nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Symmetric Closure

== Theorem ==
Let $\RR$ be a relation on a set $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
First we note that from Union of Relation with Inverse is Symmetric Relation, $\RR \cup \RR^{-1}$ is a symmetric relation.
=== Definition $(1)$ implies Definition $(2)$ ===
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$ by definition $1$.
Then by definition:
:$\RR^\leftrightarrow = \RR \cup \RR^{-1}$
Let $\QQ$ be a symmetric relation such that:
:$\RR \subseteq \QQ$
Let $\tuple {a, b} \in \RR \cup \RR^{-1}$.
By definition of set union, either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {a, b} \in \RR^{-1}$
;Case $1$
Let $\tuple {a, b} \in \RR$.
Then by definition of subset:
:$\tuple {a, b} \in \QQ$
Let $\tuple {a, b} \in \RR^{-1}$.
Then by definition of inverse relation:
:$\tuple {b, a} \in \RR$
Then by definition of subset:
:$\tuple {b, a} \in \QQ$
But $\QQ$ is symmetric.
Hence:
:$\tuple {a, b} \in \QQ$
So by Proof by Cases:
:$\tuple {a, b} \in \RR \cup \RR^{-1} \implies \tuple {a, b} \in \QQ$
Hence $\RR \cup \RR^{-1} \subseteq \QQ$
It follows that $\RR \cup \RR^{-1}$ is the smallest symmetric relation the subset ordering.
Thus $\RR^\leftrightarrow$ is the symmetric closure of $\RR$ by definition $2$.
=== Definition $(2)$ implies Definition $(1)$ ===
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$ by definition $2$.
Then by definition:
:$\RR^\leftrightarrow$ is the smallest symmetric relation on $S$ which has $\RR$ as a subset.
Hence as $\RR \cup \RR^{-1}$ is a symmetric relation :
:$\RR^\leftrightarrow \subseteq \RR \cup \RR^{-1}$
But also , for every symmetric relation $\QQ$ such that $\RR \subseteq \QQ$:
:$\RR \cup \RR^{-1} \subseteq \QQ$
Hence:
:$\RR \cup \RR^{-1} \subseteq \RR^\leftrightarrow$
Thus by set equality:
:$\RR \cup \RR^{-1} = \RR^\leftrightarrow$
That is, $\RR^\leftrightarrow$ is the symmetric closure of $\RR$ by definition $1$.
Category:Symmetric Closures
Equivalence of Definitions of Symmetric Difference

== Theorem ==
Let $S$ and $T$ be sets.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
== Proof ==
=== $(1)$ iff $(2)$ ===
=== $(1)$ iff $(3)$ ===
=== $(2)$ iff $(4)$ ===
=== $(2)$ iff $(5)$ ===
=== $(3)$ iff $(5)$ ===
Category:Symmetric Difference
Category:Set Difference
Category:Set Intersection
Category:Set Union
Category:Equivalence of Definitions of Symmetric Difference
Equivalence of Definitions of Symmetric Difference/(1) iff (2)

== Theorem ==
Let $S$ and $T$ be sets.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
== Sources ==
* : Exercise $1.1: \ 8 \ \text{(g)}$
* : $\S 1$: Sets and Functions: Problem $4 \ \text{(b)}$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \alpha$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $14$
Category:Equivalence of Definitions of Symmetric Difference
Equivalence of Definitions of Symmetric Matrices

== Theorem ==
Let $\mathbf A$ be a square matrix over a set $S$.
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
=== Definition $(1)$ implies Definition $(2)$ ===
Let $\mathbf A$ be a symmetric matrix by definition $1$.
Thus $\mathbf A$ is a symmetric matrix by definition $2$.
=== Definition $(2)$ implies Definition $(1)$ ===
Let $\mathbf A$ be a symmetric matrix by definition $2$.
Thus $\mathbf A$ is a symmetric matrix by definition $1$.
== Sources ==
* 
* 
* 
Category:Symmetric Matrices
Equivalence of Definitions of Symmetric Relation

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 3 ===
Let $\RR$ be a relation which fulfils the condition:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR$
Then:
Hence $\RR$ is symmetric by definition 3.
=== Definition 3 implies Definition 2 ===
Let $\RR$ be a relation which fulfils the condition:
:$\RR \subseteq \RR^{-1}$
Then by Inverse Relation Equal iff Subset:
:$\RR = \RR^{-1}$
Hence $\RR$ is symmetric by definition 2.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfils the condition:
:$\RR^{-1} = \RR$
Then:
Hence $\RR$ is symmetric by definition 1.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Exercise $10.6 \ \text{(b)}$
* : Chapter $\text{I}$: Sets and Functions: Relations: Theorem $3$
* : $\S 1.19$: Some Important Properties of Relations: Exercise $5$
* : $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.14 \ \text{(a)}$
Category:Symmetric Relations
Equivalence of Definitions of Synthetic Basis

== Theorem ==
Let $S$ be a set.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== 1 implies 2 ===
Let $U, V \in \BB$.
Let $x \in U \cap V$.
By hypothesis:
:$\ds \exists \AA \subseteq \BB: U \cap V = \bigcup \AA$
By definition of union, $\exists W \in\AA : x \in W$.
By Set is Subset of Union: General Result, $W \subset U \cap V$.
Therefore:
:$\ds \forall x \in A \cap B: \exists W \in \AA \subseteq \BB: x \in W \subseteq U \cap V$
=== 2 implies 1 ===
Let $U, V \in \BB$.
Define the set:
:$\ds \AA = \set {W \in \BB: W \subseteq U \cap V} \subseteq \BB$
By Union is Smallest Superset: General Result:
:$\ds \bigcup \AA \subseteq U \cap V$
By hypothesis:
:$\ds \forall x \in U \cap V: \exists W \in \AA: x \in W$
Thus $\ds U \cap V \subseteq \bigcup \AA$
By definition of set equality:
:$\ds U \cap V = \bigcup \AA$
Category:Topological Bases
Equivalence of Definitions of T0 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition by Open Sets ===
=== Definition by Limit Points ===
== Proof ==
=== Definition by Open Sets implies Definition by Limit Points ===
Let $T = \struct {S, \tau}$ be a topological space for which:
:$\forall x, y \in S$, either:
::$(1): \quad \exists U \in \tau: x \in U, y \notin U$
::$(2): \quad \exists U \in \tau: y \in U, x \notin U$
Let $x, y \in S$ such that $x \ne y$.
By the definition of limit point of a point:
:$(3): \quad x$ is a limit point of $y$ if every open set $U \in \tau$ such that $x \in U$ contains $y$.
:$(4): \quad y$ is a limit point of $x$ if every open set $U \in \tau$ such that $y \in U$ contains $x$.
Thus:
:If $(3)$ holds then $(1)$ is false.
:If $(4)$ holds then $(2)$ is false.
So both $(3)$ and $(4)$ can not both hold at once.
Hence $T = \struct {S, \tau}$ is a topological space for which no two points can be limit points of each other.
=== Definition by Limit Points implies Definition by Open Sets ===
Let $T = \struct {S, \tau}$ be a topological space for which no two points can be limit points of each other.
Let $x, y \in S$ such that $x \ne y$.
 that:
:$(1): \quad \nexists U \in \tau: x \in U, y \notin U$
and
:$(2): \quad \nexists U \in \tau: y \in U, x \notin U$
Expressing this in different terms:
:$(3): \quad$ Every open set $U \in \tau$ such that $x \in U$ contains $y$.
:$(4): \quad$ Every open set $U \in \tau$ such that $y \in U$ contains $x$.
That is, by definition:
:$x$ is a limit point of $y$ and $y$ is a limit point of $x$.
From this contradiction we deduce that:
:$\forall x, y \in S$, either:
::$\exists U \in \tau: x \in U, y \notin U$
:or:
::$\exists U \in \tau: y \in U, x \notin U$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:T0 Spaces
Equivalence of Definitions of T1 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition by Open Sets ===
=== Definition by Closed Points ===
== Proof ==
=== Definition by Open Sets implies Definition by Closed Points ===
Let $T = \struct {S, \tau}$ be a topological space for which:
:$\forall x, y \in S$, both:
::$(1): \quad \exists U \in \tau: x \in U, y \notin U$
and:
::$(2): \quad \exists U \in \tau: y \in U, x \notin U$
Let $x, y \in S$.
By the definition of limit point of a point, the above condition means:
:$(3): \quad y$ is a limit point of $x$ if every open set $U \in \tau$ such that $y \in U$ contains $x$.
Thus $(2)$ and $(3)$ give us that $y$ is not a limit point of $x$.
As $x$ and $y$ are any two points in $S$, it follows that $x$ has no limit points.
Thus it holds vacuously that all the limit point of $x$ are in $\set x$.
By Closed Set iff Contains all its Limit Points, we have that $\set x$ is a closed set in $T$.
This holds for all $x \in S$.
Hence $T = \struct {S, \tau}$ is a topological space for which all points are closed.
=== Definition by Closed Points implies Definition by Open Sets ===
$T = \struct {S, \tau}$ is a topological space for which all points are closed.
Let $x, y \in S$.
By Closed Set iff Contains all its Limit Points, we have that:
:all the limit points of $x$ are in $\set x$
:all the limit points of $y$ are in $\set y$.
It follows by definition of limit point (reversing the above argument) that:
:$\exists U \in \tau: x \in U, y \notin U$
and
:$\exists U \in \tau: y \in U, x \notin U$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
* : Problems: $\S 2: \ 7$
Category:T1 Spaces
Equivalence of Definitions of T4 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition by Open Sets ===
=== Definition by Closed Neighborhoods ===
== Proof ==
=== Definition by Open Sets implies Definition by Closed Neighborhoods ===
Let $T$ satisfy the definition by open sets of a $T_4$ space.
Let $A$ be a closed set in $T$, and let $U_A$ be an open neighborhood of $A$.
By definition of open neighborhood:
:$A \subseteq U_A$
Let $B := \relcomp S {U_A}$.
By Intersection with Complement is Empty iff Subset:
:$A \cap B = \O$
As $B$ is the complement of $U_A$ in $S$, it is by definition closed in $T$.
Thus $A$ and $B$ are disjoint closed sets in $T$.
Also, from Set Complement inverts Subsets:
:$B \subseteq \relcomp S A$
By assumption, there exist disjoint open sets $U$ and $V$ such that $A \subseteq U$ and $B \subseteq V$.
From Set Intersection Preserves Subsets:
:$B \subseteq \map \complement A \cap V$
Note that the latter set, being an intersection of open sets, is itself open.
Then, from Complement of Complement, De Morgan's Laws: Complement of Union and Set Complement inverts Subsets:
:$A \cup \map \complement V \subseteq U_A$
From Subset of Union we also have:
:$A \subseteq A \cup \map \complement V$
Since $\relcomp S A \cap V$ is open, $A \cup \relcomp S V$ is closed.
Hence we have found a closed neighborhood for $A$ in $U_A$, as desired.
Hence it is concluded that $T$ satisfies the definition by closed neighborhoods of a $T_4$ space.
=== Definition by Closed Neighborhoods implies Definition by Open Sets ===
Let $T$ satisfy the definition by closed neighborhoods of a $T_4$ space.
Let $A$ and $B$ be disjoint closed sets in $T$.
Then from Empty Intersection iff Subset of Complement, we have:
:$A \subseteq \map \complement B$
and the latter is open in $T$.
Applying the assumption, we find a closed neighborhood $C_A$ of $A$ contained in $\map \complement B$.
From Empty Intersection iff Subset of Complement and Set Complement inverts Subsets we establish:
:$A \cap \map \complement {C_A} = \O$
:$B \subseteq \map \complement {C_A}$
Similarly, we find a closed neighborhood $C_B$ of $B$ contained in $\map \complement {C_A}$.
Then from Intersection with Complement is Empty iff Subset:
:$B \cap \map \complement {C_B} = \O$
But from Set Complement inverts Subsets, we have:
:$\map \complement {C_A} \subseteq \map \complement A$
and so from Subset Relation is Transitive:
:$C_B \subseteq \map \complement A$
Finally, another application of Set Complement inverts Subsets shows:
:$A \subseteq \map \complement {C_B}$
Since $C_B \subseteq \map \complement {C_A}$, Empty Intersection iff Subset of Complement shows that $C_A$ and $C_B$ are disjoint sets.
They are also open sets, being the complement of closed sets.
Above, we established also that:
:$A \subseteq \map \complement {C_B}$
:$B \subseteq \map \complement {C_A}$
Hence it is concluded that $T$ satisfies the definition by open sets as well.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:T4 Spaces
Equivalence of Definitions of T5 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition by Open Sets ===
=== Definition by Closed Neighborhoods ===
== Proof ==
=== Definition by Open Sets implies Definition by Closed Neighborhoods ===
Suppose that:
:$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
Let $Y, A \subseteq S$.
Suppose $A \subseteq Y^\circ$ and $A^- \subseteq Y$. 
Then:
We also have:
We have by hypothesis: 
:$\exists U, V \in \tau: A \subseteq U, \relcomp S Y \subseteq V, U \cap V = \O$
Then:
So we have demonstrated that there exists a closed neighborhood $\relcomp S V$ of $A$ contained in $Y$. 
As $Y$ and $A$ are arbitrary:
:$\forall Y, A \subseteq S: \paren {A \subseteq Y^\circ \wedge A^- \subseteq Y} \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$
=== Definition by Closed Neighborhoods implies Definition by Open Sets ===
Suppose that:
:$\forall Y, A \subseteq S: \paren {A \subseteq Y^\circ \wedge A^- \subseteq Y} \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$
Let $A, B \subseteq S$ be separated sets.
Then $A^- \cap B = A \cap B^- = \O$.
Then:
We also have:
We have by hypothesis that there exists a closed neighborhood $N$ of $A$ contained in $\relcomp S B$.
Then:
:$\exists U \in \tau: A \subseteq U \subseteq N \subseteq \relcomp S B$
Then we have:
Finally:
Therefore we have:
As $A$ and $B$ are arbitrary:
:$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:T5 Spaces
Equivalence of Definitions of Tangent Vector

== Theorem == 
Let $M$ be a smooth manifold.
Let $m \in M$ be a point. 
Let $V$ be an open neighborhood of $m$. 
Let $\map {C^\infty} {V, \R}$ be defined as the set of all smooth mappings $f: V \to \R$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 2 implies Definition 1 ===
Let $\lambda \in \R$ and $f, g \in \map {C^\infty} {V, \R}$. 
Thus $X_m$ is linear. 
Hence $X_m$ satisfies the Leibniz law. 
Thus $X_m$ satisfies Definition 1.
=== Definition 1 implies Definition 2 ===
=== Lemma 1 === 
Let $X_m$ be a tangent vector at $m \in M$ according to Definition 1. 
Let $V$ be an open neighborhood of $M$. 
Let $f \in \map {C^\infty} {V, \R}$ be constant. 
Then $\map {X_m} f = 0$.
=== Proof of Lemma 1 ===
Let $\map f m = 0$. 
Then, by constancy, $f = 0$ on $V$.
Hence, by linearity, $\map {X_m} 0 = 0$. 
Let $\map f m \ne 0$. 
$f$ is constant, $\exists \lambda \in \R : f \sqbrk V = \set \lambda$ $f = \lambda$.
Let $X_m$ be a tangent vector at $m \in M$ according to Definition 1.
Denote $n := \dim M$. 
Let $f \in \map {C^\infty} {V, \R}$.
Let $\struct {U, \kappa}$ be a chart such that $\map \kappa m = 0$. 
Let $\kappa^i$ be the $i$th coordinate function of the chart $\struct {U, \kappa}$. 
By, Taylor's Theorem/n Variables :
Observe that $\ds \map {\paren {f \circ \kappa^{-1} } } 0 = \map {\paren {f \circ \kappa^{-1} } } {\map \kappa m} = \map f m$ is a constant mapping on $V$.
Define $X^i := \map {X_m} {\kappa^i}$.
Then by linearity:
:$\ds \map {X_m} f = \map {X_m} {\map f m} + \sum_{i \mathop = 1}^n \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } 0 \ X^i + \map {X_m} {\map {\OO_2} \kappa}$
=== Lemma 2 === 
:$\map {X_m} {\map {\OO_2} \kappa} = 0$
=== Proof of Lemma 2 ===
By Taylor's Theorem/n Variables, for each summand of $\map {\OO_2} \kappa$ there exists $i \in \set {1, \ldots, n}$ and an $h \in \map {C^\infty} {V, \R}$ with $\map h m = 0$ such that the summand is $\kappa^i h$.
Thus the sum $\map {\OO_2} \kappa$ vanishes. 
By Lemma 1: 
:$\map {X_m} {\map f m} = 0$
By Lemma 2: 
:$\map {X_m} {\map {\OO_2} \kappa} = 0$
Hence:
:$\ds \map {X_m} f = \sum_{i \mathop = 1}^n \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } 0 \ X^i$
Let $\set {e_i}$ be a basis of $\R^n$ such that:
:$\ds \kappa = \sum_{i \mathop = 1}^n \kappa^i e_i$
Choose a smooth curve $\gamma: I \to M$ with $0 \in I \subseteq \R$ such that $\map \gamma 0 = m$ and:
:$\map {\dfrac {\d \kappa^i \circ \gamma} {\d \tau} } 0 := X^i$
Then:
Hence $X_m$ is a tangent vector according to Definition 2. 
This proves the assertion. 
== Sources ==
* : $\S 1.4$: Proposition $1.4.7$ 
Category:Tangent Vectors
Equivalence of Definitions of Tangent of Angle

== Theorem ==
Let $\theta$ be an angle.
=== Definition from Triangle ===
=== Definition from Circle ===
== Proof ==
=== Definition from Triangle implies Definition from Circle ===
Let $\tan \theta$ be defined as $\dfrac {\text{Opposite}} {\text{Adjacent}}$ in a right triangle.
Consider the triangle $\triangle OAB$.
By construction, $\angle OAB$ is a right angle.
Thus:
That is:
: $\tan \theta = AB$
=== Definition from Circle implies Definition from Triangle ===
Let $\tan \theta$ be defined as the length of $AB$ in the triangle $\triangle OAB$.
Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above.
We have that:
:$\angle CAB = \angle BOA = \theta$
:$\angle ABC = \angle OAB$ which is a right angle
Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar.
By definition of similarity:
That is:
:$\dfrac {\text{Opposite} } {\text{Adjacent} } = \tan \theta$
Category:Tangent Function
Equivalence of Definitions of Top of Lattice

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
By definition, $\top$ is the greatest element of $S$ for all $a \in S$:
:$a \preceq \top$
By Ordering in terms of Meet, this is equivalent to:
:$a \wedge \top = a$
If this equality holds for all $a \in S$, then by definition $\top$ is an identity for $\wedge$.
The result follows.
Category:Top of Lattice
Equivalence of Definitions of Topological Group

== Theorem ==
Let $\struct {G, \odot}$ be a group.
On its underlying set $G$, let $\struct {G, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\struct {G, \odot, \tau}$ be a topological group by Definition 1.
Let $\phi: \struct {G, \tau} \to \struct {G, \tau}$ be the mapping defined as:
:$\forall x \in G: \map \phi x = x^{-1}$
By definition:
:$\odot: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$ is a continuous mapping
:$\phi: \struct {G, \tau} \to \struct {G, \tau}$ is a continuous mapping
Let $\phi': G \times G \to G \times G$ be defined by:
:$\map {\phi'} {x, y} = \tuple {x, \map \phi y}$
Let $\phi_1: G \times G \to G$ be defined as:
:$\map {\phi_1} {x, y} = x$
Let $\phi_2: G \times G \to G$ be defined as:
:$\map {\phi_2} {x, y} = y^{-1}$
Then for arbitrary open set $V \subset G$, both sets
:$\phi_1^{-1} \sqbrk V = V \times G$
and
:$\phi_2^{-1} \sqbrk V = G \times \phi^{-1} \sqbrk V$
are open in $G \times G$, thus $\phi_1$ and $\phi_2$ are continuous.
Because:
:$\map {\phi'} {x, y} = \tuple {\map {\phi_1} x, \map {\phi_2} y}$
by Continuous Mapping to Product Space, $\phi'$ is continuous.
Let the mapping $\psi: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$ be defined as:
:$\psi = \odot \circ \phi'$
where $\circ$ denotes composition of mappings.
By Composite of Continuous Mappings is Continuous, $\psi$ is continuous.
Then:
demonstrating that:
:$\map \psi {x, y} = x \odot y^{-1}$
Thus $\struct {G, \odot, \tau}$ is a topological group by Definition 2.
=== Definition 2 implies Definition 1 ===
Let $\struct {G, \odot, \tau}$ be a topological group by Definition 2.
Let $e$ be the identity of $G$.
Let the mapping $\psi: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$ be defined as:
:$\forall \tuple {x, y} \in G \times G: \map \psi {x, y} = x \odot y^{-1}$
By definition of $\struct {G, \odot, \tau}$, $\psi$ is continuous.
By Continuous Mapping to Product Space $\psi$ is continuous in each variable.
Let $\phi: \struct {G, \tau} \to \struct {G, \tau}$ be the mapping defined as:
:$\forall x \in G: \map \phi x = x^{-1}$
Since $\map \phi x = \map \psi {e, x}$, it follows that $\phi$ is continuous.
Let $\phi': G \times G \to G \times G$ be defined by:
:$\forall \tuple {x, y} \in G \times G: \map {\phi'} {x, y} = \tuple {x, \map \phi y}$
Then $\phi'$ is continuous.
$\odot$ is the composition of $\psi$ with $\phi'$.
Thus by Composite of Continuous Mappings is Continuous, $\odot$ is continuous.
Thus $\struct {G, \odot, \tau}$ is a topological group by Definition 1.
Category:Topological Groups
Equivalence of Definitions of Topologically Equivalent Metrics

== Theorem ==
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $d_1$ and $d_2$ be topologically equivalent by definition 1.
Let $\struct {B, d}$ and $\struct {C, d'}$ be metric spaces:
Let $f: B \to A$ and $g: A \to C$ be mappings.
Then by definition:
:$f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous
:$g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.
Suppose $U \subseteq A$ is $d_2$-open.
Let $h: A \to A$ be the identity mapping.
That is:
:$\forall a \in A: \map h a = a$
From Identity Mapping is Continuous, $h$ is $\tuple {d_2, d_2}$-continuous.
Hence because $h$ is $\tuple {d_1, d'}$-continuous $h$ is $\tuple {d_2, d'}$-continuous, $h$ is $\tuple {d_1, d_2}$-continuous.
From Metric Space Continuity by Open Set, it follows that $h^{-1} \sqbrk U$ is $d_1$-open.
But $h^{-1} \sqbrk U = U$, so $U$ is $d_1$-open.
A similar argument, still starting with the same proposition that $g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous, shows that if $U$ is $d_1$-open then it is $d_2$-open.
Thus $d_1$ and $d_2$ are topologically equivalent by definition 2.
=== Definition 2 implies Definition 1 ===
Let $d_1$ and $d_2$ be topologically equivalent by definition 2.
Then by definition:
:$U \subseteq A$ is $d_1$-open $U \subseteq A$ is $d_2$-open.
From Metric Space Continuity by Open Set, we have that:
:$f$ is $\tuple {d, d_1}$-continuous for every set $U \subseteq A$ which is open in $\struct {A, d_1}$, $f^{-1} \sqbrk U$ is open in $\struct {B, d}$
:$f$ is $\tuple {d, d_2}$-continuous for every set $U \subseteq A$ which is open in $\struct {A, d_2}$, $f^{-1} \sqbrk U$ is open in $\struct {B, d}$.
Hence $f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous.
Similarly:
:$g$ is $\tuple {d_1, d'}$-continuous for every set $U \subseteq C$ which is open in $\struct {C, d'}$, $g^{-1} \sqbrk U$ is open in $\struct {A, d_1}$
:$g$ is $\tuple {d_2, d'}$-continuous for every set $U \subseteq C$ which is open in $\struct {C, d'}$, $g^{-1} \sqbrk U$ is open in $\struct {A, d_2}$.
Hence $g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.
Thus $d_1$ and $d_2$ are topologically equivalent by definition 1.
== Note ==
Note that from the proposition that $g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous, we show that $U \subseteq A$ is $d_1$-open $U \subseteq A$ is $d_2$-open.
From that, we show that both:
:$f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous
:$g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.
Hence the proposition that $f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous is superfluous, as it follows directly from the proposition concerning $g$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Proposition $2.4.2$
Category:Topologically Equivalent Metrics