nyuuzyou/wikis huggingface dataset:
Metric Space Continuity by Inverse of Mapping between Open Balls

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.
Let $a \in A_1$ be a point in $A_1$.
$f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ :
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map {B_\delta} {a; d_1} \subseteq f^{-1} \sqbrk {\map {B_\epsilon} {\map f a; d_2} }$
where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$.
== Proof ==
By definition, $f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ :
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
For a mapping $f: X \to Y$ we have:
:$f \sqbrk U \subseteq V \iff U \subseteq f^{-1} \sqbrk V$
where $U \subseteq X$ and $V \subseteq Y$.
Hence the result.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.3$
Category:Open Balls
Category:Continuous Mappings on Metric Spaces
Metric Space Continuity by Neighborhood

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.
Let $a \in A_1$ be a point in $A_1$.
Then the following definitions of continuity of $f$ at $a$ with respect to $d_1$ and $d_2$ are equivalent:
=== $\epsilon$-Ball Definition ===
=== Definition by Neighborhood ===
== Proof ==
=== $\epsilon$-Ball Definition implies Definition by Neighborhood ===
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$.
Let $N'$ be a neighborhood of $\map f a$ in $M_2$.
Then:
But by Open Ball is Neighborhood of all Points Inside, $\map {B_\delta} {a; d_1}$ is a neighborhood of $a$ in $M_1$.
Thus letting $N := \map {B_\delta} {a; d_1}$:
:$f \sqbrk N \subseteq \map {B_\epsilon} {\map f a; d_2} \subseteq N'$
and so the condition for $f$ to be $\tuple {d_1, d_2}$-continuous at $a$ by Neighborhood is fulfilled.
=== Definition by Neighborhood implies $\epsilon$-Ball Definition ===
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
:for each neighborhood $N'$ of $\map f a$ in $M_2$ there exists a corresponding neighborhood $N$ of $a$ in $M_1$ such that $f \sqbrk N \subseteq N'$.
Let $\epsilon \in \R_{>0}$.
Consider the open $\epsilon$-ball $\map {B_\epsilon} {\map f a; d_2}$ of $\map f a$.
By Open Ball is Neighborhood of all Points Inside, $\map {B_\epsilon} {\map f a; d_2}$ is a neighborhood of $\map f a$ in $M_2$.
Let $N' := \map {B_\epsilon} {\map f a; d_2}$.
By definition of $\tuple {d_1, d_2}$-continuity at $a$, there exists a neighborhood $N$ of $a$ in $M_1$ such that:
:$f \sqbrk N \subseteq N'$
By definition of neighborhood:
:$\exists \delta \in \R_{>0}: \map {B_\delta} {a; d_1} \subseteq N$.
Therefore:
:$\map f {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
and so the condition for $f$ to be $\tuple {d_1, d_2}$-continuous at $a$ by open $\epsilon$-ball is fulfilled.
== Also see ==
* Metric Space Continuity by Epsilon-Delta
== Sources ==
* : $\text{III}$: $\epsilon$-Balls
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.6$
* : $2.3$: Open sets in metric spaces: Definition $2.3.6$
Category:Neighborhoods
Category:Continuous Mappings on Metric Spaces
Metric Space Continuity by Neighborhood Basis

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.
Let $a \in A_1$ be a point in $A_1$.
Let $\BB_{\map f a}$ be a basis for the neighborhood system at $\map f a$.
$f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ :
:for each neighborhood $N$ in $\BB_{\map f a}$, $f^{-1} \sqbrk N$ is a neighborhood of $a$.
== Proof ==
By definition, $\BB_{\map f a}$ be a basis for the neighborhood system at $\map f a$ :
:$\forall N_a \subseteq M_2: \exists N \in \BB_{\map f a}: N \subseteq N_a$
where $N_a$ denotes a neighborhood of $\map f a$ in $M_2$.
=== Necessary Condition ===
Let $f$ be continuous at $a$ with respect to the metrics $d_1$ and $d_2$.
Then by Metric Space Continuity by Inverse of Mapping between Neighborhoods:
:for each neighborhood $N$ of $\map f a$ in $M_2$, $f^{-1} \sqbrk N$ is a neighborhood of $a$.
In particular, let $N \in \BB_{\map f a}$.
Then $f^{-1} \sqbrk N$ is a neighborhood of $a$.
=== Sufficient Condition ===
Let $f$ be such that:
:for each neighborhood $N$ in $\BB_{\map f a}$, $f^{-1} \sqbrk N$ is a neighborhood of $a$.
Let $N'$ be any neighborhood of $\map f a$.
Then by definition:
:$\exists B \in \BB_{\map f a}: B \subseteq N'$
From Preimage of Subset is Subset of Preimage:
:$f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {N'}$
By hypothesis $f^{-1} \sqbrk B$ is a neighborhood of $a$.
From Superset of Neighborhood in Metric Space is Neighborhood it follows that $f^{-1} \sqbrk {N'}$ is a neighborhood of $a$.
As $N'$ is arbitrary, the result follows.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $3$
Category:Neighborhood Bases
Category:Continuous Mappings on Metric Spaces
Metric Space Continuity by Open Ball

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.
Let $a \in A_1$ be a point in $A_1$.
=== $\epsilon$-$\delta$ Definition ===
=== $\epsilon$-Ball Definition ===
== Proof ==
=== $\epsilon$-$\delta$ Definition implies $\epsilon$-Ball Definition ===
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$
where $\R_{>0}$ denotes the set of all strictly positive real numbers.
Let $\epsilon \in \R_{>0}$ be arbitrary.
Let $\delta$ be such that:
:$\forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$
as is known to exist by hypothesis.
Then:
As $\epsilon$ is arbitrary, it follows that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
=== $\epsilon$-Ball Definition implies $\epsilon$-$\delta$ Definition===
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$.
Let $\epsilon \in \R_{>0}$ be arbitrary.
Let $\delta$ be such that:
:$f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
as is known to exist by hypothesis.
Then:
As $\epsilon$ is arbitrary, it follows that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
== Also see ==
* Metric Space Continuity by Epsilon-Delta
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: $\varepsilon$-Balls
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.2$
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Definition $2.3.6$
Category:Open Balls
Category:Continuous Mappings on Metric Spaces
Metric Space Induced by Hausdorff Metric

== Theorem ==
Let $M = \struct {S, d}$ be a metric space.
Let $\CC$ be the set of compact subsets of $M$.
Let $\d: \CC \times \CC \to \R_{\ge 0}$ be the Hausdorff metric on $\CC$.
Then $\struct {\CC, \d}$ is a metric space.
== Proof ==
== Sources ==
* 
Category:Hausdorff Metric
Metric Space defined by Closed Sets

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then:
== Proof ==
From Metric Space is Closed in Itself, $\text C 1$ holds.
From Empty Set is Closed in Metric Space, $\text C 2$ holds.
Let $\ds \bigcup_{i \mathop = 1}^n V_i$ be the union of a finite number of closed sets of $M$.
Then from De Morgan's laws:
:$\ds A \setminus \bigcup_{i \mathop = 1}^n V_i = \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i}$
By definition of closed set, each of the $A \setminus V_i$ is by definition open in $M$.
We have that $\ds \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i}$ is the intersection of a finite number of open sets of $M$.
Therefore, by Finite Intersection of Open Sets of Metric Space is Open, $\ds \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i} = A \setminus \bigcup_{i \mathop = 1}^n V_i$ is likewise open in $M$.
Then by definition of closed set, $\ds \bigcup_{i \mathop = 1}^n V_i$ is closed in $M$.
Thus $\text C 3$ holds.
Let $I$ be an indexing set (either finite or infinite).
Let $\ds \bigcap_{i \mathop \in I} V_i$ be the intersection of a indexed family of closed sets of $M$ indexed by $I$.
Then from De Morgan's laws:
:$\ds A \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {A \setminus V_i}$
By definition of closed set, each of $A \setminus V_i$ are by definition open in $M$.
We have that $\ds \bigcup_{i \mathop \in I} \paren {A \setminus V_i}$ is the union of a family of open sets of $M$ indexed by $I$.
Therefore, by definition of a topology, $\ds \bigcup_{i \mathop \in I} \paren {A \setminus V_i} = A \setminus \bigcap_{i \mathop \in I} V_i$ is likewise open in $M$.
Then by definition of closed set, $\ds \bigcap_{i \mathop \in I} V_i$ is closed in $M$.
Thus $\text C 4$ holds.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.11$
Category:Metric Spaces
Category:Closed Sets
Metric Space fulfils all Separation Axioms

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$, considered as a topological space, fulfils all separation axioms:
:$M$ is a $T_0$ (Kolmogorov) space
:$M$ is a $T_1$ (Fréchet) space
:$M$ is a $T_2$ (Hausdorff) space
:$M$ is a semiregular space
:$M$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space 
:$M$ is a $T_3$ space 
:$M$ is a regular space
:$M$ is an Urysohn space
:$M$ is a $T_{3 \frac 1 2}$ space
:$M$ is a Tychonoff (completely regular) space
:$M$ is a $T_4$ space
:$M$ is a normal space
:$M$ is a $T_5$ space 
:$M$ is a completely normal space
:$M$ is a perfectly $T_4$ space
:$M$ is a perfectly normal space
== Proof ==
We have that:
:A metric space is a $T_2$ (Hausdorff) space.
:A metric space is a perfectly $T_4$ space.
:A metric space is a $T_5$ space.
‎
From Sequence of Implications of Separation Axioms we then have:
:$T_5$ space is $T_4$ space.
:$T_2$ (Hausdorff) Space is $T_1$ (Fréchet) Space.
:$T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space.
By definition, a perfectly normal space is:
:a perfectly $T_4$ space
:a $T_1$ (Fréchet) space
So $M$ is a perfectly normal space.
By definition, a completely normal space is:
:a $T_5$ space
:a $T_1$ (Fréchet) space
So $M$ is a completely normal space.
Then from Sequence of Implications of Separation Axioms we can complete the chain:
:Completely Normal Space is Normal Space.
:Normal Space is Tychonoff (Completely Regular) Space.
:Completely Regular (Tychonoff) implies $T_{3 \frac 1 2}$ by definition.
:Completely Regular (Tychonoff) Space is Regular Space.
:Completely Regular (Tychonoff) Space is Urysohn Space.
:Regular implies $T_3$ by definition.
:Regular Space is Completely Hausdorff Space.
:Regular Space is Semiregular Space.
There are other chains of implications which can be used, but the above are sufficient to prove the hypothesis.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Separation Axioms
Metric Space generates Uniformity

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\UU$ be the set of sets defined as:
:$\UU := \set {u \in A \times A: \exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u}$
where:
:$\R_{>0}$ is the set of strictly positive real numbers
:$u_\epsilon$ is defined as:
::$u_\epsilon := \set {\tuple {x, y}: \map d {x, y} < \epsilon}$
Then $\UU$ is a uniformity on $A$ which generates a uniform space with the same topology as the topology induced by $d$.
== Proof ==
From definition it is clear that a metric space is a pseudometric space.
The result then follows from Pseudometric Space generates Uniformity.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Metric Uniformities
Category:Metric Spaces
Category:Uniformities
Metric Space is Closed in Itself

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $A$ is closed in $M$.
== Proof ==
From Empty Set is Open in Metric Space, $\O$ is open in $M$.
But:
:$A = \relcomp A \O$
where $\complement_A$ denotes the set complement relative to $A$.
The result follows by definition of closed set.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets
Category:Metric Spaces
Category:Closed Sets
Metric Space is Compact iff Countably Compact

== Theorem ==
A metric space is compact it is countably compact.
== Proof ==
Follows directly from:
:Compact Space is Countably Compact
:Countably Compact Metric Space is Compact
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Compact Spaces
Category:Countably Compact Spaces
Metric Space is Completely Normal

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a completely normal space.
== Proof ==
By definition, a topological space is completely normal space it is:
:a $T_5$ space
:a $T_1$ (Fréchet) space.
We have that:
:a Metric Space is $T_5$
:a Metric Space is $T_2$ (Hausdorff)
:a $T_2$ (Hausdorff) Space is a $T_1$ (Fréchet) Space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Completely Normal Spaces
Metric Space is Countably Compact iff Sequentially Compact

== Theorem ==
Let $M$ be a metric space.
Then $M$ is countably compact $M$ is sequentially compact.
== Proof ==
This follows directly from the results:
:Countably Compact Metric Space is Sequentially Compact
:Sequentially Compact Space is Countably Compact
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Countably Compact Spaces
Category:Sequentially Compact Spaces
Metric Space is First-Countable

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is first-countable.
== Proof ==
Let $x \in A$.
Let:
:$\BB = \set {\map {B_{1/n} } x: n \in \N_{>0} }$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$ in $M$.
By Surjection from Natural Numbers iff Countable, we have that $\BB$ is countable.
By the definition of a first-countable space, it suffices to show that $\BB$ is a local basis at $x$.
By Open Ball of Metric Space is Open Set, every element of $\BB$ is an open neighborhood of $x$.
Let $U$ be an open neighborhood of $x$.
By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$.
By the Axiom of Archimedes, there exists a natural number $n > \dfrac 1 \epsilon$.
That is, $\dfrac 1 n < \epsilon$, and so:
:$\map {B_{1/n} } x \subseteq \map {B_\epsilon} x \subseteq U$
The result follows from Subset Relation is Transitive.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:First-Countable Spaces
Metric Space is Fully Normal

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a fully normal space.
== Proof ==
We have that a metric space is fully $T_4$.
We also have that a metric space is a $T_1$ (Fréchet) space.
Hence the result, by definition of a fully normal space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Fully Normal Spaces
Metric Space is Hausdorff

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a Hausdorff space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $4$: The Hausdorff condition: $4.2$: Separation axioms: Example $4.2.3$
Category:Metric Spaces
Category:Hausdorff Spaces
Category:Metric Space is Hausdorff
Metric Space is Hausdorff/Proof 1

== Theorem ==
== Proof ==
Let $x, y \in A: x \ne y$.
Then from Distinct Points in Metric Space have Disjoint Open Balls, there exist open $\epsilon$-balls $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ which are disjoint open sets containing $x$ and $y$ respectively.
Hence the result by the definition of Hausdorff space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Space is Hausdorff
Metric Space is Hausdorff/Proof 2

== Theorem ==
== Proof ==
 $M$ is not Hausdorff.
That is, there are $x, y \in A: x \ne y$ such that:
:$\forall \epsilon \in \R_{>0}: \exists z \in \map {B_\epsilon} x \cap \map {B_\epsilon} y$
where $\map {B_\epsilon} x$ denote the open $\epsilon$-ball of $x$ in $M$.
Let $r = \dfrac {\map d {x, y} } 2$.
Let $z \in \map {B_r} x \cap \map {B_r} y$.
Then:
This contradicts .
Thus, $M$ has to be Hausdorff.
Category:Metric Space is Hausdorff
Metric Space is Lindelöf iff Second-Countable

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is Lindelöf $M$ is second-countable.
== Proof ==
=== Sufficient Condition ===
We have from Second-Countable Space is Lindelöf that second-countability implies Lindelöf in all topological spaces, regardless of whether they are metric spaces or not.
=== Necessary Condition ===
Suppose $M = \struct {A, d}$ is Lindelöf.
Let us define the open covers on $A$:
:$\CC_k = \set {\map {N_{1/k} } x: x \in S}$
for all $k \in \N_{>0}$.
As $M$ is Lindelöf, each one of these has a countable subcover.
The union of all these subcovers is a countable basis for the topology on $A$.
Hence the result, by definition of second-countable space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Lindelöf Spaces
Category:Second-Countable Spaces
Metric Space is Open and Closed in Itself

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $A$ is both open and closed in $M$.
== Proof ==
From Metric Space is Open in Itself, $A$ is open in $M$.
From Metric Space is Closed in Itself, $A$ is closed in $M$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets
Category:Metric Spaces
Category:Open Sets (Metric Spaces)
Category:Closed Sets (Metric Spaces)
Metric Space is Open in Itself

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then the set $A$ is an open set of $M$.
== Proof ==
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
Let $x \in A$.
An open ball of $x$ in $M$ is by definition a subset of $A$.
Hence the result.
== Examples ==
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.4$
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.10$
Category:Metric Space is Open in Itself
Category:Metric Spaces
Category:Open Sets (Metric Spaces)
Metric Space is Open in Itself/Examples/Closed Real Interval

== Examples of Use of Metric Space is Open in Itself ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $\closedint a b \subset \R$ be a closed interval of $\R$.
Then from Closed Real Interval is not Open Set, $\closedint a b$ is not an open set of $\R$.
However, if $\closedint a b$ is considered as a subspace of $\R$, then it is seen that $\closedint a b$ is an open set of $\closedint a b$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.10$
Category:Metric Space is Open in Itself
Metric Space is Paracompact/Proof 2

== Theorem ==
== Proof ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {C_\alpha}$ be an open cover of $M$ indexed by ordinals.
Let $\map {B_r} x$ be the open $r$-ball in $M$ around $x$.
For $n \in \N$, let $D_{\alpha n}$ denote the union of all the open $r$-balls $\map {B_{2^{-n} } } x$ such that:
:$(1): \quad \alpha$ is the smallest ordinal such that $x \in C_\alpha$
:$(2): \quad x \notin D_{\beta j}$ for $j < n$
:$(3): \quad \map {B_{3 \times 2^{-n} } } x \subseteq C_\alpha$
It is to be shown that $\sequence {D_{\alpha n} }$ is a locally finite open refinement of $\sequence {C_\alpha}$ which covers $M$.
By construction, $D_{\alpha n}$ is a refinement of $\sequence {C_\alpha}$.
Let $x \in A$.
Then there exists a smallest ordinal $\alpha$ such that $x \in C_\alpha$.
Then there exists $n$ sufficiently large such that $(3)$ holds.
Then by $(2)$:
:$x \in D_{\beta j}$
for some $j \le n$.
Thus $\sequence {D_{\alpha n} }$ covers $M$.
Now assume $x \in A$.
Let $\alpha$ be the smallest ordinal such that $x \in D_{\alpha n}$ for some $n$.
Choose $j \in \N$ such that $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$.
Let $i \ge n + j$.
As $i > n$, every one of the open balls of radius $2^{-i}$ used to define $D_{\beta i}$ has its center $y$ outside $D_{\alpha n}$
And because $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$, we have that $\map d {x, y} \ge 2^{-j}$.
But $i \ge j + 1$ and $n + j \ge j + 1$.
So:
:$\map {B_{2^{-n - j} } } x \cap \map {B_{2^{-i} } } y = \O$
Thus if $i \ge n$, then $\map {B_{2^{-n - j} } } x$ intersects no $D_{\beta i}$.
Let $i < n+j$.
Let us choose:
:$p \in D_{\beta i}$
:$q \in D_{\gamma i}$
:$\beta < \gamma$
There exist points $y$ and $z$ such that:
:$p \in \map {B_{2^{-i} } } y \subseteq D_{\beta i}$
:$q \in \map {B_{2^{-i} } } z \subseteq D_{\gamma i}$
By $(3)$:
:$\map {B_{3 \times 2^{-i} } } y \subseteq C_\beta$
But by $(1)$:
:$z \notin C_\beta$
So:
:$\map d {y, z} \ge 3 \times 2^{-i}$
and:
:$\map d {p, q} > 2^{-i} \ge 2^{-n - j + 1}$
Thus if $i < n + j$, then $\map {B_{2^{-n - j} } } x$ intersects $D_{\beta i}$ for at most one $\beta$.
Thus $D_{\alpha n}$ is locally finite.
Then $D_{\alpha n}$ is a locally finite open refinement of $\sequence {C_\alpha}$ which covers $M$.
Hence by definition $M$ is paracompact.
== Sources ==
* 
Category:Metric Space is Paracompact
Metric Space is Perfectly Normal

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a perfectly normal space.
== Proof ==
By definition, a topological space is perfectly normal space it is:
:a perfectly $T_4$ space
:a $T_1$ (Fréchet) space.
We have that:
:a Metric Space is Perfectly $T_4$
:a Metric Space is $T_2$ (Hausdorff)
:a $T_2$ (Hausdorff) Space is a $T_1$ (Fréchet) Space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Perfectly Normal Spaces
Metric Space is Perfectly T4

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a perfectly $T_4$ space.
== Proof ==
We have that a metric space is $T_4$.
We also have that every closed set in a metric space is a $G_\delta$ set.
Hence the result, by definition of a perfectly $T_4$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Perfectly T4 Spaces
Metric Space is Separable iff Second-Countable

== Theorem ==
A metric space is separable it is second-countable.
== Proof ==
Follows directly from:
:Separable Metric Space is Second-Countable
:Second-Countable Space is Separable
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 37$
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Second-Countable Spaces
Category:Separable Spaces