avewright/proofwiki-math huggingface dataset:
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 Power Set of Finite Set is Finite, the power set of $S$ is finite.
Since the topology $\tau$ is by definition a set of subsets of $S$, it follows that $\tau$ is finite as well.
Let $\VV$ be an open cover of $S$.
By definition $\VV \subseteq \tau$ and so is also a finite set.
From Set is Subset of Itself, $\VV \subseteq \VV$.
Thus by definition $\VV$ is a finite subcover of $\VV$.
The result follows by definition of compact.
{{qed}}
Theorem: Finite Space is Second-Countable

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

Proof:
Let $T = \struct {S, \tau}$ be a topological space where $S$ is finite.
As $S$ is finite, then so is its power set $\powerset S$.
So is $\tau \subseteq \powerset S$.
Now $\tau$ is a basis for itself, as every set in $\tau$ is a union of sets from $\tau$ (just one).
But $\tau$ is finite, and so countable.
The result follows from definition of a second-countable space.
{{qed}}
Category:Second-Countable Spaces
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Theorem: Finite Discrete Space satisfies all Compactness Properties

Let $T = \struct {S, \tau}$ be a finite discrete topological space.
Then $T$ satisfies the following compactness properties:
:$T$ is compact.
:$T$ is Sequentially Compact.
:$T$ is Countably Compact.
:$T$ is Weakly Countably Compact.
:$T$ is a Lindelöf Space
:$T$ is Pseudocompact.
:$T$ is $\sigma$-Compact.
:$T$ is Locally Compact.
:$T$ is Weakly Locally Compact.
:$T$ is Strongly Locally Compact.
:$T$ is $\sigma$-Locally Compact.
:$T$ is Weakly $\sigma$-Locally Compact.
:$T$ is Fully Normal.
:$T$ is Fully $T_4$.
:$T$ is Paracompact.
:$T$ is Countably Paracompact.
:$T$ is Metacompact.
:$T$ is Countably Metacompact.

Proof:
A finite discrete space is by definition a topology on a finite set.
A Discrete Space is Fully Normal.
A fully normal space is fully $T_4$ by definition.
The rest of the results follow directly from Finite Space Satisfies All Compactness Properties.
Theorem: Finite Space Satisfies All Compactness Properties

Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.
Then $T$ satisfies the following compactness properties:
:$T$ is compact.
:$T$ is sequentially compact.
:$T$ is countably compact.
:$T$ is weakly countably compact.
:$T$ is a Lindelöf space.
:$T$ is pseudocompact.
:$T$ is $\sigma$-compact.
:$T$ is strongly locally compact.
:$T$ is $\sigma$-locally compact.
:$T$ is weakly $\sigma$-locally compact.
:$T$ is locally compact.
:$T$ is weakly locally compact.
:$T$ is paracompact.
:$T$ is countably paracompact.
:$T$ is metacompact.
:$T$ is countably metacompact.

Proof:
We have that:
:A Finite Topological Space is Compact.
:A Finite Space is Sequentially Compact.
The remaining properties are demonstrated in:
:Sequence of Implications of Global Compactness Properties
:Sequence of Implications of Local Compactness Properties
:Sequence of Implications of Paracompactness Properties
{{qed}}
Category:Finite Topological Spaces
Category:Compact Topological Spaces
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Theorem: Discrete Space is Non-Meager

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

Proof:
Let $U \subseteq S$ such that $U \ne \O$.
From Interior Equals Closure of Subset of Discrete Space, we have:
:$U^\circ = U = U^-$
where $U^\circ$ is the interior and $U^-$ the closure of $U$.
So:
:$\paren {U^-}^\circ = U \ne \O$
Thus, by definition, no non-empty subset of $S$ is nowhere dense.
So $S$ can not be the union (countable or otherwise) of nowhere dense subsets.
So by definition $S$ cannot be meager.
Hence the result.
{{qed}}
Theorem: Discrete Space is Non-Meager

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

Proof:
Let $x \in S$ be an arbitrary point of $T$.
From Set in Discrete Topology is Clopen it follows that $\set x$ is open in $T$.
The result follows from Space with Open Point is Non-Meager.
{{qed}}
Theorem: Discrete Space is Complete Metric Space

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

Proof:
Let $d: S \times S \to \R$ be the standard discrete metric.
From Standard Discrete Metric induces Discrete Topology we have that $\struct {S, d}$ is a metric space whose induced topology is $\tau$.
Consider now a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$.
By the definition of Cauchy sequence:
:$\forall \epsilon > 0: \exists N \in \N$ such that $\forall n, m > N: \map d {x_n, x_m} < \epsilon$
Take $\epsilon = \dfrac 1 2$.
Then:
:$\exists N \in \N: \forall n, m > N: \map d {x_n, x_m} < \epsilon = \dfrac 1 2$
From the definition of standard discrete metric:
:$\exists N \in \N: \forall n, m > N: \map d {x_n, x_m} = 0 \implies x_n = x_m$
Thus the sequence $\sequence {x_n}_{n \mathop \in \N}$ is constant from some $N \in \N$.
From Eventually Constant Sequence is Convergent:
:$x_n \to x \in S$
Thus every Cauchy sequence in $S$ converges.
The result follows by definition of complete metric space.
{{qed}}
Theorem: Discrete Space is not Dense-In-Itself

Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is not dense-in-itself.

Proof:
By definition, $T$ is dense-in-itself {{iff}} it contains no isolated points.
The result follows from Topological Space is Discrete iff All Points are Isolated.
{{qed}}
Theorem: Non-Trivial Discrete Space is not Connected

Let $T = \struct {S, \tau}$ be a non-trivial discrete topological space.
Then $T$ is not connected.

Proof:
Let $T = \struct {S, \tau}$ be a non-trivial discrete space.
Let $a \in S$.
Let $A = \set a$ and $B = \relcomp S {\set a}$, where $\relcomp S {\set a}$ is the complement of $A$ in $S$.
As $T$ is not trivial, $B \ne \O$.
Then from Set in Discrete Topology is Clopen, $A$ and $B$ are both open.
So $A \mid B$ is a separation of $S$.
It follows, by definition, that $T$ is not connected.
{{qed}}
Theorem: Discrete Space is Locally Path-Connected

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

Proof:
From Set in Discrete Topology is Clopen, $\set a$ is open in $T$.
From Basis for Discrete Topology, the set:
:$\BB := \set {\set x: x \in S}$
is a basis for $T$.
Let $\set x \in \BB$.
From Point is Path-Connected to Itself, it follows that $\set x$ is path-connected.
Hence $T$ has a basis consisting entirely of path-connected sets.
So by definition $T$ is locally path-connected.
{{qed}}
Theorem: Point is Path-Connected to Itself

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $a \in S$.
Then $a$ is path-connected to itself.

Proof:
Consider the constant mapping on the closed unit interval $\mathbb I = \left[{0 \,.\,.\, 1}\right]$:
:$\forall x \in \mathbb I: f_a \left({x}\right) = a$
Thus, in particular:
:$f_a \left({0}\right) = a$
:$f_a \left({1}\right) = a$
As a Constant Mapping is Continuous, it follows that $f_a$ is a path in $X$.
Thus $a$ is path-connected to itself.
{{qed}}
Category:Path-Connected Spaces
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Theorem: Discrete Uniformity is Uniformity

Let $S$ be a set.
Let $\UU$ be the discrete uniformity on $S$
Then $\UU$ is indeed a uniformity.

Proof:
We examine the uniformity axioms in turn:
;$\text U 1$: $\forall u \in \UU: \Delta_S \subseteq u$
This follows by definition of the discrete uniformity:
:$\UU := \set {u \subseteq S \times S: \Delta_S \subseteq u}$
;$\text U 2$: $\forall u, v \in \UU: u \cap v \in \UU$
We have that $\forall u, v \subseteq S \times S: \Delta_S \subseteq u, \Delta_S \subseteq v$
So $\Delta_S \subseteq u \cap v$ from Intersection is Largest Subset.
So $u \cap v \in \UU$, and $\text U 2$ holds.
;$\text U 3$: $u \in \UU, u \subseteq v \subseteq S \times S \implies v \in \UU$
Let $u \subseteq v \subseteq S \times S$.
We have that $\Delta_S \subseteq u$ and so $\Delta_S \subseteq v$ from Subset Relation is Transitive.
So $v \in \UU$ by definition of $\UU$, and $\text U 3$ holds.
;$\text U 4$: $\forall u \in \UU: \exists v \in \UU: v \circ v \subseteq u$ where $\circ$ is defined as:
::$u \circ v := \set {\tuple {x, z}: \exists y \in S: \tuple {x, y} \in v, \tuple {y, z} \in u}$
We have by definition of $\UU$ that $\Delta_S \in \UU$.
We note from Diagonal Relation is Equivalence that $\Delta_S$ is an equivalence relation and so by definition transitive.
Therefore from Relation contains Composite with Self iff Transitive we have that $\Delta_S \circ \Delta_S \subseteq \Delta_S$.
So:
:$\forall u \in \UU: \exists \Delta_S \in \UU: \Delta_S \circ \Delta_S \subseteq \Delta_S \subseteq u$
and so $\text U 4$ holds.
;$\text U 5$: $\forall u \in \UU: \exists u^{-1} \in \UU$ where $u^{-1}$ is defined as:
::$u^{-1} := \set {\tuple {y, x}: \tuple {x, y} \in u}$
Let $u \in \UU$, so $\Delta_S \subseteq u \subseteq S \times S$.
Thus:
:$\forall x \in u: \tuple {x, x} \in u \implies \tuple {x, x} \in u^{-1}$
and so $\Delta_S \subseteq u^{-1}$
Thus $u^{-1} \in \UU$.
{{qed|lemma}}
All of the uniformity axioms are satisfied, and $\UU$ is demonstrated to be a uniformity.
{{qed}}
Theorem: Discrete Uniformity generates Discrete Topology

Let $S$ be a set.
Let $\UU$ be the discrete uniformity on $S$.
Then the topology generated by $\UU$ is the discrete topology on $S$.
The diagonal relation $\Delta_S$ generates the basis for this discrete topology.

Proof:
From the construction, let $\tau \subseteq \powerset S$ be a subset of the power set of $S$, created from $\UU$ by:
:$\tau := \set {\map u x: u \in \UU, x \in X}$
where:
:$\forall x \in X: \map u x = \set {y: \tuple {x, y} \in u}$
We need to show that $\tau$ is the discrete topology.
Consider $\Delta_S \in \UU$.
Let $x \in S$.
Then the set:
:$U_x := \set {y: \tuple {x, y} \in \Delta_S}$
We have that:
$\forall x \in S: U_x = \set x \in \UU$
Thus we have:
:$\BB := \set {U_x: x \in S} = \set {\set x: x \in S}$
From Basis for Discrete Topology, we have that $\BB$ is a basis for the discrete topology on $S$.
{{qed}}
Theorem: Indiscrete Topology is Topology

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

Proof:
Let $T = \struct {S, \set {\O, S} }$ be the indiscrete space on $S$.
Confirming the criteria for $T$ to be a topology:
:$(1): \quad$ Trivially, by definition, $\O \in \tau$ and $S \in \tau$.
:$(2): \quad \O \cup \O = \O \in \tau$, $\O \cup S = S \in \tau$ and $S \cup S = S \in \tau$ from Union with Empty Set and Set Union is Idempotent.
:$(3): \quad \O \cap \O = \O \in \tau$, $\O \cap S = \O \in \tau$ and $S \cap S = S \in \tau$ from Intersection with Empty Set and Set Intersection is Idempotent.
{{qed}}
Theorem: Indiscrete Topology is Coarsest Topology

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

Proof:
Let $\phi$ be any topology on $S$.
Then by definition of topology, $\O \in \phi$ and $S \in \phi$
Hence by definition of subset, $\tau \subseteq \phi$. 
Hence by definition of coarser topology, $\tau$ is coarser than $\phi$.
{{qed}}
Theorem: Limit Points of Indiscrete Space

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space consisting of at least two points.
Let $H$ be a subset of $T$ such that $H \ne \O$.
Then every point of $T$ is a limit point of $H$.

Proof:
First we dispose of the case where $S$ is a singleton.
From Trivial Topological Space is Indiscrete, such a $T = \struct {S, \set {\O, S} }$ is a trivial topological space.
From Trivial Topological Space has no Limit Point, such a $T$ has no limit point.
{{Recall|Limit Point of Set/Definition from Open Neighborhood|limit point of $H$}}
{{:Definition:Limit Point of Set/Definition from Open Neighborhood}}
From Open Sets in Indiscrete Topology, there is only one open set of $T$ that contains any points at all, and that is $S$.
So as $S$ contains more than one point, it follows that every point of $T$ is a limit point of $H$.
{{qed}}
Theorem: Sequence in Indiscrete Space converges to Every Point

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $\sequence {s_n}$ be a sequence in $T$.
Then $\sequence {s_n}$ converges to every point of $S$.

Proof:
Let $\alpha \in S$.
By definition, $\sequence {s_n}$ converges to $\alpha$ if every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\sequence {s_n}$.
But as $T$ has only one open set containing any points at all, '''every''' point of $\sequence {s_n}$ is contained in every open set in $T$ containing $\alpha$.
Hence the result.
{{qed}}
Theorem: Subset of Indiscrete Space is Dense-in-itself

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ be a subset of $S$ containing more than one point.
Then $H$ is dense-in-itself.

Proof:
Let $x \in H$.
Then as $H$ is not singleton, $\exists y \in H: y \ne x$.
Then every neighborhood of $x$ contains $y$, as the only open set of $T$ is $S$, which also contains both $x$ and $y$.
Hence $x$ is not isolated by definition.
$x$ is general, so all points in $H$ are similarly not isolated.
Hence the subset $H$ is dense-in-itself by definition.
{{qed}}
Theorem: Indiscrete Space is Non-Meager

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is non-meager.

Proof:
Let $H \subseteq S$ such that $H \ne \O$.
From Limit Points of Indiscrete Space, every point of $S$ is a limit point of $H$.
So, by definition:
:$H^- = S$
where $H^-$ is the closure of $H$ in $T$.
Now the interior of $S$ is $S$ itself (trivially, by definition).
So:
:$\paren {H^-}^\circ = S \ne \O$
where $\paren {H^-}^\circ$ is the interior of the closure of $H$.
So, by definition, $H$ is not nowhere dense.
(Note that if $H = \O$, then Empty Set is Nowhere Dense applies, which is why the stipulation above that $H \ne \O$.)
It follows that $H$ can not be the countable union of nowhere dense subsets of $S$.
So $H$ is, by definition, not meager, so must be non-meager.
{{qed}}
Theorem: Empty Set is Nowhere Dense

Let $T = \struct {S, \tau}$ be a topological space.
Then the empty set $\O$ is nowhere dense in $T$.

Proof:
From Empty Set is Closed in Topological Space, $\O$ is closed in $T$.
From Closed Set Equals its Closure:
:$\O^- = \O$
where $\O^-$ is the the closure of $\O$.
From Empty Set is Element of Topology, $\O$ is open in $T$.
From the definition (trivially) we also have that:
:$\O^\circ = \O$
where $\O^\circ$ is the interior of $\O$.
So:
:$\struct {\O^-}^\circ = \O$
and so by definition $\O$ is nowhere dense in $T$.
{{qed}}
Theorem: Interior of Open Set

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $U \subseteq T$ be open in $T$.
Then:
: $U^\circ = U \iff U \in \tau$
where $U^\circ$ is the interior of $U$.
That is, a subset of $S$ is open in $T$ {{iff}} it equals its interior.

Proof:
=== Necessary Condition ===
Let $U \subseteq T$ be open in $T$.
From Set Interior is Largest Open Set:
: $U \subseteq U^\circ$
But by definition of interior:
: $U^\circ \subseteq U$
Hence $U^\circ = U$.
{{qed|lemma}}
Theorem: Interior of Subset of Indiscrete Space

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subsetneqq S$ (that is, let $H$ be a proper subset of $S$).
Then:
:$H^\circ = H^{\circ -} = H^{\circ - \circ} = \O$
where:
:$H^\circ$ denotes the interior of $H$
:$H^-$ denotes the closure of $H$.

Proof:
As $H \subsetneqq S$, it follows that $H \ne S$.
So the only open subset of $H$ is $\O$.
So by definition:
:$H^\circ = \O$
From Empty Set is Closed in Topological Space, $\O$ is closed in $T$.
From Closed Set Equals its Closure:
:$\O^- = \O$
From Empty Set is Element of Topology, $\O$ is open in $T$.
From Interior of Open Set:
:$\O^\circ = \O$
The result follows.
{{qed}}
Theorem: Closure of Subset of Indiscrete Space

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $\O \subsetneq H \subseteq S$ (that is, let $H$ be a non-empty subset of $T$).
Then:
:$H^- = H^{- \circ} = H^{- \circ -} = S$
where:
:$H^\circ$ denotes the interior of $H$
:$H^-$ denotes the closure of $H$.

Proof:
From Limit Points of Indiscrete Space, every point in $S$ is a limit point of $H$.
So from the definition of closure, $H \ne \O \implies H^-= S$.
From the open set axioms, $S$ is open in $T$.
From Interior of Open Set, $S^\circ = S$.
From Underlying Set of Topological Space is Clopen, $S$ is closed in $T$.
From Closed Set Equals its Closure, $S^- = S$.
The result follows.
{{qed}}
Theorem: Boundary of Subset of Indiscrete Space

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $\O \subsetneq H \subsetneq S$ (that is, let $H$ be a non-empty proper subset of $T$).
Then:
:$\partial H = S$
where $\partial H$ denotes the boundary of $H$.
If $H = \O$ or $H = S$ then $\partial H = \O$.

Proof:
From Closure of Subset of Indiscrete Space, $H^- = S$, where $H^-$ denotes set closure.
From Interior of Subset of Indiscrete Space, $H^\circ = \O$, where $H^\circ$ denotes set interior.
By definition:
:$\partial H = H^- \setminus H^\circ = S \setminus \O = S$
From Open and Closed Sets in Topological Space, $\O$ and $S$ are both closed and open in $T$.
So from Set Clopen iff Boundary is Empty:
:$\partial H = \O$
{{qed}}
Theorem: Boundary of Boundary of Subset of Indiscrete Space

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
Then:
:$\map \partial {\partial H} = \O$
where $\partial H$ denotes the boundary of $H$.

Proof:
From Boundary of Subset of Indiscrete Space, either $\partial H = S$ or $\partial H = \O$, depending on whether $H = \O$ or $H = S$ or not.
From Open and Closed Sets in Topological Space, $\O$ and $S$ are both closed and open in $T$.
So from Set Clopen iff Boundary is Empty:
:$\map \partial {\partial H} = \O$
{{qed}}
Theorem: Subset of Indiscrete Space is Everywhere Dense

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ such that $H \ne \O$.
Then $H$ is everywhere dense.

Proof:
From Limit Points of Indiscrete Space, every point of $T$ is a limit point of $H$.
Hence $H$ is everywhere dense by definition.
{{qed}}
Theorem: Indiscrete Space is Second-Countable

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then  $T$ is a second-countable space.

Proof:
The only basis for $T$ is $\set S$ which is trivially countable.
Hence the result.
{{qed}}
Theorem: Mapping to Indiscrete Space is Continuous

Let $T_1 = \struct {S_1, \tau_1}$ be any topological space.
Let $T_2 = \struct {S_2, \tau_2}$ be the indiscrete topological space on $S_2$.
Let $\phi: S_1 \to S_2$ be a mapping.
Then $\phi$ is continuous.

Proof:
From the definition of continuous:
:$U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$
The only elements of $\tau_2$ are $S_2$ and $\O$, from which:
:$\phi^{-1} \sqbrk {S_2} = S_1 \in \tau_1$
:$\phi^{-1} \sqbrk \O = \O \in \tau_1$
{{qed}}
Theorem: Indiscrete Space is Path-Connected

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is path-connected.

Proof:
Let $a, b \in S$.
Consider any mapping $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$.
From Mapping to Indiscrete Space is Continuous, we have that $f$ is continuous.
The result follows by definition of path-connectedness.
{{qed}}
Theorem: Indiscrete Space is Connected

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is connected.

Proof:
We have that an Indiscrete Space is Path-Connected.
Then we have that a Path-Connected Space is Connected.
{{qed}}
Theorem: Indiscrete Space is Injectively Path-Connected iff Uncountable

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is injectively path-connected {{iff}} $S$ is an uncountable set.

Proof:
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space such that $S$ is uncountable.
Let $a, b \in S$.
Consider an injection $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$.
This can always be found because $S$ is itself uncountable.
From Mapping to Indiscrete Space is Continuous, we have that $f$ is continuous.
Thus $T$ is injectively path-connected.
Now suppose $S$ is an indiscrete topological space which is injectively path-connected.
Then there exists an injection $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$.
This can only exist if $S$ is uncountable.
{{qed}}
Theorem: Indiscrete Space is Irreducible

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is irreducible.

Proof:
There is only one non-empty open set in $T$.
So there can be no two open sets in $T$ which are disjoint.
Hence (trivially) $T$ is irreducible.
{{qed}}
Theorem: Indiscrete Space is Ultraconnected

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is ultraconnected.

Proof:
There is only one non-empty closed set in $T$.
So there can be no two closed sets in $T$ which are disjoint.
Hence (trivially) $T$ is ultraconnected.
{{qed}}
Theorem: Indiscrete Non-Singleton Space is not T0

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space which has more than one element.
Then $T$ is not a $T_0$ space.

Proof:
Let $a, b \in S$.
By definition of indiscrete space, $S$ is the only open set in $T$ which is non-empty.
So (trivially) there is no open set in $T$ containing $a$ and not $b$, or $b$ and not $a$.
Hence the result, by definition of $T_0$ space.
{{qed}}
Theorem: Indiscrete Space is T5

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is a $T_5$ space.

Proof:
By definition, the only two separated sets in $T$ are $S$ and $\O$.
Then there exist two disjoint open sets $S$ and $\O$ containing $S$ and $\O$ respectively.
Hence (trivially) $T$ is a $T_5$ space.
{{qed}}
Theorem: Indiscrete Space is T4

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is a $T_4$ space.

Proof:
We have that an indiscrete space is a $T_5$ space.
Then we have that a $T_5$ Space is $T_4$.
{{qed}}
Theorem: Indiscrete Space is T3

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space .
Then $T$ is a $T_3$ space.

Proof:
Let $F \subseteq S$ be a closed set in $S$.
Let $y \in S$ such that $y \notin F$.
The only way this can happen is if $F = \O$.
So there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
That is, $U = \O$ and $V = S$.
Hence (trivially) the result.
{{qed}}
Theorem: Indiscrete Space is Pseudometrizable

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is pseudometrizable.

Proof:
Let $d: S \times S \to \R$ be the mapping defined as:
:$\forall x, y \in S: \map d {x, y} = 0$
Then clearly $d$ is a pseudometric.
Let $\struct {S, \tau_{\struct {S, d} } }$ be the topological space induced by $d$.
Since $\struct {S, \tau_{\struct {S, d} } }$ is a topological space, by Empty Set is Element of Topology we have that $\O$ is open.
For any $\epsilon \in \R_{>0}$ we have:
:$\map {B_\epsilon} a := \set {x \in S: \map d {x, a} < \epsilon} = S$
where $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$.
Let $U\subset S$ be a non-empty open set.
By Open Sets in Pseudometric Space we see that for every $x \in U$, there must exist an $\epsilon > 0$ such that $\map {B_\epsilon} x \subset U$.
However, $S \subset \map {B_\epsilon} x$.
Thus $U = S$.
Hence it is seen that $\O$ and $S$ are the only open sets of this resulting pseudometric space.
{{qed}}
Theorem: Partition Topology is Topology

Let $S$ be a set.
Let $\PP$ be a partition of $S$.
Let $\tau$ be the set of subsets of $S$ defined as:
:$a \in \tau \iff a$ is the union of sets of $\PP$
Then $\tau$ is a topology on $S$.

Proof:
From Basis for Partition Topology, we have that $\PP$ is a basis for the partition topology.
The result follows.
{{qed}}