==
* 
* 
Category:Topological Order Theory
Mapping Preserves Finite and Directed Suprema

== Theorem ==
Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ be join semilattices.
Let $f: S_1 \to S_2$ be a mapping.
Let $f$ preserve finite suprema and preserve directed suprema.
Then $f$ also preserves all suprema
== Proof ==
This follows by mutatis mutandis of the proof of Mapping Preserves Finite and Filtered Infima.
== Sources ==
* 
* 
Category:Order Theory
Category:Join and Meet Semilattices
Mapping Preserves Infima implies Mapping is Continuous in Lower Topological Lattice

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ and $Q = \struct {X, \preceq', \tau'}$ be complete topological lattices with lower topologies.
Let $f: S \to X$ be a mapping such that
:$f$ preserves all infima.
Then $f$ is continuous mapping.
== Proof ==
By assumption:
:for all non-empty subsets $Y$ of $S$: $f$ preserves the infimum of $Y$.
Thus by Mapping Preserves Non-Empty Infima implies Mapping is Continuous in Lower Topological Lattice:
:$f$ is continuous mapping.
== Sources ==
* 
Category:Topological Order Theory
Mapping at Element is Supremum implies Mapping is Increasing

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $\struct {T, \vee_2, \wedge_2, \precsim}$ be a complete lattice.
Let $f: S \to T$ be a mapping such that:
:$\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$
Then $f$ is an increasing mapping.
== Proof ==
Let $x, y \in S$ such that:
:$x \preceq y$
By Preceding implies Way Below Closure is Subset of Way Below Closure:
:$x^\ll \subseteq y^\ll$
By definitions of image of set and way below closure:
:$f \sqbrk {x^\ll} = \set {\map f w: w \in S \land w \ll x}$
and
:$f \sqbrk {y^\ll} = \set {\map f w: w \in S \land w \ll y}$
where $f \sqbrk {x^\ll}$ denotes the image of $x^\ll$ under $f$.
By Image of Subset under Mapping is Subset of Image:
:$f \sqbrk {x^\ll} \subseteq f \sqbrk {y^\ll}$
Thus
== Sources ==
* 
Category:Increasing Mappings
Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below

== Theorem ==
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.
Let $\left({T, \vee_2, \wedge_2, \precsim}\right)$ be a continuous complete lattice.
Let $f: S \to T$ be a mapping such that
:$\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
Let $x \in S, y \in T$.
Then
:$y \ll f\left({x}\right) \iff \exists w \in S: w \ll x \land y \ll f\left({w}\right)$
== Proof ==
By Mapping at Element is Supremum implies Mapping is Increasing:
:$f$ is an increasing mapping.
=== Sufficient Condition ===
Let $y \ll f\left({x}\right)$
By Way Below Closure is Directed in Bounded Below Join Semilattice:
:$x^\ll$ is directed
where $x^\ll$ denotes the way below closure of $x$.
By Image of Directed Subset under Increasing Mapping is Directed:
:$f\left[{x^\ll}\right]$ is directed
where $f\left[{x^\ll}\right]$ denotes the image of $x^\ll$ under $f$.
By assumption:
:$f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
By definitions of image of set and way below closure:
:$f\left[{x^\ll}\right] = \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
By Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element:
:$\exists z \in f\left[{x^\ll}\right]: y \ll z$
By definition of image of set:
:$\exists w \in x^\ll: f\left({w}\right) = z$
Thus by definition of way below closure:
:$w \ll x$
Thus $y \ll f\left({w}\right)$
=== Necessary Condition ===
Let
:$\exists w \in S: w \ll x \land y \ll f\left({w}\right)$
By Way Below implies Preceding:
:$w \preceq x$
By definition of increasing mapping:
:$f\left({w}\right) \precsim f\left({x}\right)$
Thus by Preceding and Way Below implies Way Below:
:$y \ll f\left({x}\right)$
== Sources ==
* 
Category:Way Below Relation
Category:Continuous Lattices
Mapping at Element is Supremum of Compact Elements implies Mapping at Element is Supremum that Way Below

== Theorem ==
Let $\struct{S, \vee_1, \wedge_1, \preceq_1}$ and $\struct{T, \vee_2, \wedge_2, \preceq_2}$ be complete lattices.
Let $f: S \to T$ be a mapping such that
:$\forall x \in S: \map f x = \sup \leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset{}$
Then
:$\forall x \in S: \map f x = \sup \set{ \map f w: w \in S \land w \ll x}$
== Proof ==
Let $x \in S$.
Define $X = \leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset{}$
Define $Y = \set { \map f w: w \in S \land w \ll x}$
We will prove that
:$X \subseteq Y$
Let $b \in X$.
By definition of $X$:
:$\exists w \in S: b = \map f w \land w \preceq_1 x \land w$ is compact.
By definition of compact element:
:$w \ll w$
By Preceding and Way Below implies Way Below:
:$w \ll x$
Thus by definition of $Y$:
:$b \in Y$
We will prove that
:$\map f x$ is upper bound for $Y$.
Let $b \in Y$.
By definition of $Y$:
:$\exists w \in S: b = \map f w \land w \ll x$
By Way Below implies Preceding:
:$w \preceq_1 x$
By Mapping at Element is Supremum implies Mapping is Increasing:
:$f$ is an increasing mapping.
Thus by definition of increasing mapping:
:$b \preceq_2 \map f x$
We will prove that
:$\forall b \in T: b$ is upper bound for $Y \implies \map f x \preceq_2 b$
Let $b \in T$ such that
:$b$ is upper bound for $Y$.
By Upper Bound for Subset:
:$b$ is upper bound for $X$.
By assumption:
:$\map f x = \sup X$
Thus by definition of supremum:
:$\map f x \preceq_2 b$
Thus by definition of supremum:
:$\map f x = \sup Y$
== Sources ==
* 
* 
Category:Way Below Relation
Mapping at Element is Supremum of Compact Elements implies Mapping is Increasing

== Theorem ==
Let $\struct {S, \vee_1, \wedge_1, \preceq_1}$ be a lattice.
Let $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be a complete lattice.
Let $f: S \to T$ be a mapping such that
:$\forall x \in S: \map f x = \sup \leftset {\map f w : w \in S \land w \preceq_1 x \land w}$ is compact$\rightset{}$
Then $f$ is increasing.
== Proof ==
Let $x, y \in S$ such that
:$x \preceq_1 y$
By Compact Closure is Increasing:
:$x^{\mathrm {compact} } \subseteq y^{\mathrm {compact} }$
By Image of Subset under Mapping is Subset of Image:
:$f \sqbrk {x^{\mathrm {compact} } } \subseteq f \sqbrk {y^{\mathrm {compact} } }$
By assumption:
:$\map f x = \sup \leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset{}$
and
:$\map f y = \sup \leftset {\map f w: w \in S \land w \preceq_1 y \land w}$ is compact$\rightset{}$
By definitions of image of set and compact closure:
:$\map f x = \map \sup {f \sqbrk {x^{\mathrm {compact} } } }$
and
:$\map f y = \map \sup {f \sqbrk {y^{\mathrm {compact} } } }$
Thus by Supremum of Subset and definition of complete lattice:
:$\map f x \preceq_2 \map f y$
== Sources ==
* 
* 
Category:Way Below Relation
Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping Preserves Directed Suprema

== Theorem ==
Let $\struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be complete lattices.
Let $f: S \to T$ be a mapping such that
:for all directed sets $\struct {D, \precsim}$ and nets $N:D \to S$ in $S$: $\map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$
Then $f$ preserves directed suprema.
== Proof ==
Let $D$ be a directed subset of $S$.
Assume that
:$D$ admits a supremum.
Thus by definition of complete lattice:
:$f \sqbrk D$ admits a supremum.
Thus by Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Supremum of Image is Mapping at Supremum of Directed Subset:
:$\map \sup {f \sqbrk D} = \map f {\sup D}$
== Sources ==
* 
* 
Category:Order Theory
Category:Limits Inferior of Nets
Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping is Increasing

== Theorem ==
Let $\struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be lattices.
Let $f: S \to T$ be a mapping such that:
:for all directed set $\struct {D, \precsim}$ and nets $N:D \to S$ in $S$: $\map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$
Then $f$ is an increasing mapping.
== Proof ==
Let $a, b \in S$ such that
:$a \preceq_1 b$
Define $M = \struct {\N, \le}$ being an ordered set.
We will prove that:
:$M$ is a directed set.
Let $x, y \in \N$.
Thus by definition of max operation:
:$\max \set {x, y} \in \N$
Thus by definition of max operation:
:$x \le \max \set {x, y}$ and $y \le \max \set {x, y}$
Define $g = \sequence {c_i}_{i \mathop \in \N} = \tuple {a, b, a, b, \dots}: \N \to S$ being a net in $S$.
By Limit Inferior of Repetition Net:
:$\liminf \sequence {c_i}_{i \mathop \in \N} = a \wedge_1 b$
By Preceding iff Meet equals Less Operand:
:$\map f {\liminf \sequence {c_i}_{i \mathop \in \N} } = \map f a$
By definition of composition of mappings:
:$f \circ g = \sequence {\map f {c_i} }_{i \mathop \in \N} = \tuple {\map f a, \map f b, \map f a, \map f b, \dots}$
By Limit Inferior of Repetition Net:
:$\map \liminf {f \circ g} = \map f a \wedge_2 \map f b$
By assumption:
:$\map f a \preceq_2 \map f a \wedge_2 \map f b$
By Meet Precedes Operands:
:$\map f a \wedge_2 \map f b \preceq_2 \map f a$
By definition of antisymmetry:
:$\map f a = \map f a \wedge_2 \map f b$
Thus by Preceding iff Meet equals Less Operand:
:$\map f a \preceq_2 \map f b$
== Sources ==
* 
Category:Increasing Mappings
Category:Limits Inferior of Nets
Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Supremum of Image is Mapping at Supremum of Directed Subset

== Theorem ==
Let $\struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be up-complete lattices.
Let $f: S \to T$ be a mapping such that
:for all directed set $\struct {D, \precsim}$ and net $N: D \to S$ in $S: \map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$
Let $D$ be a directed subset of $S$.
Then: 
:$\map \sup {f \sqbrk D} = \map f {\sup D}$
where $f \sqbrk D$ denotes the image of $D$ under $f$.
== Proof ==
By Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping is Increasing:
:$f$ is an increasing mapping.
By Image of Directed Subset under Increasing Mapping is Directed:
:$f \sqbrk D$ is directed.
By definition of up-complete:
:$D$ and $f \sqbrk D$ admit suprema.
By Subset and Image Admit Suprema and Mapping is Increasing implies Supremum of Image Precedes Mapping at Supremum:
:$\map \sup {f \sqbrk D} \preceq_2 \map f {\sup D}$
By Limit Inferior of Inclusion Net is Supremum of Directed Subset:
:$\sup D = \liminf i_D$
By assumption:
:$\map f {\sup D} \preceq_2 \map \liminf {f \circ i_D}$
By Composition of Mapping and Inclusion is Restriction of Mapping:
:$f \circ i_D = f {\restriction} D$
By Limit Inferior of Restriction Net is Supremum of Image of Directed Subset:
:$\map \sup {f \sqbrk D} = \map \liminf {f \circ i_D}$
Then
:$\map f {\sup D} \preceq_2 \map \sup {f \sqbrk D}$
Thus by definition of antisymmetry:
:$\map \sup {f \sqbrk D} = \map f {\sup D}$
== Sources ==
* 
* 
Category:Order Theory
Category:Limits Inferior of Nets
Mapping between Euclidean Spaces Measurable iff Components Measurable

== Theorem ==
Let $\R^n$ and $\R^m$ be Euclidean spaces.
Denote by $\BB^n$ and $\BB^m$ their respective Borel $\sigma$-algebras.
Denote with $\BB$ the Borel $\sigma$-algebra on $\R$.
Let $f: \R^n \to \R^m$ be a mapping, and write:
:$\map f {\mathbf x} = \begin{bmatrix} \map {f_1} {\mathbf x} \\ \vdots \\ \map {f_m} {\mathbf x} \end{bmatrix}$
with, for $1 \le i \le m$, $f_i: \R^n \to \R$.
Then $f$ is $\BB^n \, / \, \BB^m$-measurable :
:$\forall i: f_i: \R^n \to \R$ is $\BB^n \, / \, \BB$-measurable
== Proof ==
=== Necessary Condition ===
Suppose that $f$ is $\BB^n \, / \, \BB^m$-measurable.
It is to be shown that for $1 \le i \le m$, $f_i: \R^n \to \R$ is $\BB^n \, / \, \BB$-measurable.
By Mapping Measurable iff Measurable on Generator and Characterization of Euclidean Borel Sigma-Algebra, it suffices to show that:
:$\map {f_i^{-1} } {\openint a b} \in \BB^n$
for every open real interval $\openint a b$.
Since $\map f {\mathbf x} = \mathbf y$ , for all $i$, $\map {f_i} {\mathbf x} = y_i$, it follows that:
:$\mathbf x \in \map {f^{-1} } {\mathbf y} \iff \forall i: \mathbf x \in \map {f_i^{-1} } {y_i}$
That is, $\ds \mathbf x \in \bigcap_{i \mathop = 1}^m \map {f_i^{-1} } {y_i}$.
By Preimage of Union under Mapping, it follows that for any $B \in \BB^m$:
:$\mathbf x \in \map {f^{-1} } B \iff \exists \mathbf y \in B: \forall i: \mathbf x \in \map {f_i^{-1} } {y_i}$
In particular, consider the open set $B = \set {\mathbf y \in \R^n: y_i \in \openint a b}$.
Then by construction of $B$, $\mathbf x \in \map {f_i^{-1} } {\openint a b}$ $\map f x \in B$.
Therefore, $\map {f_i^{-1} } {\openint a b} = \map {f^{-1} } B$.
Since $B$ is open, it is also measurable by Characterization of Euclidean Borel Sigma-Algebra.
Thus $\map {f_i^{-1} } {\openint a b} \in \BB^n$ as $f$ was assumed $\BB^n \, / \, \BB^m$-measurable.
=== Sufficient Condition ===
Suppose that all $f_i$ are $\BB^n \, / \, \BB$-measurable.
It is to be shown that $f$ is $\BB^n \, / \, \BB^m$-measurable.
By Mapping Measurable iff Measurable on Generator and Characterization of Euclidean Borel Sigma-Algebra, it suffices to show that:
:$\map {f^{-1} } {\horectr {\mathbf a} {\mathbf b}} \in \BB^n$
for every half-open $m$-rectangle $\horectr {\mathbf a} {\mathbf b}$.
Now observe, for all $\mathbf x \in \R^n$:
Thus:
:$\ds \map {f^{-1} } {\horectr {\mathbf a} {\mathbf b} } = \bigcap_{i \mathop = 1}^m \map {f_i^{-1} } {\hointr {a_i} {b_i} }$
Now the $f_i^{-1} \hointr {a_i} {b_i}$ are measurable sets since the $f_i$ are $\BB^n \, / \, \BB$-measurable.
Since $\BB^n$ is a $\sigma$-algebra, Sigma-Algebra Closed under Countable Intersection applies, and it follows that:
:$\map {f^{-1} } {\horectr {\mathbf a} {\mathbf b} } \in \BB^n$
== Sources ==
* : $\S 7$: Problem $5$
Category:Measure Theory
Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a homeomorphism.
Let $H \subseteq S_1$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace of $T_1$ under the subspace topology $\tau_H$ induced by $\tau_1$.
Let $K = f \sqbrk H$ be the image of $H$ under $f$.
Let $T_K = \struct {K, \tau_K}$ be the topological subspace of $T_2$ under the subspace topology $\tau_K$ induced by $\tau_2$.
Let $g: H \to K$ be the restriction of $f$ to $H$.
Then $g$ is a homeomorphism.
=== Corollary ===
== Proof ==
Let $U \in \tau_K$ be open in $K$.
Then either:
:$U \in \tau_2$
or:
:$U = K \cap V$
where $V \in \tau_2$.
Suppose $U \in \tau_2$.
Because $f$ is continuous:
:$f^{-1} \sqbrk U \in \tau_1$
As $U \subseteq K$ we have that $g^{-1} \sqbrk U \subseteq H$.
So by definition of subspace topology, $g^{-1} \sqbrk U$ is open in $H$.
Now suppose $U = K \cap V$ for some $V \in \tau_2$.
Then:
But as $f$ is continuous:
:$f^{-1} \sqbrk V \in \tau_1$
As $f^{-1} \sqbrk K$ it follows that:
:$f^{-1} \sqbrk U = H \cap f^{-1} \sqbrk U$
Hence:
:$g^{-1} \sqbrk U \in \tau_H$
and the result follows.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 17$
Category:Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism
Category:Homeomorphisms (Topological Spaces)
Category:Topological Subspaces
Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism/Corollary

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a homeomorphism.
Let $H \subseteq S_1$.
Let $\overline H$ denote the complement $S_1 \setminus H$ of $H$ relative to $S_1$.
Let $T_{\overline H} = \struct {\overline H, \tau_{\overline H} }$ be the topological subspace of $T_1$ under the subspace topology $\tau_{\overline H}$ induced by $\tau_1$.
Let $\overline K = f \sqbrk {\overline H}$ be the image of $\overline H$ under $f$.
Let $T_{\overline K} = \struct {\overline K, \tau_{\overline K} }$ be the topological subspace of $T_2$ under the subspace topology $\tau_{\overline K}$ induced by $\tau_2$.
Let $h: \overline H \to \overline K$ be the restriction of $f$ to $\overline H$.
Then $h$ is a homeomorphism.
== Proof ==
By definition of relative complement, $\overline H \subseteq S_1$.
As $f$ is a homeomorphism, it is a fortiori a bijection.
Hence from Image of Relative Complement under Bijection is Relative Complement of Image:
:$f \sqbrk {\overline H} = \overline K$
Hence Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism applies directly, using $\overline H$ for $H$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 17$
Category:Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism
Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $g \in G$.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\phi_g: \struct {\Z, +} \to \struct {G, \circ}$ be the mapping defined as:
:$\forall k \in \Z: \map {\phi_g} k = g^k$
Then $\phi_g$ is a (group) homomorphism.
== Proof ==
Let $k, l \in \Z$.
thus proving that $\phi_g$ is a homomorphism as required.
== Sources ==
* : $\S 7.1$. Homomorphisms: Example $129$
Category:Additive Group of Integers
Category:Examples of Group Homomorphisms
Mapping from Cartesian Product under Chebyshev Distance to Real Number Line is Continuous

== Theorem ==
Let $M = \struct {A, d'}$ be a metric space.
Let $\ds \AA = A \times A$ be the cartesian product of $A$ with itself.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max \set {\map {d'} {x_1, y_1}, \map {d'} {x_2, y_2} }$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Then the mapping:
:$d': \struct {A \times A, d_\infty} \to \struct {\R, d}$
where $d$ is the usual metric, is continuous.
== Proof ==
From definition of continuous mapping:
We just need to show that:
:$\forall \tuple {a, b} \in A \times A: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall \tuple {x, y} \in A \times A: \map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \delta \implies \map d {\map {d'} {x, y}, \map {d'} {a, b}} < \epsilon$
Let $\tuple {a, b} \in A \times A$ and $\epsilon \in \R_{>0}$.
Let $\tuple {x, y} \in A \times A$.
Suppose $\map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \dfrac \epsilon 2$.
Then:
Therefore $d'$ is continuous.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $9$
Category:Continuous Mappings
Category:Chebyshev Distance
Mapping from Discrete Space is Continuous

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be the discrete topological space on $S_1$.
Let $T_2 = \struct {S_2, \tau_2}$ be any other topological space.
Let $\phi: S_1 \to S_2$ be a mapping.
Then $\phi$ is continuous.
== Proof ==
From the definition of continuous:
:$U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$
But as $\phi^{-1} \sqbrk U \subseteq S_1$ it follows from the definition of discrete space that $\phi^{-1} \sqbrk U \in \tau_1$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 6$
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $5$
Category:Discrete Topology