avewright/proofwiki-math huggingface dataset:
Theorem: Metric Space Completeness is Preserved by Isometry

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
If $M_1$ is complete then so is $M_2$.

Proof:
Let $\epsilon \in \R_{>0}$.
Let $\sequence {b_n}$ be a Cauchy sequence in $A_2$.
Thus:
:$\exists N_1 \in \N: \map {d_2} {b_n, b_m} < \epsilon$
whenever $n, m \ge N_1$ and $b_n, b_m \in A_2$.
We have that $M_1$ is isometric to $M_2$.
Isometry is Equivalence Relation and so in particular symmetric.
Hence $M_2$ is isometric to $M_1$, via $\phi^{-1}$.
Thus:
:$\map {d_1} {\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } {b_m} } = \map {d_2} {b_n, b_m} < \epsilon$
whenever $n, m \ge N_1$ and $\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } {b_m} \in A$.
So $\sequence {\map {\phi^{-1} } {b_n} }$ is Cauchy in $A_1$.
Since $A_1$ is complete, $\sequence {\map {\phi^{-1} } {b_n} }$ converges in $A_1$ to, say, $a$.
By definition of isometry, $\phi^{-1}$ is a bijection, and in particular surjective.
Thus there exists some $b \in A_2$ such that $\map {\phi^{-1} } b = a$.
Since $\sequence {\map {\phi^{-1} } {b_n} }$ converges to $\map {\phi^{-1} } b$, there exists some $N_2 \in \N$ such that:
:$\map {d_1} {\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } b} < \epsilon$
whenever $n \ge N_2$ and $\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } b \in A_1$.
Since $M_2$ is  isometric to $M_1$, we have:
:$\map {d_1} {\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } b} = \map {d_2} {b_n, b}$
and so:
:$\map {d_2} {b_n, b} < \epsilon$
whenever $n \ge N_2$ and $b_n, b \in A_2$.
Thus $\sequence {b_n}$ converges in $A_2$.
Since $\sequence {b_n}$ was an arbitrary Cauchy sequence, we have that $M_2$ is complete, as required.
{{qed}}
Theorem: Urysohn's Metrization Theorem

Let $T = \struct {S, \tau}$ be a topological space which is regular and second-countable.
Then $T$ is metrizable.

Proof:
From Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube:
:$T$ is homeomorphic to a subspace of the Hilbert cube $\struct {I^\omega, d_2}$ 
where $d_2$ is the metric defined as:
:$\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in I^\omega: \map {d_2} {x, y} := \paren {\sum_{k \mathop \in \N_{>0} } \paren {x_k - y_k}^2}^{\frac 1 2}$
{{Recall|Metrizable Space|metrizable space}}:
:$\struct {I^\omega, \tau_{d_2}}$ is metrizable
where $\tau_{d_2}$ is the topology induced by $d_2$.
From Subspace of Metrizable Space is Metrizable Space:
:$T$ is homeomorphic to a metrizable space
It follows that $T$ is metrizable by definition.
{{qed}}
Theorem: Uniformity iff Quasiuniformity has Symmetric Basis

Let $S$ be a set.
Let $\UU$ be a quasiuniformity on $S$.
Then $\UU$ is a uniformity {{iff}} $\UU$ has a symmetric filter basis.

Proof:
Let $\UU$ be a quasiuniformity on $S$ which has a symmetric filter basis $\BB$.
From the definition of filter basis, all the elements of $\UU$ can be formed from intersections of elements of $\BB$.
But from Intersection of Symmetric Relations is Symmetric, it follows that all elements of $\UU$ are symmetric.
Now suppose $\UU$ is a uniformity.
If $\BB$ is a filter basis of $\UU$ then all the elements of $\BB$ are also elements of $\UU$.
Hence $\BB$ is a symmetric filter basis of $\UU$.
{{qed}}
Theorem: Topological Space is Quasiuniformizable

Every topological space is quasiuniformizable.

Proof:
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be defined as:
:$\BB := \set {u_G: u_G = \paren {G \times G} \cup \paren {\paren {S \setminus G} \times G}, G \in \tau}$
Then $\BB$ is a filter sub-basis for a quasiuniformity on $S$ such that $\struct {\struct {S, \UU}, \tau}$ is a quasiuniform space.
{{finish|Prove that above assertion.}}
Theorem: Pseudometric Space generates Uniformity

Let $P = \struct {A, d}$ be a pseudometric 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:
We check whether the Uniformity Axioms are satisfied.
Theorem: Metric Space generates Uniformity

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.
{{qed}}
Theorem: Multiplication is Superfunction

The function $f: \C \to \C$, defined as:
:$\map f z = z \times c$
is a superfunction for any complex number $c$.

Proof:
Define $h: \C \to \C$ by $\map h z = z + c$.
Then:
{{begin-eqn}}
{{eqn | l = \map h {\map f z}
      | r = \map h {z \times c}
      | c = 
}}
{{eqn | r = z \times c + c
      | c = 
}}
{{eqn | r = \paren {z + 1} \times c
      | c = 
}}
{{eqn | r = \map f {z + 1}
      | c = 
}}
{{end-eqn}}
Thus $\map f z = z \times c$ is a superfunction and $\map h z = z + c$ is the corresponding transfer function.
{{qed}}
Category:Superfunctions
cab3ee5qz40q2bqeshg3oe1373qvkwn
Theorem: Discrete Topology is Topology

:$\tau$ is a topology on $S$.

Proof:
Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Then by definition $\tau = \powerset S$, that is, is the power set of $S$.
We confirm the criteria for $T$ to be a topology:
:$(1): \quad$ By definition of power set, $\O \in \powerset S$ and $S \in \powerset S$.
:$(2): \quad$ From Power Set with Union is Monoid, $\powerset S$ is closed under set union.
:$(3): \quad$ From Power Set with Intersection is Monoid, $\powerset S$ is closed under set intersection.
{{qed}}
Theorem: Discrete Topology is Finest Topology

:$\tau$ is the finest topology on $S$.

Proof:
Let $\phi$ be any topology on $S$.
Let $U \in \phi$.
Then, by the definition of topology, $U \subseteq S$.
Then, by the definition of discrete topological space, $U \in \tau$.
Hence by definition of subset, $\phi \subseteq \tau$.
Hence by definition of finer topology, $\tau$ is finer than $\phi$.
{{qed}}
Theorem: Set in Discrete Topology is Clopen

:$\forall U \subseteq S: U$ is both closed and open in $\struct {S, \tau}$.

Proof:
Let $U \subseteq S$.
By definition of discrete topological space, $U \in \tau$.
By definition of closed set, $\relcomp S U$ is closed in $T$, where $\relcomp S U$ is the relative complement of $U$ in $S$. 
But from Set Difference is Subset:
:$\relcomp S U = S \setminus U \subseteq S$
and so:
:$\relcomp S U \in \tau$
That is, $\relcomp S U$ is both closed and open in $T$. 
Then by Relative Complement of Relative Complement:
:$\relcomp S {\relcomp S U} = U$
which is seen to be both closed and open in $T$.
{{qed}}
Theorem: Topological Space is Discrete iff All Points are Isolated

:$\tau$ is the discrete topology on $S$ {{iff}} all points in $S$ are isolated points of $T$.

Proof:
=== Necessary Condition ===
Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Then by definition $\tau = \powerset S$, that is, $\tau$ is the power set of $S$.
Let $x \in S$.
Then from Set in Discrete Topology is Clopen it follows that $\set x$ is open in $T$.
Thus by definition $x \in S$ is an isolated point of $T$.
{{qed|lemma}}
Theorem: Point in Discrete Space is Adherent Point

Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $U \subseteq S$.
Then $x$ is an adherent point of $U$ {{iff}} $x \in U$.

Proof:
Let $U \subseteq S$.
From Set in Discrete Topology is Clopen it follows that $U$ is open in $T$.
Let $x \in U$.
Then $U$ is an open neighborhood of $x$ containing (trivially) an element of $U$, that is, $x$.
So, by definition, $x$ is an adherent point of $U$.
Now suppose $x \notin U$.
Then $\set x$ is an open neighborhood of $x$ such that $\set x \cap U = \O$.
So $x$ can not be an adherent point of $U$.
{{qed}}
Theorem: Interior Equals Closure of Subset of Discrete Space

Let $A \subseteq S$.
Then:
:$A = A^\circ = A^-$
where:
:$A^\circ$ is the interior of $A$
:$A^-$ is the closure of $A$.

Proof:
Let $A \subseteq S$.
Then from Set in Discrete Topology is Clopen it follows that $A$ is both open and closed in $T$.
From Closed Set Equals its Closure we have that $A = A^-$.
From Set Interior is Largest Open Set, we have that $A^\circ = A$.
{{qed}}
Theorem: Boundary of Subset of Discrete Space is Empty Set

Let $A \subseteq S$.
Then:
:$\partial A = \O$
where:
:$\partial A$ is the boundary of $A$ in $T$.

Proof:
Let $A \subseteq S$.
Then from Set in Discrete Topology is Clopen it follows that $A$ is both open and closed in $T$.
The result follows from Set Clopen iff Boundary is Empty.
{{qed}}
Theorem: Mapping from Discrete Space is Continuous

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 another arbitrary 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 $\phi^{-1} \sqbrk U \in \tau_1$.
{{qed}}
Theorem: Exponential Function is Superfunction

The function $f : \C \to \C$, defined as:
:$\map f z = c^z$
is a superfunction for any complex number $c$.

Proof:
Define $h: \C \to \C$ by $\map h z = z \times c$.
Then:
{{begin-eqn}}
{{eqn | l = \map h {\map f z}
      | r = \map h {c^z}
      | c =
}}
{{eqn | r = c^z \times c
      | c =
}}
{{eqn | r = c^{z + 1}
      | c =
}}
{{eqn | r = \map f {z + 1}
      | c =
}}
{{end-eqn}}
Thus $\map f z = c^z$ is a superfunction and $\map h z = z \times c$ is the corresponding transfer function.
{{qed}}
Category:Superfunctions
ks4u15csxyuzvz8m1rtehu9rmi5euo0
Theorem: Standard Discrete Metric induces Discrete Topology

Let $M = \struct {A, d}$ be the (standard) discrete metric space on $A$.
Then $d$ induces the discrete topology on $A$.
Thus the discrete topology is metrizable.

Proof:
Let $a \in A$.
From Subset of Standard Discrete Metric Space is Open, a set $U \subseteq A$ is open in $M$.
So, in particular, $\set a$ is open in $\struct {A, d}$.
This holds for all $a \in A$.
From Metric Induces Topology it follows that $\set a$ is an open set in $\struct {A, \tau_{A, d} }$.
The result follows from Basis for Discrete Topology.
{{qed}}
Theorem: Discrete Space satisfies all Separation Properties

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ fulfils all separation axioms.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
Then we use Metric Space fulfils all Separation Axioms.
{{qed}}
Theorem: Point in Discrete Space is Neighborhood

Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $x \in S$.
Then $\set x$ is a neighborhood of $x$ in $T$.

Proof:
By definition, a neighborhood $N_x$ of $x$ is any subset of $S$ containing an open set which itself contains $x$.
That is:
:$\exists U \in \tau: x \in U \subseteq N_x \subseteq S$
From Set in Discrete Topology is Clopen we have that $\set x$ is open set in $S$.
So by Set is Subset of Itself, $\set x$ is a subset of $S$ containing an open set $\set x$ which itself contains $x$.
Hence the result.
{{qed}}
Theorem: Discrete Space is Strongly Locally Compact

Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is strongly locally compact.

Proof:
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$ of $T$.
Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure in $T$.
From Singleton Set in Discrete Space is Compact, we have that $\set x$ is compact in $T$.
Hence the result by definition of strongly locally compact.
{{qed}}
Theorem: Discrete Space is First-Countable

Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is first-countable.

Proof:
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$.
From the definition of local basis, it is clear that $\set {\set x}$ is (trivially) a local basis at $x$.
That is, that every open set of $S$ containing $x$ also contains at least one of the sets of $\set {\set x}$.
Equally trivially, we have that $\set {\set x}$ is countable.
Hence the result by definition of first-countable.
{{qed}}
Theorem: Discrete Space has Open Locally Finite Cover

Let $T = \struct {S, \tau}$ be a discrete topological space.
Consider the set $\CC$ of all singleton subsets of $S$:
:$\CC := \set {\set x: x \in S}$
Then $\CC$ is an open cover of $T$ which is locally finite.
This cover is the finest cover on $S$.
That is, if $\VV$ is a cover of $T$, then $\CC$ is a refinement of $\VV$.

Proof:
We have that:
:$\forall x \in S: \exists \set x \in \CC: x \in \set x$
and so $\CC$ is a cover for $S$.
Then from Set in Discrete Topology is Clopen, it follows that $\CC$ is an open cover of $T$.
From Point in Discrete Space is Neighborhood, every point $x \in S$ has a neighborhood $\set x$.
This neighborhood $\set x$ intersects exactly one element of $\CC$, that is: $\set x$ itself.
As $1$ is a finite number, the result follows from definition of locally finite.
Now consider any cover $\VV$ of $S$.
By definition:
:$\forall x \in S: \exists V \in \VV: x \in V$
That is:
:$\forall x \in S: \exists V \in \VV: \set x \subseteq V$
That is, every element of $\CC$ is contained in some element of $\VV$.
Thus by definition, $\CC$ is a refinement of $\VV$.
{{qed}}
Theorem: Discrete Space is Paracompact

Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is paracompact.

Proof:
Let $\VV$ be any open cover of $S$.
Consider the set $\CC$ of all singleton subsets of $S$:
:$\CC := \set {\set x: x \in S}$
From Discrete Space has Open Locally Finite Cover, $\CC$ is an open cover which is locally finite.
This result also shows that $\CC$ is the finest cover on $T$.
So $\CC$ is an open refinement of $\VV$ which is locally finite.
So $T$ is paracompact, by definition.
{{qed}}
Theorem: Singleton Set in Discrete Space is Compact

Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Let $x \in S$.
Then $\set x$ is compact.

Proof:
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$.
Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure.
As $\set {\set x}$ is (trivially) an open cover of $\set x$, it follows by definition that $\set x$ is compact.
{{qed}}
Theorem: Singleton Set in Discrete Space is Compact

Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Let $x \in S$.
Then $\set x$ is compact.

Proof:
Follows directly from Finite Topological Space is Compact.
{{qed}}
Theorem: Countable Discrete Space is Sigma-Compact

Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is $\sigma$-compact.

Proof:
We have that Singleton Set in Discrete Space is Compact.
We also have that $S$ is the union of all its singleton sets:
:$\ds S = \bigcup_{x \mathop \in S} \set x$
As $S$ is countable, it is the union of countably many compact sets.
Hence the result, by definition of $\sigma$-compact.
{{qed}}
Theorem: Countable Discrete Space is Sigma-Compact

Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is $\sigma$-compact.

Proof:
A direct application of Countable Space is Sigma-Compact.
Theorem: Countable Discrete Space is Lindelöf

Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is a Lindelöf space.

Proof:
We have:
:Countable Discrete Space is $\sigma$-Compact
:$\sigma$-Compact Space is Lindelöf Space
So if $S$ is countable, $T$ is a Lindelöf space.
{{qed}}
Theorem: Basis for Discrete Topology

Let $S$ be a set.
Let $\tau$ be the discrete topology on $S$.
Let $\BB$ be the set of all singleton subsets of $S$:
:$\BB := \set {\set x: x \in S}$.
Then $\BB$ is a basis for $T$.

Proof:
Let $T = \struct {S, \tau}$ be the discrete space on $S$. 
Let $U \in \tau$.
Then:
:$\ds U = \bigcup_{x \mathop \in U} \set x$
Hence:
:$\forall x \in U: \exists \set x \in \BB: \set x \subseteq U$
Thus $U$ is the union of elements of $\BB$.
Hence by definition $\BB$ is a basis for $T$.
{{qed}}
Category:Discrete Topologies
Category:Examples of Topological Bases
st7vgz824totwiqa4g72uh6xesv7g37
Theorem: Countable Discrete Space is Second-Countable

Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is second-countable.

Proof:
From Basis for Discrete Topology, the set:
:$\BB := \set {\set x: x \in S}$
is a basis for $T$.
There is a trivial one-to-one correspondence $\phi: S \leftrightarrow \BB$ between $S$ and $\BB$:
:$\forall x \in S: \map \phi x = \set x$
Let $S$ be countable.
Then $\BB$ is also countable by definition of countability.
So we have that $T$ has a countable basis, and so is second-countable by definition.
{{qed}}
Theorem: Countable Discrete Space is Separable

Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is separable.

Proof:
Let $T = \left({S, \tau}\right)$ be a countable discrete topological space.
From Countable Discrete Space is Second-Countable:
:$T$ is second-countable.
From Second-Countable Space is Separable:
:$T$ is separable.
{{qed}}
Theorem: Countable Discrete Space is Separable

Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is separable.

Proof:
Follows immediately from Countable Space is Separable.
{{qed}}
Theorem: Uncountable Discrete Space is not Separable

Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not separable.

Proof:
{{Recall|Separable Space|separable space}}
{{:Definition:Separable Space}}
Let $H \subseteq S$ be everywhere dense in $T$.
Then by definition of everywhere dense:
:$H^- = S$
where $H^-$ denotes the closure of $H$.
We have {{hypothesis}} that $T$ is a discrete space.
Hence from Interior Equals Closure of Subset of Discrete Space:
:$H^- = H$
So:
:$H^- = S \implies H = S$
But $S$ is uncountable.
So there exists no $H \subseteq S$ such that $H$ is both countable and everywhere dense.
Hence by definition of separable space, if $T$ is an uncountable discrete space it can not be separable.
{{qed}}
Theorem: Uncountable Discrete Space is not Second-Countable

Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not second-countable.

Proof:
We have that an Uncountable Discrete Space is not Separable.
From Second-Countable Space is Separable, it follows that $T$ cannot be second-countable.
{{qed}}
Theorem: Uncountable Discrete Space is not Lindelöf

Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not a Lindelöf space.

Proof:
Consider the set $\CC$ of all singleton subsets of $S$:
:$\CC := \set {\set x: x \in S}$
From Discrete Space has Open Locally Finite Cover, $\CC$ is an open cover of $S$ which is finer than any other open cover of $S$.
That is, $\CC$ is an open cover of $S$ which is uncountable and has no countable subcover.
(Note that a subcover is a refinement of a cover.)
So by definition $T$ cannot be a Lindelöf space.
{{qed}}
Theorem: Uncountable Discrete Space is not Sigma-Compact

Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not a $\sigma$-compact space.

Proof:
We have that an Uncountable Discrete Space is not Lindelöf.
But a $\sigma$-compact space is a Lindelöf space.
So an uncountable discrete space cannot be $\sigma$-compact.
{{qed}}
Theorem: Discrete Space is Compact iff Finite

Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Then $T$ is compact {{iff}} $S$ is a finite set, thereby making $\tau$ the finite discrete topology on $S$.

Proof:
=== Necessary Condition ===
Let $T$ be a compact discrete space.
{{AimForCont}} $T$ is infinite.
Let $\CC$ be the cover for $S$ defined as:
:$\CC = \set {\set x: x \in S}$
As $S$ is an infinite set then so is $\CC$.
Let $\CC'$ be a proper subset of $\CC$.
Then:
:$\exists y \in S: \set y \notin \CC'$
and so $\CC'$ is not a cover for $S$.
So by definition $\CC'$ is not a subcover of $\CC$.
So $\CC$ can have no finite subcover.
Hence by definition $T$ can not be compact.
By Proof by Contradiction it follows that if $T$ is compact then $S$ must be finite.
{{qed|lemma}}
Theorem: Finite Space is Sequentially Compact

Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.
Then $T$ is sequentially compact.

Proof:
We have:
:Finite Topological Space is Compact
:Compact Space is Countably Compact
:Finite Space is Second-Countable
:Second-Countable Space is First-Countable
:First-Countable Space is Sequentially Compact iff Countably Compact.
Hence the result.
{{qed}}
Category:Sequentially Compact Spaces
Category:Finite Topological Spaces
9ri8lk23csukgbmryg9cz43fvkt1331
Theorem: Discrete Space is Fully Normal

Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is fully normal.

Proof:
We have that a Discrete Space is fully $T_4$.
Then we note that from Discrete Space satisfies all Separation Properties, a discrete space is a $T_1$ space.
Therefore, by definition, $T$ is fully normal.
{{qed}}
Category:Discrete Topologies
Category:Examples of Fully Normal Spaces
g87snaed81aaupu98m4ycsnw3udifqp
Theorem: Discrete Space is Fully T4

Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is fully $T_4$.

Proof:
Consider the set $\CC$ of all singleton subsets of $S$:
:$\CC := \set {\set x: x \in S}$
From Discrete Space has Open Locally Finite Cover:
:$\CC$ is an open cover of $T$ which is locally finite.
Let $x \in S$.
The star of $x$ with respect to $\CC$ is defined as:
:$\ds x^* := \bigcup \set {U \in \CC: x \in U}$
That is, the union of all sets in $\CC$ which contain $x$.
But here the only set of $\CC$ containing $x$ is $\set x$.
Now let $\VV$ be an open cover of $T$.
From the above, it follows that $\CC$ is a cover for $S$ such that:
:$\forall x \in S: \exists U \in \VV: x^* \subseteq U$
where $x^* = \set x$ is the star of $x$ with respect to $\CC$.
That is, $\CC$ is a barycentric refinement of $\VV$.
Therefore, by definition, $T$ is fully $T_4$.
{{qed}}
Category:Discrete Topologies
Category:Examples of Fully T4 Spaces
9xdaud3vmdf2k0cdrvy0fq4lbhwtw7a
Theorem: Finite Topological Space is Compact

Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.
Then $T$ is compact.

Proof:
From Cardinality of Empty Set, the empty set is finite.
Suppose $S$ is empty.
Then by Empty Set is Compact, $T$ is compact.
Suppose $S$ is non-empty.
Let $\VV$ be an open cover of $T$.
For each $x \in S$, define $\VV_x$ to be $\set {V \in \VV : x \in V}$
Since $S$ is finite, and since by definition of a cover, each $x\in S$ is contained in at least one $V$ in $\VV$, we have that $\set {\VV_x : x \in S}$ is a finite collection of nonempty sets.
From Principle of Finite Choice, there is a choice function which selects one $V_x$ from $\VV_x$ for each $x \in S$.  By definition of $\VV_x$, such a $V_x$ contains $x$.
Since there were only finitely many $\VV_x$, this provides finitely many open sets $V_x \in \VV$ such that $\ds S \subseteq \bigcup_{x \mathop \in S} V_x$.
Thus $\set {\VV_x : x \in S}$ is a finite subcover of $\VV$.
The result follows by definition of compact.
{{qed}}