avewright/proofwiki-math huggingface dataset:
Theorem: Basis for Partition Topology

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

Proof:
Checking the criteria for $\PP$ to be a synthetic basis for $\tau$:
We have that $\ds S = \bigcup \PP$ from the definition of a partition.
Therefore, $\ds S \subseteq \bigcup \PP$ and $\PP$ is a cover for $S$.
Next, let $B_1, B_2 \in \PP$.
Then as $\PP$ is a partition of $S$, we have that $B_1 \cap B_2 = \O$.
But from Union of Empty Set we have that $\O$ is the (vacuous) union of sets of $\PP$.
Hence $\PP$ is a synthetic basis for $\tau$.
{{qed}}
Theorem: Subset of Partition Space is Open iff Closed

Let $T = \struct {S, \tau}$ be a topological space.
Then $T$ is a partition space {{iff}}:
:$\forall U \subseteq S: U \in \tau \iff \relcomp S U \in \tau$
That is, a topological space is a partition space {{iff}}:
:all open sets are closed
:all closed sets are also open.
That is, all open sets and closed sets of a partition space are clopen.

Proof:
Let $T = \struct {S, \tau}$ be a topological space.
Theorem: Open Set in Partition Topology is Component

Let $T = \struct {S, \tau}$ be a partition topological space.
Then each of its open sets is a component of $T$.

Proof:
Let the partition $\PP$ be a basis of $T$.
From Subset of Partition Space is Open iff Closed, open sets are in fact clopen.
So the elements of $\PP$ are clopen.
The result follows from the definition of components.
{{qed}}
Theorem: Quotient Topology of Partition Topology is Discrete Space

Let $\PP$ be a partition of a set $S$.
Let $T = \struct {S, \tau}$ be the partition space formed from $\PP$.
Let $S / \PP$ be the quotient set of $S$ by $\PP$.
Then the quotient topology $\tau_{S / \PP}$ is a discrete topology.

Proof:
Let $\BB$ be the set defined as:
:$\BB = \set {\set A: A \in S / \PP}$
From Basis for Partition Topology, $\BB$ forms a basis for a partition space on $S$.
From Basis for Discrete Topology, $\BB$ forms a basis for the discrete topology on $S / \PP$.
Hence the result, by definition of quotient topology.
{{qed}}
Theorem: Singleton Partition yields Indiscrete Topology

Let $S$ be a set which is not empty.
Let $\PP$ be the (trivial) singleton partition $\set S$ on $S$.
Then the partition topology on $\PP$ is the indiscrete topology.

Proof:
By definition, the partition topology on $\PP$ is the set of all unions from $\PP$.
This is (trivially, and from Union of Empty Set) $\set {\O, S}$ which is the indiscrete topology on $S$ by definition.
{{qed}}
Theorem: Partition of Singletons yields Discrete Topology

Let $S$ be a set which is non-empty.
Let $\PP$ be the (trivial) partition of singletons on $S$:
:$\PP = \set {\set x: x \in S}$
Then the partition topology on $\PP$ is the discrete topology.

Proof:
From Basis for Discrete Topology it is shown that $\PP$ as defined here forms the basis of the discrete topology.
{{qed}}
Theorem: Partition Topology is not T0

Let $S$ be a set and let $\PP$ be a partition on $S$ which is not the (trivial) partition of singletons.
Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$.
Then $T$ is not a $T_0$ space.

Proof:
As $\PP$ is not the partition of singletons, there exists some $H \in \PP$ such that $a, b \in H: a \ne b$.
Any union of sets from $\PP$ which includes $H$ will therefore contain both $a$ and $b$.
Therefore, any element of $\tau$ containing $a$ will also contain $b$, and similarly, any element of $\tau$ containing $b$ will also contain $a$.
So 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}}
Note the reason for the exception in the statement as given.
If $\PP$ is the partition of singletons:
:$\PP = \set {\set x: x \in S}$
then $T$ as constructed is the discrete topology on $S$.
From Discrete Space satisfies all Separation Properties, in that case $T$ ''does'' satisfy the $T_0$ property.
Theorem: Partition Topology is T5

Let $S$ be a set and let $\PP$ be a partition on $S$ which is not the (trivial) partition of singletons.
Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$.
Then $T$ is a $T_5$ space.

Proof:
Let $A$ and $B$ be subsets of $S$ such that $A^- \cap B = A \cap B^- = \O$.
From Subset of Partition Space is Open iff Closed, we get that $A^-$ and $B^-$ are both clopen. 
From Set is Subset of its Topological Closure:
:$A \subseteq A^-$
From $A^- \cap B = \O$ it follows from Intersection with Complement is Empty iff Subset that $B \subseteq S \setminus \paren {A^-}$.
Since $A^-$ is clopen, it follows that $A^-$ and $S \setminus \paren {A^-}$ are both open.
Also, they are disjoint by definition.
Thus we have $A^-$, an open set containing $A$, and $S \setminus \paren {A^-}$, an open set containing $B$.
Thus, by definition, $T$ is a $T_5$ space.
{{qed}}
Theorem: Partition Space is T3.5

Let $S$ be a set and let $\PP$ be a partition on $S$.
Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$.
Then $T$ is a $T_{3 \frac 1 2}$ space.

Proof:
Let $F \subseteq S$ be closed.
Denote by $S \setminus F$ the relative complement of $F$ in $S$.
Let $x \in S \setminus F$.
Define a mapping $f: S \to \closedint 0 1$ as:
:<nowiki>$\map f s := \begin{cases}
 1 & : \text { if } s \in F \\
 0 & : \text { if } s \in S \setminus F
\end{cases}$</nowiki>
Then $f$ is identically $1$ on $F$, and identically $0$ on $\set x$.
Now if $f$ is continuous, it will be a Urysohn function for $F$ and $\set y$, and $T$ will be a $T_{3 \frac 1 2}$ space.
Now for any $V \subseteq \closedint 0 1$, we have:
:<nowiki>$f^{-1} \sqbrk V = \begin{cases}
 \O            & : \text{ if } 0, 1 \notin V \\
 F             & : \text{ if } 0 \notin V \text { and } 1 \in V \\
 S \setminus F & : \text{ if } 0 \in V \text { and } 1 \notin V \\
 S             & : \text{ if } 0, 1 \in V
\end{cases}$</nowiki>
We have {{hypothesis}} that $F$ is closed in $T$.
Hence by definition $S \setminus F$ is open in $T$.
By Subset of Partition Space is Open iff Closed, $F$ is also open in $T$.
Thus, the preimage of any subset $V$ of $\closedint 0 1$ is open in $T$.
In particular, this holds for the open sets of $\closedint 0 1$.
It follows that $f$ is a continuous mapping, and so a Urysohn function.
Hence $T$ is $T_{3 \frac 1 2}$ space.
{{qed}}
Theorem: Odd-Even Topology is Second-Countable

Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is second-countable.

Proof:
From Basis for Partition Topology, the set:
:$\BB := \set {\set {2 k - 1, 2 k}: k \in \Z, k > 0}$
is a basis for $T$.
There is an obvious one-to-one correspondence $\phi: \Z_{>0} \leftrightarrow \BB$ between $\Z_{>0}$ and $\BB$:
:$\forall x \in \Z_{>0}: \map \phi x = \set {2 x - 1, 2 x}$
But $\Z_{>0} \subseteq \Z$, and Integers are Countably Infinite.
So from Subset of Countably Infinite Set is Countable, $\Z_{>0}$ is countable.
Thus $\BB$ 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: Odd-Even Topology is Weakly Countably Compact

Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is weakly countably compact.

Proof:
Let $H \subseteq \Z_{>0}$ such that $H$ is infinite.
Let $x \in H$.
By definition, the odd-even topology is a partition topology.
So $U$ is a union of sets of the form $\set {2 k - 1, 2 k}$.
Now if $x \in U$, it will be of the form $2 k - 1$ or $2 k$.
So there will exist $y \in U$ of the form $2 k$ or $2 k - 1$.
So, by definition, $x$ is a limit point of $H$.
So, by definition, $T$ is weakly countably compact.
{{qed}}
Theorem: Odd-Even Topology is not Countably Compact

Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is not countably compact.

Proof:
By definition, the odd-even topology is a partition topology.
Let $\PP$ be the partition which is the basis for $T$:
:$\PP = \set {\set {2 k - 1, 2 k}: k \in \Z_{>0} }$
Then $\PP$ is a countable open cover of $S$ which has no finite subcover.
Hence the result.
{{qed}}
Theorem: Countable Discrete Space is not Weakly Countably Compact

Let $T = \struct {S, \tau}$ be a countable discrete space.
Then $T$ is not weakly countably compact.

Proof:
Let $A \subseteq S$ be an infinite subset of $S$.
Then as Set in Discrete Topology is Clopen, $A$ is closed in $T$.
From Closed Set Equals its Closure, $A = A^-$ where $A^-$ is the closure of $A$.
From the definition, $x$ is a limit point of $A$ if it belongs to the closure of $A$ but is not an isolated point of $A$.
But then we have Topological Space is Discrete iff All Points are Isolated.
So $A$ has no limit points in $T$.
So $T$ is not weakly countably compact.
{{qed}}
Theorem: Weak Countable Compactness is not Preserved under Continuous Maps

Let $T_A = \struct {S_A, \tau_A}$ be a topological space which is weakly countably compact.
Let $T_B = \struct {S_B, \tau_B}$ be another topological space.
Let $\phi: T_A \to T_B$ be a continuous mapping.
Then $T_B$ is not necessarily weakly countably compact.

Proof:
Let $\Z_{>0}$ be the strictly positive integers:
:$\Z_{>0} = \set {1, 2, 3, \ldots}$
Let $T_A = \struct {\Z_{>0}, \tau_A}$ be the odd-even topology.
Let $T_B = \struct {\Z_{>0}, \tau_B}$ be the discrete topology on $\Z_{>0}$.
Let $\phi: T_A \to T_B$ be the mapping:
:$\map \phi {2 k} = k, \map \phi {2 k - 1} = k$
Then:
:$\map {\phi^{-1} } k = \set {2 k, 2 k - 1} \in \tau_A$
demonstrating that $\phi$ is continuous.
Now we have that the Odd-Even Topology is Weakly Countably Compact.
But we also have that a Countable Discrete Space is not Weakly Countably Compact.
Hence the result.
{{qed}}
Theorem: Deleted Integer Topology is not Countably Compact

Let $S = \R_{\ge 0} \setminus \Z$, and let $\tau$ be the deleted integer topology on $S$.
Then the topological space $T = \struct {S, \tau}$ is not countably compact.

Proof:
By definition, the deleted integer topology is a partition topology.
Let $\PP$ be the partition which is the basis for $T$, that is:
:$\PP = \set {\openint {n - 1} n: n \in \Z_{> 0} }$
Then $\PP$ is a countable open cover of $S$ which has no finite subcover.
Hence the result.
{{qed}}
Theorem: Deleted Integer Topology is Second-Countable

Let $S = \R_{\ge 0} \setminus \Z$.
Let $\tau$ be the deleted integer topology on $S$.
Then the topological space $T = \struct {S, \tau}$ is second-countable.

Proof:
Let $\Z_{>0}$ be understood as the set of strictly positive integers:
:$\Z_{>0} = \set {x \in \Z: x > 0} = \set {1, 2, 3, \ldots}$
From Basis for Partition Topology, the set:
:$\BB = \set {\openint {n - 1} n: n \in Z_{>0} }$
is a basis for $T$.
There is an obvious one-to-one correspondence $\phi: \Z_{> 0} \leftrightarrow \BB$ between $\Z_{> 0}$ and $\BB$:
:$\forall x \in \Z_{>0}: \map \phi x = \openint {x - 1} x$
But $\Z_{>0} \subseteq \Z$ and Integers are Countably Infinite.
So from Subset of Countably Infinite Set is Countable, $\Z_{>0}$ is countable.
Thus $\BB$ 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: Deleted Integer Topology is Weakly Countably Compact

Let $S = \R_{\ge 0} \setminus \Z$.
Let $\tau$ be the deleted integer topology on $S$.
Then the topological space $T = \struct {S, \tau}$ is weakly countably compact.

Proof:
Let $A \subseteq S$ such that $A$ is infinite.
Let $x \in A$.
Let $n \in \Z$ such that $n < x < n + 1$.
By the definition of the deleted integer topology, $\openint n {n + 1}$ is open in $T$.
We have that $\openint n {n + 1}$ is infinite.
Take some $y \in \openint n {n + 1}$ such that $x \ne y$.
Now we claim that $y$ is a limit point of $A$.
Let $N_y$ be an open neighbourhood of $y$.
To be open in $T$, $N_y$ needs to be the union of open intervals of the form $\openint k {k + 1}$ for some $k \in \Z_{\ge 0}$.
Therefore to contain $y$ and be open at the same time, $\openint n {n + 1} \subseteq N_y$.
Thus $x \in N_y$ and $y \ne x \in A \cap N_y \ne \O$.
This is the definition of a limit point.
So, by definition, $T$ is weakly countably compact.
{{qed}}
Theorem: Pseudometric induces Topology

Let $S \ne \O$ be a non-empty set.
Consider a pseudometric space $\struct {S, d}$ where $d: S \times S \to \R_{\ge 0}$ is a pseudometric.
Then $\struct {S, d}$ gives rise to a topological space $\struct {S, \tau_d}$ whose topology $\tau_d$ is '''defined''' (or '''induced''') by $d$.

Proof:
Let $\tau_d$ be the set of all $X \subseteq S$ which are open in the sense that:
:$\forall y \in X: \exists \epsilon > 0: \map {B_\epsilon} y \subseteq X$
where $\map {B_\epsilon} y$ is the open $\epsilon$-ball of $y$.
Equivalently:
:$\forall x \in X: \exists \epsilon \in \R_{>0}: \forall y \in S: \map d {x, y} < \epsilon \implies y \in X$
We need to show that $\tau_d$ forms a topology on $S$.
We examine each of the open set axioms in turn.
Theorem: Open Sets in Pseudometric Space

Let $P = \struct {A, d}$ be a pseudometric space.
Then $\O$ and $A$ are both open in $P$.

Proof:
From the definition of open set:
: $\forall y \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} y \subseteq U$
where $U$ is an open set of $P$.
That is:
:''One cannot get out of $U$ by moving an arbitrarily small distance from any point in $U$.''
Take the case where $U = \O$.
The empty set $\O$ is open by dint of the fact that there are no points $y$ in $\O$ for which the condition is false.
Thus for $\O$:
: $\forall y \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} y \subseteq U$ is vacuously true.
When $U = A$, there are no points in $A$ to which one could get to by leaving $U$, an arbitrarily short distance or no, because there are no points in $A$ that ''are'' outside of $U$.
{{qed}}
Category:Pseudometric Spaces
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Theorem: Partition Space is Pseudometrizable

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

Proof:
Let $\PP$ be the partition which is the basis for $T$.
Let us define $d: S^2 \to \R$ by:
:<nowiki>$\forall x, y \in S: \map d {x, y} = \begin{cases}
0 & : \exists U \in \PP: x, y \in U \\
1 & : \text{otherwise} \end{cases}$</nowiki>
That is, $\map d {x, y} = 0$ when $x$ and $y$ are in the same set in the partition $\PP$, and $1$ otherwise.
This is easily established as being a pseudometric on $S$.
Hence the result.
{{qed}}
Theorem: Double Pointed Space is not T0

Let $T_S = \struct {S, \tau_S}$ be a topological space.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete space on an arbitrary doubleton $A = \set {a, b}$.
Let $T = \struct {T_S \times D, \tau}$ be the double pointed topological space on $T_S$.
Then $T$ is not a $T_0$ space.

Proof:
By definition, the double pointed topology $\tau$ on $T_S$ is the product topology on $T_S \times D$.
Let $x \in S$, and consider the point $\tuple {x, a} \in S \times A$.
Then:
{{begin-eqn}}
{{eqn | q = \forall U \in \tau
      | l = \tuple {x, a} \in U
      | o = \implies
      | r = \tuple {x, b} \in U
}}
{{eqn | q = \forall U \in \tau
      | l = \tuple {x, b} \in U
      | o = \implies
      | r = \tuple {x, a} \in U
}}
{{end-eqn}}
as $D$ is an indiscrete space.
Hence the result, by definition of $T_0$ space.
{{qed}}
Theorem: Double Pointed Space is not T0

Let $T_S = \struct {S, \tau_S}$ be a topological space.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete space on an arbitrary doubleton $A = \set {a, b}$.
Let $T = \struct {T_S \times D, \tau}$ be the double pointed topological space on $T_S$.
Then $T$ is not a $T_0$ space.

Proof:
By definition, the double pointed topology $\tau$ on $T_S$ is the product topology on $T_S \times D$.
By definition, $D$ is the indiscrete space on a doubleton.
{{AimForCont}} $T$ is a $T_0$ space.
Then from Product Space is $T_0$ iff Factor Spaces are $T_0$ it follows that $D$ is also a $T_0$ space.
But from Indiscrete Non-Singleton Space is not $T_0$, $D$ is not a $T_0$ space.
The result follows by Proof by Contradiction.
{{qed}}
Theorem: Double Pointed Discrete Real Number Space is Weakly Countably Compact

Let $T_\R = \struct {\R, \tau_\R}$ be the (uncountable) discrete space on the set of real numbers.
Let $T_D = \struct {D, \tau_D}$ be the indiscrete topology on the doubleton $D = \set {a, b}$.
Let $T = T_\R \times T_D$ be the double pointed (uncountable) discrete space which is the product space of $T_\R$ and $T_D$.
Then $T$ is weakly countably compact.

Proof:
We have that $T$ is a partition topology, whose basis $\PP$ is defined as:
:$\PP = \set {\set {\tuple {s, a}, \tuple {s, b} }: s \in \R}$
Let $A \subseteq \R \times D$ such that $A$ is infinite.
Let $x \in A$.
Let $U$ be the union of sets of the form $\set {\tuple {s, a}, \tuple {s, b} }$, and hence open in $T$.
Now if $x \in U$, it will be an element in some $\set {\tuple {s, a}, \tuple {s, b} }$.
So there will exist $y \in U$ which will also be an element in that $\set {\tuple {s, a}, \tuple {s, b} }$.
So, by definition, $x$ is a limit point of $A$.
So, by definition, $T$ is weakly countably compact.
{{qed}}
Theorem: Double Pointed Discrete Real Number Space is not Lindelöf

Let $T_\R = \struct {\R, \tau_\R}$ be the (uncountable) discrete space on the set of real numbers.
Let $T_D = \struct {D, \tau_D}$ be the indiscrete topology on the doubleton $D = \set {a, b}$.
Let $T = T_\R \times T_D$ be the double pointed (uncountable) discrete space which is the product space of $T_\R$ and $T_D$.
Then $T$ is '''not''' a Lindelöf space.

Proof:
We have that $T$ is a partition topology, whose basis $\PP$ is defined as:
:$\PP = \set {\set {\paren {s, a}, \paren {s, b} }: s \in \R}$
We have that $\PP$ is an open cover of $T$.
But $\PP$ has no countable subcover.
Hence the result, by definition of Lindelöf space.
{{qed}}
Theorem: Particular Point Topology is Topology

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $\tau_p$ is a topology on $S$, and $T$ is a topological space.

Proof:
We have by definition that $\O \in \tau_p$, and as $p \in S$ we have that $S \in \tau_p$.
Now let $U_1, U_2 \in \tau_p$.
By definition $p \in U_1$ and $p \in U_2$, and so $p \in U_1 \cap U_2$ by definition of set intersection.
So $U_1 \cap U_2 \in \tau_p$.
Now let $\UU \subseteq \tau_p$.
We have that $\forall U \in \UU: p \in U$.
Hence from Subset of Union $p \in \bigcup \UU$.
So all the properties are fulfilled for $\tau_p$ to be a topology on $S$.
{{qed}}
Theorem: Accumulation Points of Sequence of Distinct Terms in Infinite Particular Point Space

Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Let $U \in \tau_p$ be a countably infinite open set of $T$.
Let the elements of $U$ be arranged into a sequence $\sequence {a_i}$ of distinct terms in $T$.
Then while every element $x$ of $U$ such that $x \ne p$ is a limit point of $U$, there exists no $x \in U$ such that $x$ is an accumulation point of $\sequence {a_i}$.

Proof:
We note from Limit Points in Particular Point Space that every $x \in U$ such that $x \ne p$ is a limit point of $U$.
Let $\sequence {a_i}$ be the (infinite) sequence in $T$ whose terms are the elements of $U$.
{{AimForCont}} $\beta \in S$ is an accumulation point of $\sequence {a_i}$.
Then from Accumulation Points for Sequence in Particular Point Space, $\beta$ is such that an infinite number of terms of $\sequence {a_i}$ are equal either to $\beta$ or to $p$.
But this contradicts the assertion that $\sequence {a_i}$ is a sequence of distinct terms.
Hence no element of $U$ is an accumulation point of $\sequence {a_i}$.
{{qed}}
Theorem: Closure of Open Set of Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $U \in \tau_p$ be open in $T$ such that $U \ne \O$.
Then:
:$U^- = S$
where $U^-$ denotes the closure of $U$.

Proof:
Follows directly from:
:Particular Point Topology is Closed Extension Topology of Discrete Topology
:Closure of Open Set of Closed Extension Space
{{qed}}
Theorem: Interior of Closed Set of Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $V \subseteq S$ be closed in $T$ such that $V \ne S$.
Then:
:$V^\circ = \O$
where $V^\circ$ denotes the interior of $V$.

Proof:
By definition:
:$\forall U \in \tau_p, U \ne \O: p \in U$
Thus if $V$ is closed in $T$:
:$\exists U \subseteq T: V = \relcomp S U$
So $p \notin V$.
Hence no open set of $T$ can be a subset of $V$ unless it is $\O$.
Hence the result, by definition of interior.
{{qed}}
Theorem: Point in Open Set of Particular Point Space is not Omega-Accumulation Point

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $x \in S$ such that $x \ne p$.
Let $H \subseteq S$ such that $p \in H$.
Then $x$ is not an $\omega$-accumulation point of $H$.

Proof:
Let $x \in S, x \ne p$.
By Limit Points in Particular Point Space, $x$ is a limit point of $H$.
Consider the set $U = \set {x, p} \subseteq S$.
By definition of the particular point topology, $U$ is open in $T$.
But as $U$ contains only $x$ and $p$, it is clear that $U$ does not contain an infinite number of points of $H$.
Hence, by definition, $x$ cannot be an $\omega$-accumulation point of $H$.
{{qed}}
Theorem: Particular Point Space is T0

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is a $T_0$ space.

Proof:
Let $T$ be a trivial space, so that $S = \set p$.
Then the result holds vacuously as there are no two distinct points in $T$.
Now suppose $T$ is not trivial.
Then $\exists x \in S: x \ne p$.
Now we have that $\set p \subseteq T$ is open in $T$ such that $p \in \set p$ but $x \notin \set p$.
Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$.
Then we have that $\set {p, x} \subseteq T$ is open in $T$ such that $x \in \set {p, x}$ but $y \notin \set {p, x}$.
Hence the result.
{{qed}}
Theorem: Particular Point Space is T0

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is a $T_0$ space.

Proof:
We have:
:Particular Point Topology is Closed Extension Topology of Discrete Topology
:Discrete Space satisfies all Separation Properties (including being a $T_0$ space)
Then by Condition for Closed Extension Space to be $T_0$ Space, as a discrete space is $T_0$ then so is its closed extension.
{{qed}}
Theorem: Non-Trivial Particular Point Space is not T1

Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
Then $T$ is not a $T_1$ space.

Proof:
Follows directly from:
:Particular Point Topology is Closed Extension Topology of Discrete Topology
:Closed Extension Topology is not $T_1$.
{{qed}}
Theorem: Particular Point Topology with Three Points is not T4

Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton or a doubleton.
That is, such that $S$ has more than two distinct elements.
Then $T$ is not a $T_4$ space.

Proof:
We have that there are at least three elements of $S$.
So, consider $x, y, p \in S: x \ne y, x \ne p, y \ne p$.
Then $X = \set x, Y = \set y$ are closed in $T$ and $X \cap Y = \O$.
Suppose $U, V \in \tau_p$ are open sets in $T$ such that $X \subseteq U, Y \subseteq V$.
But as $p \in U, p \in V$ we have that $U \cap V \ne \O$.
So $T$ is not a $T_4$ space.
{{qed}}
Theorem: Non-Trivial Particular Point Topology is not T3

Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
Then $T$ is not a $T_3$ space.

Proof:
We have that there are at least two distinct elements of $S$.
So, consider $x, p \in S: x \ne p$.
Then $X = \set x$ is closed in $T$ and $p \notin X$.
Suppose $U \in \tau_p$ is an open set in $T$ such that $X \subseteq U$.
We have that $\set p \in \tau_p$ such that $p \in \set p$.
But as $p \in U, p \in \set p$ we have that $U \cap \set p \ne \O$.
So $T$ is not a $T_3$ space.
{{qed}}
Theorem: Compact Space in Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $\set p$ is compact in $T$.

Proof:
Any open cover of $\set p$ has a finite subcover: an arbitrary single set that contains $p$ is a cover for $\set p$.
So $\set p$ is compact in $T$.
{{qed}}