avewright/proofwiki-math huggingface dataset:
Theorem: Condition for Open Extension Space to be T0

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a $T_0$ space {{iff}} $T$ is.

Proof:
By definition:
:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
Let $T = \struct {S, \tau}$ be a $T_0$ space.
Let $\forall x, y \in S$ such that $x \ne y$.
First, suppose $x \ne p \ne y$.
Then by definition of $T$ as a $T_0$ space:
:$\exists U \in \tau: x \in U, y \notin U$
or:
:$\exists U \in \tau: y \in U, x \notin U$
Then as $U \in \tau^*_{\bar p}$, it follows that:
:$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$
or:
:$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$
Now suppose that $y = p$.
Then by definition:
:$\forall U \in \tau^*_{\bar p}: x \in U, p \notin U$
unless $U = S^*_p$.
So we have shown that:
:$\forall x, y \in S^*_p$ such that $x \ne y$, either:
::$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$
:or:
::$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$
and so $T^*_{\bar p}$ is a $T_0$ space.
Now suppose that $T = \struct {S, \tau}$ is '''not''' a $T_0$ space.
Let $x, y \in S$ such that:
:$\forall U \in \tau: x \in U \implies y \in U$
:$\forall U \in \tau: y \in U \implies x \in U$
As $U \in \tau^*_{\bar p}$ it follows that:
:$\forall U \in \tau^*_{\bar p}: x \in U \implies y \in U$
:$\forall U \in \tau^*_{\bar p}: y \in U \implies x \in U$
and if $U = S^*_p$ the same applies.
Hence $T^*_{\bar p}$ is not a $T_0$ space.
{{qed}}
Theorem: Non-Trivial Excluded Point Topology is not T1

Let $T = \struct {S, \tau_{\bar p} }$ be a excluded point space such that $S$ is not a singleton.
Then $T$ is not a $T_1$ space.

Proof:
Follows directly from:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Open Extension Topology is not $T_1$
{{qed}}
Theorem: Limit Points in Open Extension Space

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Let $x \in S$.
Then $p$ is a limit point of $x$.

Proof:
Every open set of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by definition.
So every open set $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such open set) contains $x$.
So by definition of the limit point of a point, $p$ is a limit point of $x$.
{{qed}}
Category:Limit Points in Open Extension Space
Category:Open Extension Topologies
Category:Examples of Limit Points
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Theorem: Limit Points in Excluded Point Space

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Let $x \in S$ such that $x \ne p$.
Then $p$ is the only limit point of $x$.
Similarly, let $U \subseteq S$.
Then $p$ is the only limit point of $U$.

Proof:
Let $U \subseteq S$.
Let $x \in S$ such that $x \ne p$.
From:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Limit Points in Open Extension Space
it follows that:
:$p$ is a limit point of $U$
:$p$ is a limit point of $x$.
Now suppose $y \in S$ such that $y \ne p$.
We have by definition of excluded point space that $\set y$ is open in $T$.
So there is no $z \in \set y: z \ne y, z \in U$. 
Hence $y$ can not be a limit point of $U$.
Similarly $x \notin \set y$.
So $y$ can not be a limit point of $x$.
Hence the result.
{{qed}}
Theorem: Excluded Point Space is T5

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_5$ space.

Proof:
Let $A, B \subseteq S$ such that:
:$A^- \cap B = A \cap B^- = \O$
where $A^-$ denotes the closure of $A$.
Every closed set of $T$ contains $\set p$.
From Limit Points in Excluded Point Space and the definition of closure:
:$A^- = A \cup \set p, B^- = B \cup \set p$
So if $A^- \cap B = A \cap B^- = \O$ then $p \in A^-$ and $p \in B^-$.
Thus, $p \notin A$ and $p \notin B$ and so by definition $A$ and $B$ are open sets in $T$.
It follows that $A \cap B = \O$.
Hence the result, as we have shown that:
:$\forall A, B \subseteq X, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau_{\bar p}: A \subseteq U, B \subseteq V, U \cap V = \O$
where $U = A$ and $V = B$.
{{qed}}
Theorem: Excluded Point Space is Compact

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a compact space.

Proof:
We have:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Open Extension Space is Compact
{{qed}}
Theorem: Excluded Point Space is Compact

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a compact space.

Proof:
By definition of excluded point space, the only open set of $T$ which contains $p$ is $S$.
So any open cover $\CC$ of $T$ must have $S$ in it.
So $\set S$ will be a subcover of $\CC$, whatever $\CC$ may be.
And $\set S$ (having only one set in it) is trivially a finite cover of $T$.
Hence the result, by definition of compact topological space.
{{qed}}
Theorem: Open Extension Space is Compact

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a compact space.

Proof:
By definition of open extension space, the only open set of $T^*_{\bar p}$ containing $p$ is $S^*_p$.
So any open cover $\CC$ of $T^*_{\bar p}$ must have $S^*_p$ in it.
So $\set {S^*_p}$ will be a subcover of $\CC$, whatever $\CC$ may be.
And $\set {S^*_p}$ (having only one set in it) is trivially a finite cover of $T^*_{\bar p}$.
Hence the result, by definition of compact space.
{{qed}}
Theorem: Open Extension Space is Connected

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a connected space.

Proof:
The only open set of $T$ which contains $p$ is $S^*_p$.
Therefore it is impossible to set up a separation of $T$, as $S^*_p$ will always need to be an element of such a separation.
{{qed}}
Theorem: Excluded Point Space is Connected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T^*_{\bar p}$ is a connected space.

Proof:
We have:
: Excluded Point Topology is Open Extension Topology of Discrete Topology
: Open Extension Space is Connected
{{qed}}
Theorem: Excluded Point Space is Connected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T^*_{\bar p}$ is a connected space.

Proof:
The only open set of $T$ which contains $p$ is $S$.
Therefore it is impossible to set up a separation of $T$, as $S$ will always need to be an element of such a separation.
{{qed}}
Theorem: Excluded Point Space is Connected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T^*_{\bar p}$ is a connected space.

Proof:
: Excluded Point Space is Ultraconnected
: Ultraconnected Space is Path-Connected
: Path-Connected Space is Connected
{{qed}}
Theorem: Open Extension Space is Ultraconnected

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is ultraconnected.

Proof:
Apart from $S^*_p$, every open set of $T^*_{\bar p}$ does not contain $p$, by definition of open extension space.
So, apart from $\O$, every closed set of $T$ does contain $p$, by definition of closed set.
So every pair of closed sets of $T$ has an intersection which contains at least $p$.
So there are no non-empty disjoint closed sets of $T$.
Hence the result, by definition of ultraconnected.
{{qed}}
Theorem: Excluded Point Space is Ultraconnected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is ultraconnected.

Proof:
: Excluded Point Topology is Open Extension Topology of Discrete Topology
: Open Extension Space is Ultraconnected
{{qed}}
Theorem: Excluded Point Space is Ultraconnected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is ultraconnected.

Proof:
Apart from $S$, every open set of $T$ does not contain $p$, by definition of excluded point space.
So, apart from $\O$, every closed set of $T$ does contain $p$, by definition of closed set.
So every pair of closed sets of $T$ has an intersection which contains at least $p$.
So there are no non-empty disjoint closed sets of $T$.
Hence the result, by definition of ultraconnected.
{{qed}}
Theorem: Excluded Point Space is not Irreducible

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space with at least three points.
Then $T^*_{\bar p}$ is not irreducible.

Proof:
By definition, open sets of $S$ are precisely the open sets of $S \setminus \set p$ under the discrete topology.
Let $x, y \in S \setminus \set p: x \ne y$.
Then $\set x$ and $\set y$ are both open sets of $T$ such that $\set x \cap \set y = \O$.
Hence the result, by definition of irreducible.
{{qed}}
Theorem: Excluded Point Space is Path-Connected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T^*_{\bar p}$ is path-connected.

Proof:
: Excluded Point Topology is Open Extension Topology of Discrete Topology
: Open Extension Space is Path-Connected
{{qed}}
Theorem: Excluded Point Space is Path-Connected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T^*_{\bar p}$ is path-connected.

Proof:
:Excluded Point Space is Ultraconnected
:Ultraconnected Space is Path-Connected
{{qed}}
Theorem: Open Extension Space is Path-Connected

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is path-connected.

Proof:
:Open Extension Space is Ultraconnected
:Ultraconnected Space is Path-Connected
{{qed}}
Category:Open Extension Topologies
Category:Examples of Path-Connected Spaces
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Theorem: Excluded Point Space is not Injectively Path-Connected

Let $S$ be a set with at least two distinct elements.
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is not injectively path-connected.

Proof:
Let $q \in S$ be such that $q \ne p$.
Let $f: \closedint 0 1 \to T$ be an injection such that:
{{begin-eqn}}
{{eqn | l = \map f 0
      | r = p
}}
{{eqn | l = \map f 1
      | r = q
}}
{{end-eqn}}
Because $f$ is an injection, it must be that:
:$f^{-1} \sqbrk {\set q} = \set 1$
where $f^{-1}$ denotes the preimage under $f$.
Now since $p \notin \set q$, $\set q$ is open in $\tau_{\bar p}$, by definition of the excluded point topology.
By Closed Real Interval is not Open Set, $\set 1 = \closedint 1 1$ is not open.
Thus we have exhibited an open set whose preimage under $f$ is not open.
Hence $f$ is not continuous.
Since $f$ was an arbitrary injection, no injective path between $p$ and $q$ can exist.
Hence $T$ is not injectively path-connected.
{{qed}}
Theorem: Basis for Excluded Point Space

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Consider the set $\BB$ defined as:
:$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$
Then $B$ is a basis for $S$.

Proof:
Let $H \in \tau_{\bar p}$ be open in $T$.
If $H = S$ then trivially $H$ is the union of elements of $\BB$.
So suppose $H \ne S$.
Then by definition $p \notin H$ and so:
:$\forall y \in H: \exists \set y \in \BB$
Thus:
:$\ds H = \bigcup_{y \mathop \in H} \set y$
So $\BB$ is an analytic basis for $T$.
{{qed}}
It could equally well be shown that $\BB$ is also a synthetic basis for $T$.
Category:Excluded Point Topologies
Category:Topological Bases
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Theorem: Excluded Point Space is Locally Path-Connected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is locally path-connected.

Proof:
Consider the set $\BB$ defined as:
:$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$
Then $\BB$ is a basis for $T$.
Let $H \in \BB$.
Then $\exists x \in S: H = \set x$.
From Point is Path-Connected to Itself we have that $H$ is path-connected.
Now consider the open set $S \in \BB$.
Let $a, b \in S$.
Let $\mathbb I$ be the closed unit interval in $\R$.
Let $f: \mathbb I \to S$ be the mapping defined as:
:<nowiki>$\forall x \in \mathbb I: \map f x = \begin {cases}
a & : x \in \hointr 0 1 \\
b & : x = 1
\end {cases}$</nowiki>
Then $f^{-1} \sqbrk S = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
As $a$ and $b$ are any two points of $S$ it follows that $S$ is path-connected.
So we have shown that all elements of a basis of $T$ are path-connected.
Hence by definition $T$ is locally path-connected.
{{qed}}
Theorem: Excluded Point Space is not Locally Injectively Path-Connected

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is not locally injectively path-connected.

Proof:
Let $\BB \subseteq \tau_{\bar p}$ be a basis for $\tau_{\bar p}$.
Since $\BB$ covers $S$, there must be an open set $B \in \BB$ such that $p \in B$.
By definition of the excluded point topology, the only open set containing $p$ is $S$ itself.
Hence necessarily $S \in \BB$.
But by Excluded Point Space is not Injectively Path-Connected, $S$ is not injectively path-connected.
Hence $\BB$ does not consist only of injectively path-connected sets.
Because $\BB$ was arbitrary, there cannot exist a basis for $\tau_{\bar p}$ comprising only injectively path-connected sets.
Hence, by definition, $T$ is not locally injectively path-connected.
{{qed}}
Theorem: Partition Topology is Zero Dimensional

Let $T = \struct {S, \tau}$ be a partition space.
Then $T$ is zero dimensional.

Proof:
Let $\PP$ be the partition which is the basis for $T$.
From Subset of Partition Space is Open iff Closed, all the elements of $\PP$ are both closed and open.
Hence the result, by definition of zero dimensional space
{{qed}}
Category:Partition Topologies
Category:Examples of Zero Dimensional Spaces
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Theorem: Discrete Space is Zero Dimensional

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

Proof:
We have from Partition of Singletons yields Discrete Topology that a discrete space is a partition space.
The result follows from Partition Topology is Zero Dimensional.
{{qed}}
Theorem: Discrete Space is Zero Dimensional

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

Proof:
Let $\BB$ be the set:
:$\BB := \set {\set x: x \in S}$
From Basis for Discrete Topology, $\BB$ is a basis for $T$.
From Set in Discrete Topology is Clopen, all the elements of $\BB$ are both closed and open.
Hence the result, by definition of zero dimensional space
{{qed}}
Theorem: Discrete Space is Scattered

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

Proof:
We have that Topological Space is Discrete iff All Points are Isolated.
So, by definition, no subset $H \subseteq S$ of $T$ such that $H \ne \O$ is dense-in-itself.
So, again, by definition, $T$ is scattered.
{{qed}}
Theorem: Discrete Space is Extremally Disconnected Hausdorff

Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Then $T$ is an extremally disconnected Hausdorff space.

Proof:
First we note that as Discrete Space satisfies all Separation Properties, $T$ is a $T_2$ (Hausdorff) space.
Then from Interior Equals Closure of Subset of Discrete Space, it follows directly that the closure of every open set of $T$ is open.
Hence, by definition, $T$ is extremally disconnected Hausdorff.
{{qed}}
Theorem: Totally Separated Space is Totally Disconnected

Let $T = \struct {S, \tau}$ be a topological space which is totally separated.
Then $T$ is totally disconnected.

Proof:
{{Recall|Totally Disconnected Space|totally disconnected space}}
{{:Definition:Totally Disconnected Space}}
Let $T = \struct {S, \tau}$ be a totally separated space.
{{Recall|Totally Separated Space|totally separated space}}
{{:Definition:Totally Separated Space/Definition 1}}
{{AimForCont}} $T$ were not totally disconnected.
Then $\exists H \subseteq S: x, y \in U$ such that $H$ is connected.
But by definition of connected, there can be no separation $U \mid V$ of $T$ such that $x \in U, y \in V$.
This contradicts our assertion that $T$ is totally separated.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Open Extension Topology is not Perfectly T4

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is not a perfectly $T_4$ space.

Proof:
By definition:
:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
We have that $S$ is an open set in $T$ and so open set in $T^*_{\bar p}$.
So $\set p = S^*_p \setminus S$ is closed in $T^*_{\bar p}$.
The only open set in $T^*_{\bar p}$ which contains $p$ is $S^*_p$.
So $\set p$ can not be the intersection of open sets of $T^*_{\bar p}$, whether that intersection be countable or not.
So by definition $T^*_{\bar p}$ is not a perfectly $T_4$ space.
{{qed}}
Category:Open Extension Topologies
Category:Examples of Perfectly T4 Spaces
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Theorem: Excluded Point Space is not Perfectly T4

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is not a perfectly $T_4$ space.

Proof:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Open Extension Topology is not Perfectly T4
{{qed}}
Theorem: Subset of Excluded Point Space is not Dense-in-itself

Let $T = \struct {S, \tau_{\bar p} }$ be a excluded point space such that $S$ is not a singleton.
Let $H \subseteq S$.
Then $H$ is not dense-in-itself.

Proof:
From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$.
So by definition, all points of $H$ are isolated in $H$ except $p$.
So if  $H \ne \set p$, $H$ contains at least one point which is isolated in $H$.
As for $p$ itself, from Singleton Point is Isolated we have that $p$ is itself isolated in $\set p$.
So if  $H = \set p$, $H$ also contains one point which is isolated in $H$.
Hence the result, by definition of dense-in-itself.
{{qed}}
Theorem: Excluded Point Space is Scattered

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a scattered space.

Proof:
We have that Subset of Excluded Point Space is not Dense-in-itself.
So, by definition, $T$ is scattered.
{{qed}}
Theorem: Dispersion Point of Excluded Point Space

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $p$ is a dispersion point of $T$.

Proof:
We have that the Excluded Point Topology is Open Extension Topology of Discrete Topology.
So $S \setminus \set p$ is a discrete space.
Then a discrete space is totally disconnected.
The result follows from definition of dispersion point.
{{qed}}
Theorem: Excluded Point Space is First-Countable

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is first-countable.

Proof:
Let $x \in S$ such that $x \ne p$.
By definition, the set $\set x$ is open in $T$.
Let $U \in \tau_{\bar p}: x \in U$.
Then $\set x \subseteq U$ and so $\set {\set x}$ is a local basis at $x$ which is trivially countable.
Now if $x = p$ there is only one $U \in \tau_{\bar p}: p \in U$, and that is $S$.
So $\set S$ is a local basis at $x$ which is trivially countable.
Hence the result, by definition of first-countable.
{{qed}}
Theorem: Excluded Point Space is Sequentially Compact

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a sequentially compact space.

Proof:
:Excluded Point Space is Compact
:Compact Space is Countably Compact
:Excluded Point Space is First-Countable
:First-Countable Space is Sequentially Compact iff Countably Compact
{{qed}}
Theorem: Countable Excluded Point Space is Second-Countable

Let $T = \struct {S, \tau_{\bar p} }$ be a countable excluded point space.
Then $T$ is a second-countable space.

Proof:
Consider the set $\BB$ defined as:
:$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$
From Basis for Excluded Point Space, $\BB$ is a basis for $T$, and trivially has the same cardinality as $S$.
So by definition, if $S$ is countable, then $T$ is second-countable.
{{qed}}
Theorem: Countable Excluded Point Space is Second-Countable

Let $T = \struct {S, \tau_{\bar p} }$ be a countable excluded point space.
Then $T$ is a second-countable space.

Proof:
We have:
: Countable Discrete Space is Second-Countable
: Excluded Point Topology is Open Extension Topology of Discrete Topology
The result follows from Condition for Open Extension Space to be Second-Countable.
{{qed}}
Theorem: Uncountable Excluded Point Space is not Second-Countable

Let $T = \struct {S, \tau_{\bar p} }$ be an uncountable excluded point space.
Then $T$ is not second-countable.

Proof:
Let $H = S \setminus \left\{{p}\right\}$ where $\setminus$ denotes set difference.
By definition, $H$ is an uncountable discrete space.
The result follows from Uncountable Discrete Space is not Second-Countable.
{{qed}}
Theorem: Uncountable Excluded Point Space is not Second-Countable

Let $T = \struct {S, \tau_{\bar p} }$ be an uncountable excluded point space.
Then $T$ is not second-countable.

Proof:
We have:
: Uncountable Discrete Space is not Second-Countable
: Excluded Point Topology is Open Extension Topology of Discrete Topology
The result follows from Condition for Open Extension Space to be Second-Countable
{{qed}}
Theorem: Countable Excluded Point Space is Separable

Let $T = \struct {S, \tau_{\bar p} }$ be a countable excluded point space.
Then $T$ is a separable space.

Proof:
The closure of the set $S$ (trivially) equals $S$.
That is, $S$ is everywhere dense in $T$.
But as $S$ is countable it follows by definition that $T$ is separable.
{{qed}}
Theorem: Uncountable Excluded Point Space is not Separable

Let $T = \struct {S, \tau_{\bar p} }$ be an uncountable excluded point space.
Then $T$ is not separable.

Proof:
Let $H \subseteq S$ such that $H$ is countable.
Then $H \ne S$ as $S$ is uncountable by hypothesis.
From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$.
So, by definition, the closure of $H$ is $H \cup \set p$.
From Countable Union of Countable Sets is Countable we have that $H \cup \set p$ is countable.
So $H \cup \set p \ne S$.
So $H$ is not everywhere dense in $T$.
Thus $T$ can have no countable subset of $S$ which is everywhere dense in $T$.
Hence the result by definition of separable space.
{{qed}}
Theorem: Excluded Set Topology is Topology

Let $T = \struct {S, \tau_{\bar H} }$ be an excluded set space.
Then $\tau_{\bar H}$ is a topology on $S$, and $T$ is a topological space.

Proof:
We have by definition that $S \in \tau_{\bar H}$.
Also, as $H \cap \O = \O$, we have that $\O \in \tau_{\bar H}$.
Now let $U_1, U_2 \in \tau_{\bar H}$.
By definition:
:$H \cap U_1 = \O$
and:
:$H \cap U_2 = \O$
and so by definition of set intersection:
:$H \cap \paren {U_1 \cap U_2} = \O$
So:
:$U_1 \cap U_2 \in \tau_{\bar H}$
Now let $\UU \subseteq \tau_{\bar H}$.
We have that:
:$\forall U \in \UU: H \cap U = \O$
Hence from Subset of Union:
:$H \cap \bigcup \UU = \O$
So all the conditions are fulfilled for $\tau_{\bar H}$ to be a topology on $S$.
{{qed}}
Theorem: Excluded Set Topology is not T0

Let $T = \struct {S, \tau_{\bar H} }$ be an excluded set space where $H$ has at least two distinct points.
Then $T$ is not a $T_0$ space.

Proof:
Let $x, y \in H$ such that $x \ne y$.
Then $x, y \in S$, but by definition $x$ and $y$ are in no other open sets of $T$.
Hence there is no $U \in \tau_{\bar H}$ such that $x \in U, y \notin U$ or $y \in U, x \notin U$.
Hence the result, by definition of $T_0$ space.
{{qed}}
Theorem: Either-Or Topology is Topology

Let $T = \struct {S, \tau}$ be the either-or topology.
Then $\tau$ is a topology on $T$.

Proof:
{{Recall|Either-Or Topology|either-or topology}}
{{:Definition:Either-Or Topology}}