nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Integer Modulo Addition

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $+_m$ be the operation of addition modulo $m$ by definition $1$.
Then $+_m$ is defined as:
:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
By the Division Theorem:
Hence $p \in \eqclass a m$
Thus:
Thus:
:$p + q \in \eqclass {a + b} m$
By the Division Theorem:
:$p + q = s + k m$
for some $k \in \Z$ and $0 \le s < m$
Then:
:$s \in \eqclass {p + q} m$
while $s$ equals the remainder after $p + q$ has been divided by $m$.
That is:
:$p +_m q$ equals the remainder after $p + q$ has been divided by $m$.
Thus $+_m$ is the operation of addition modulo $m$ by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $+_m$ be the operation of addition modulo $m$ by definition $2$.
Then by definition:
:$\forall a, b \in \Z_m: a +_m b$ equals the remainder after $a + b$ has been divided by $m$
where:
:$\Z_m = \set {0, 1, \ldots, m - 1}$
Thus by definition of residue class:
Let $s$ equal the remainder after $a + b$ has been divided by $m$:
:$a + b = s + k m$
where $k \in \Z$.
Indeed, in this context $k = 0$ or $k = 1$.
Hence:
:$s \in \eqclass {a + b} m$
So as Modulo Addition is Well-Defined:
:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
Thus $+_m$ is the operation of addition modulo $m$ by definition $1$.
=== $(2)$ implies $(3)$ ===
Let $+_m$ be the operation of addition modulo $m$ by definition $2$.
Then by definition:
:$\forall a, b \in \Z_m: a +_m b$ equals the remainder after $a + b$ has been divided by $m$
where:
:$\Z_m = \set {0, 1, \ldots, m - 1}$
Let $s$ equal the remainder after $a + b$ has been divided by $m$.
That is, there exists $k \in \Z$ such that:
:$a + b = s + k m$
where $0 \le s < m$.
It follows that:
:$k m \le a + b$
but:
:$\paren {k + 1} m > a + b$
Thus we have:
where $k$ is the largest integer such that $k m \le a + b$.
Thus $+_m$ is the operation of addition modulo $m$ by definition $3$.
=== $(3)$ implies $(2)$ ===
Let $+_m$ be the operation of addition modulo $m$ by definition $3$.
Then by definition:
:$a +_m b := a + b - k m$
where $k$ is the largest integer such that $k m \le a + b$.
Let $s$ be defined as:
:$s = a + b - k m$
By definition:
:$0 \le s < m$
That is, $s$ equals the remainder after $a + b$ has been divided by $m$.
Thus $+_m$ is the operation of addition modulo $m$ by definition $2$.
Category:Modulo Addition
Equivalence of Definitions of Integral Dependence

== Theorem ==
Let $A$ be an extension of a commutative ring with unity $R$.
For $x \in A$, the following are equivalent:
== Proof ==
=== $(1) \implies (2)$ ===
By hypothesis, there exist $r_0, \ldots, r_{n - 1} \in R$ such that:
:$x^n + r_{n - 1} x^{n - 1} + \cdots + r_1 x + r_0 = 0$
So the powers $x^k$, $k \ge n$ can be written as an $R$-linear combination of:
:$\set {1, \ldots, x^{n - 1} }$
Therefore this set generates $R \sqbrk x$.
=== $(2) \implies (3)$ ===
$B = R \sqbrk x$ trivially satisfies the required conditions.
=== $(3) \implies (4)$ ===
By $(3)$ we have an $R$-module $B$ such that $R \subseteq B$, $B$ is finitely generated over $R$.
Also, $x \in B$, so $x B \subseteq B$ as required.
=== $(4) \implies (5)$ ===
By $(4)$ we have an $R \sqbrk x$-module $B$ that is finitely generated over $R$.
Let $y$ lie in the annihilator $\map {\operatorname {Ann}_{R \sqbrk x} } B$
We have that $1 \in B$.
Then in particular $y \cdot 1 = 0$, and $y = 0$.
Therefore, $B$ is faithful over $R \sqbrk x$.
=== $(5) \implies (1)$ ===
Let $B$ be as in $(5)$, say generated by $m_1, \ldots, m_n \in B$.
Then there are $r_{i j} \in R$, $i, j = 1,\ldots, n$ such that:
:$\ds x \cdot m_i = \sum_{j \mathop = 1}^n r_{i j} m_j$
Let $b_{i j} = x \delta_{i j} - r_{i j}$ where $\delta_{i j}$ is the Kronecker delta.
Then:
:$\ds \sum_{j \mathop = 1}^n b_{i j} m_j = 0, \quad i = 1, \ldots, n$
So, let $M = \sqbrk {b_{i j} }_{1 \le i, j \le n}$.
Then by Cramer's Rule:
:$\paren {\det M} m_i = 0$, $i = 1, \ldots, n$
Since $\det M \in R \sqbrk x$, also $\det M \in \map {\operatorname {Ann}_{R \sqbrk x} } B$.
So $\det M = 0$ by hypothesis.
But $\det M = 0$ is a monic polynomial in $x$ with coefficients in $R$.
Thus $x$ is integral over $R$.
== Sources ==
* : Chapter $2$ Ring of Integers: Dedekind’s proof that the integral elements form a ring: Proposition $2.4$
Category:Algebraic Number Theory
Category:Commutative Algebra
Equivalence of Definitions of Integral Domain

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\struct {D, +, \circ}$ be an integral domain in sense $1$.
As $\struct {D, +, \circ}$ is already a commutative ring, it remains to show that $\struct {D^*, \circ}$ is a monoid.
Because $\circ$ is a ring product, and $\struct {D, +, \circ}$ has no zero divisors, we conclude Closure and Associativity.
Furthermore, $\struct {D, +, \circ}$ is non-null, hence $0_D \ne 1_D$, and we conclude $1_D \in D^*$.
Therefore, we also have an Identity for $\struct {D^*, \circ}$, and hence it is a monoid.
It remains to show that all elements of $\struct {D^*, \circ}$ are cancellable.
As $\struct {D, +, \circ}$ has no zero divisors, this follows from Ring Element is Zero Divisor iff not Cancellable.
=== $(2)$ implies $(1)$ ===
Let $\struct {D, +, \circ}$ be an integral domain in sense $2$.
$\struct {D, +, \circ}$ is already a commutative ring.
Furthermore, as $\struct {D^*, \circ}$ is a monoid, it is nonempty.
Also, we conclude that $\struct {D, +, \circ}$ is a non-null ring with unity.
It remains to show that $\struct {D, +, \circ}$ has no zero divisors.
We know all elements of $\struct {D^*, \circ}$ are cancellable.
From Ring Element is Zero Divisor iff not Cancellable, we conclude that $\struct {D, +, \circ}$ cannot have zero divisors.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.2$
Category:Integral Domains
Equivalence of Definitions of Integral Element of Algebra

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $f : A \to B$ be a commutative $A$-algebra.
Let $b\in B$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Assume $b$ is a root of a monic polynomial in $A \sqbrk x$.
That is, there are $n \in \N_{>0}$ and $a_1, \ldots , a_{n-1} \in A$ be such that:
:$b^n + a_{n - 1} b^{n-1} + \cdots + a_1 b + a_0 = 0$
That is:
:$b^n = - a_0 - a_1 b - \cdots - a_{n-1} b^{n-1}$
Therefore $\set {1, b, \ldots , b^{n-1} }$ generates $A \sqbrk b$.
=== Definition 2 implies Definition 3 ===
Choose $C = A \sqbrk b$.
=== Definition 3 implies Definition 4 ===
Let $C$ be a finitely generated $A$-module such that:
:$A \sqbrk b \subseteq C$
Then $C$ is also $A \sqbrk b$-module.
Clearly, $C$ is faithful over $A \sqbrk b$, since:
:$1 \in A \sqbrk b \subseteq C$.
=== Definition 4 implies Definition 1 ===
Let $M$ be a faithful $A \sqbrk b$-module whose restriction of scalars to $A$ is finitely generated.
Define an $A$-module endomorphism $\phi : M \to M$ by:
:$ \map \phi m := b m$
Let $\mathfrak a := A$.
Then:
:$\phi \sqbrk M \subseteq M = {\mathfrak a} M$
By Cayley-Hamilton Theorem for Finitely Generated Modules:
:$\phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$
for some $a_i \in {\mathfrak a} = A$.
That is, for all $m \in M$:
as $\map {\phi^i} m = b^i m$.
As $M$ is faithful as an $A \sqbrk b$-module, we have:
:$b^n + a_{n - 1} b^{n-1} + \cdots + a_1 b + a_0 = 0$
Category:Integral Elements
Equivalence of Definitions of Interior (Topology)

== Theorem ==
Let $\struct {T, \tau}$ be a topological space.
Let $H \subseteq T$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition $1$ is equivalent to Definition $2$ ===
Let $\mathbb K$ be defined as:
:$\mathbb K := \set {K \in \tau: K \subseteq H}$
That is, let $\mathbb K$ be the set of all open sets of $T$ contained in $H$.
Then from definition 1 of the interior of $H$, we have:
:$\ds H^\circ = \bigcup_{K \mathop \in \mathbb K} K$
That is, $H^\circ$ is the union of all the open sets of $T$ contained in $H$.
Let $K \subseteq T$ such that $K$ is open in $T$ and $K \subseteq H$.
That is, let $K \in \mathbb K$.
Then from Subset of Union it follows directly that $K \subseteq H^\circ$.
So any open set in $T$ contained in $H$ is a subset of $H^\circ$, and so $H^\circ$ is the largest open set of $T$ contained in $H$.
That is, $H^\circ$ is also the interior of $H$ by definition 2.
Hence both definitions are equivalent.
=== Definition $1$ is equivalent to Definition $3$ ===
Therefore:
:$\ds \bigcup_{K \mathop \in \mathbb K} K = \set {x \mid \text {$x$ is an interior point of $H$} }$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Interiors
Equivalence of Definitions of Internal Group Direct Product

== Theorem ==
=== Definition $1$: Definition by Isomorphism ===
=== Definition $2$: Definition by Subset Product ===
=== Definition $3$: Definition by Unique Expression ===
== Proof ==
Let $e$ denote the identity element of $\struct {G, \circ}$.
=== $(1)$ $(2)$ ===
This is demonstrated in the Internal Direct Product Theorem.
=== $(1)$ implies $(3)$ ===
Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$ by definition $1$.
Then by definition the mapping $\phi: H \times K \to G$ defined as:
:$\forall h \in H, k \in K: \map \phi {h, k} = h \circ k$
is a group isomorphism from $\struct {H, \circ {\restriction_H} } \times \struct {K, \circ {\restriction_K} }$ onto $\struct {G, \circ}$.
From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection:
:for all $g \in G$: there exists a unique $\tuple {h, k} \in H \times K$ such that $h \circ k = g$.
We now need to show that $H$ and $K$ are normal subgroups of $G$.
This is demonstrated in Internal Group Direct Product Isomorphism.
Thus we have shown that $\struct {G, \circ}$ is the internal group direct product of $H$ and $K$ by definition $3$.
=== $(3)$ implies $(2)$ ===
Criterion $(1)$ is common to both definitions.
Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$ by definition $3$.
Then by definition:
:$(1): \quad \struct {H, \circ {\restriction_H} }$ and $\struct {K, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$
:$(2): \quad$ every element of $G$ can be expressed uniquely in the form:
::::$g = h \circ k$
:::where $h \in H$ and $k \in K$.
It remains to be shown that:
:$G$ is the subset product of $H$ and $K$, that is: $G = H \circ K$
:$H \cap K = \set e$, where $e$ is the identity element of $G$.
Indeed, from $(2)$:
Suppose $x \in H \cap K$.
Recall that $e$ denotes the identity element of $\struct {G, \circ}$.
We have:
Because of uniqueness of representation:
Thus:
:$H \cap K = \set e$
Thus $\struct {G, \circ}$ is the internal group direct product of $H$ and $K$ by definition $2$.
Category:Internal Group Direct Products
Equivalence of Definitions of Inverse Image Mapping of Mapping

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Consider the mapping defined as:
:$\forall Y \in \powerset T: \map {f^\gets} Y = \set {s \in S: \exists t \in Y: \map f s = t}$
Let $\Img f \cap Y = \O$.
Then:
:$\forall t \in T: \neg \exists t \in \Img f \cap Y$
and so:
:$\set {s \in S: \exists t \in Y: \map f s = t} = \O$
and so:
:$\forall Y \in \powerset T: \map {f^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \map f s = t} & : \Img f \cap Y \ne \O \\ \O & : \Img f \cap Y = \O \end {cases}$
Category:Inverse Image Mappings
Equivalence of Definitions of Inverse Image Mapping of Relation

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Consider the mapping defined as:
:$\forall Y \in \powerset T: \map {\RR^\gets} Y = \set {s \in S: \exists t \in Y: \tuple {t, s} \in \RR^{-1} }$
Let $\Img \RR \cap Y = \O$.
Then:
:$\forall t \in T: \neg \exists t \in \Img \RR \cap Y$
and so:
:$\set {s \in S: \exists t \in Y: \tuple {s, t} \in \RR} = \O$
and so:
:$\forall Y \in \powerset T: \map {\RR^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \tuple {t, s} \in \RR^{-1} } & : \Img f \cap Y \ne \O \\ \O & : \Img f \cap Y = \O \end {cases}$
Category:Inverse Image Mappings
Equivalence of Definitions of Inverse Mapping

== Theorem ==
Let $S$ and $T$ be sets.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $f^{-1}: T \to S$ be an inverse mapping of $f: S \to T$ by definition 1.
From Mapping is Injection and Surjection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection.
By Composite of Bijection with Inverse is Identity Mapping:
: $f^{-1} \circ f = I_S$
: $f \circ f^{-1} = I_T$
Thus $f^{-1}: T \to S$ is an inverse mapping of $f: S \to T$ by definition 2.
=== Definition 2 implies Definition 1 ===
Let $g: T \to S$ be an inverse mapping of $f: S \to T$ by definition 2.
From Left and Right Inverse Mappings Implies Bijection it follows that both $f$ and $g$ are bijections.
From Mapping is Injection and Surjection iff Inverse is Mapping it follows that $g = f^{-1}$.
Thus $g: T \to S$ is an inverse mapping of $f: S \to T$ by definition 1.
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations
Category:Inverse Mappings
Equivalence of Definitions of Involutive Mapping

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $f: A \to A$ be an involutive mapping by definition 1.
Then by definition:
:$(1): \quad \forall x \in A: \map f {\map f x} = x$
Let $\map f x = y$.
Then by substituting $y$ for $\map f x$ into $(1)$:
:$\forall x \in A: \map f y = x$
Thus $f: A \to A$ is an involutive mapping by definition 2.
=== $(2)$ implies $(1)$ ===
Let $f: A \to A$ be an involutive mapping by definition 2.
Then by definition:
:$\forall x, y \in A: \map f x = y \implies \map f y = x$
and so substituting $\map f x$ for $y$ we have:
:$\map f {\map f x} = x$
Thus $f: A \to A$ is an involutive mapping by definition 1.
=== $(1)$ iff $(3)$ ===
This is demonstrated in Mapping is Involution iff Bijective and Symmetric.
Category:Involutions
Equivalence of Definitions of Irreducible Element of Ring

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
=== $(1)$ implies $(2)$ ===
Let $x$ be an irreducible element of $\struct {D, +, \circ}$ by definition 1.
By definition:
:$x$ has no non-trivial factorization in $D$.
Let $x = y \circ z$ for some $y, z \in D$.
By definition, it cannot be the case that neither $y$ nor $z$ are units of $D$.
So either $y$ or $z$ is a unit of $D$.
, suppose $y$ is a unit of $D$.
Then by definition $z$ is an associate of $x$.
Contrariwise, suppose $z$ is a unit of $D$.
Then by definition $y$ is an associate of $x$.
Thus both $y$ and $z$ are either a unit of $D$ or an associate of $x$.
$x = y \circ z$ is an arbitrary factorization of $x$ in $D$.
It follows that the only divisors of $x$ are its associates and the units of $D$.
Thus $x$ is an irreducible element of $\struct {D, +, \circ}$ by definition 2.
=== $(2)$ implies $(1)$ ===
Let $x$ be an irreducible element of $\struct {D, +, \circ}$ by definition 2.
Then by definition:
:the only divisors of $x$ are its associates and the units of $D$.
Let $x = y \circ z$.
Then either:
:$y$ is an associate of $x$ and $z$ is a unit of $D$
or:
:$z$ is an associate of $x$ and $y$ is a unit of $D$.
In either case, $y \circ z$ is a trivial factorization of $x$.
Thus $x$ is an irreducible element of $\struct {D, +, \circ}$ by definition 1.
Category:Irreducible Elements of Rings
Equivalence of Definitions of Irreducible Space

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
=== Definition 6 ===
=== Definition 7 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $T$ be an irreducible space by Definition 1.
That is, $T$ is not the union of any two proper closed sets.
 $T$ admits a finite cover by proper closed sets.
By the Well-Ordering Principle, there exists a minimal natural number $n \in \N$ such that $T$ has a cover by $n$ proper closed sets, say $F_1, \ldots, F_n$.
By definition of proper subset, it must be the case that $n > 1$.
By Finite Union of Closed Sets is Closed in Topological Space, $F_{n - 1} \cup F_n$ is closed.
By Union is Associative:
:$S = F_1 \cup \cdots \cup \paren {F_{n - 1} \cup F_n}$
By minimality of $n$, $F_{n - 1} \cup F_n$ is not a proper subset.
Thus $F_{n - 1} \cup F_n = S$.
This contradicts the definition of an irreducible space by Definition 1.
Thus $T$ has no finite cover by proper closed sets.
That is, $T$ is an irreducible space by Definition 2.
=== $(2)$ implies $(1)$ ===
Let $T$ be an irreducible space by Definition 2.
Then $T$ has no finite cover by proper closed sets.
A cover by two sets is a finite cover.
Hence it immediately follows that $T$ has no finite cover by $2$ proper closed sets.
That is, $T$ is an irreducible space by Definition 1.
=== $(1)$ implies $(3)$ ===
Let $T$ be an irreducible space by Definition 1.
That is, $T$ is not the union of any two proper closed sets.
Let $U$ and $V$ be two arbitrary non-empty open sets of $T$.
 $U$ and $V$ have an empty intersection.
By De Morgan's Laws, their complements are proper closed sets whose union is $T$.
This contradicts the definition of an irreducible space by Definition 1.
Thus:
:$U \cap V \ne \O$
As $U$ and $V$ are arbitrary, it follows that $T$ is an irreducible space by Definition 3.
=== $(3)$ implies $(1)$ ===
Let $T$ be an irreducible space by Definition 3.
Let $F$ and $G$ be two arbitrary proper closed sets of $T$
 their union is $T$.
By De Morgan's Laws, their complements are non-empty open sets whose intersection is empty.
This contradicts the definition of an irreducible space by Definition 3.
Thus $F \cup G \ne T$.
As $F$ and $G$ are arbitrary, it follows that $T$ is an irreducible space by Definition 1.
=== $(3)$ iff $(4)$ ===
By definition of everywhere dense subset, an open set is dense it has non-empty intersection with any other open set.
=== $(4)$ iff $(5)$ ===
Follows from Everywhere Dense iff Interior of Complement is Empty.
=== $(3)$ iff $(6)$ ===
=== $(3)$ iff $(7)$ ===
Category:Irreducible Spaces
Equivalence of Definitions of Irreducible Space/3 iff 6

== Theorem ==
=== Definition 3 ===
=== Definition 6 ===
== Proof ==
Let $T = \struct {S, \tau}$ be an irreducible space by Definition 3.
That is, let every non-empty open set in $T$ have a non-empty intersection with every other non-empty open set.
Let $U \in \tau$ be open in $T$ such that $U \ne \O$.
If $U = S$ then $U^- = S$ trivially.
So, let $U \ne S$.
Let $x \in S$ such that $x \ne U$.
Let $V$ be any non-empty open set of $T$ such that $x \in V$.
By hypothesis, $U \cap V \ne \O$.
Therefore $V$ contains some point of $U$ distinct from $x$.
$V$ is arbitrary, so it follows that every open set $V$ of $T$ such that $x \in V$ contains some point of $U$ distinct from $x$.
That is, $x$ is a limit point of $U$.
So every point of $T$ is a limit point of $U$.
Thus, by definition, every point of $T$ is in the closure of $U$.
The argument reverses directly.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Irreducible Spaces
Equivalence of Definitions of Irreducible Space/3 iff 7

== Theorem ==
=== Open Sets Intersect ===
=== Open Sets are Connected ===
== Proof ==
=== Open Sets Intersect implies Open Sets are Connected ===
Let $T = \struct {S, \tau}$ be irreducible in the sense that:
:no two non-empty open sets of $T$ are disjoint.
Let $U \subseteq S$ be an open set of $T$.
 $U$ is not connected.
Then there exist non-empty open sets $V, W$ of $U$ that are disjoint and whose union is $U$.
By Open Set in Open Subspace, $V$ and $W$ are open sets of $T$.
Because $V \cap W = \O$, $T$ is not irreducible.
This is a contradiction.
Thus $U$ is connected.
Thus $T = \struct {S, \tau}$ is irreducible in the sense that:
:every open set of $T$ is connected..
=== Open Sets are Connected implies Open Sets Intersect ===
Let $T = \struct {S, \tau}$ be irreducible in the sense that:
:every open set of $T$ is connected.
Let $V$ and $W$ be open sets of $T$.
By definition of topology, their union $V \cup W$ is open in $T$.
By hypothesis, $V \cup W$ is connected.
By Open Set in Open Subspace, $V$ and $W$ are open sets of $V \cup W$.
Because $V \cup W$ is connected, $V \cap W$ is non-empty.
Because $V$ and $W$ were arbitrary, $T$ is irreducible.
Thus $T = \struct {S, \tau}$ is irreducible in the sense that:
:no two non-empty open sets of $T$ are disjoint.
== Also see ==
* Open Subset of Irreducible Space is Irreducible
* Space is Ultraconnected iff Closed Sets are Connected
Category:Irreducible Spaces
Category:Connected Spaces
Equivalence of Definitions of Isolated Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $x$ be an isolated point of $H$ by definition 1.
Then by definition:
:$\exists U \in \tau: U \cap H = \set x$
Thus we have an open set in $T$ such that $x \in U$ contains no other point of $H$ than $x$.
Thus, by definition, $x$ is not a limit point of $H$.
Thus $x$ is an isolated point of $H$ by definition 2.
=== Definition 2 implies Definition 1 ===
Let $x$ be an isolated point of $H$ by definition 2.
 $x$ is a limit point of $H$.
Then by definition every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.
That is:
:$\forall U \in \tau: x \in U \implies \exists y \in S, y \ne x: y \in U \cap H$
That is:
:$\not \exists U \in \tau: U \cap H = \set x$
because all $U$ with $x$ in them are such that there is at least one point in $U \cap H$ apart from $x$.
Thus by Proof by Contradiction $x$ is not a limit point of $H$.
That is, $x$ is an isolated point of $H$ by definition 1.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Limit Points
Category:Isolated Points
Equivalence of Definitions of Isometry of Metric Spaces

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces or pseudometric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
Then its inverse $\phi^{-1}: M_2 \to M_1$ is likewise an isometry.
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\phi: M_1 \to M_2$ be an isometry by definition 1.
Then by definition:
:$\phi$ is a bijection such that:
::$\forall a, b \in A_1: \map {d_1} {a, b} = \map {d_2} {\map \phi a, \map \phi b}$
By Inverse of Bijection is Bijection, $\phi^{-1}$ is also a bijection.
Thus $\forall a, b \in A_2$ there exists $\map {\phi^{-1} } a$ and $\map {\phi^{-1} } b$ in $A_1$.
From Inverse of Inverse of Bijection:
:$\forall a, b \in A_2: \map {d_2} {\map \phi {\map {\phi^{-1} } a}, \map \phi {\map {\phi^{-1} } b} } = \map {d_1} {\map {\phi^{-1} } a, \map {\phi^{-1} } b}$
and so:
:$\forall a, b \in A_2: \map {d_2} {a, b} = \map {d_1} {\map {\phi^{-1} } a, \map {\phi^{-1} } b}$
Thus $M_1$ and $M_2$ are isometric by definition 2.
=== $(2)$ implies $(1)$ ===
Let $M_1$ and $M_2$ be isometric by definition 2.
Then by definition:
:there exist inverse mappings $\phi: A_1 \to A_2$ and $\phi^{-1}: A_2 \to A_1$ such that:
::$\forall a, b \in A_1: \map {d_1} {a, b} = \map {d_2} {\map \phi a, \map \phi b}$
:and:
::$\forall u, v \in A_2: \map {d_2} {u, v} = \map {d_1} {\map {\phi^{-1} } u, \map {\phi^{-1} } v}$
By Inverse of Bijection is Bijection, $\phi$ and $\phi^{-1}$ are both bijections.
Thus $\phi: M_1 \to M_2$ is a isometry by definition 1.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S $: Subspaces and Equivalence of Metric Spaces: Theorem $7.4$
Category:Isometries (Metric Spaces)