nyuuzyou/wikis huggingface dataset:
Partial Fractions Expansion of Cotangent/Proof 2

== Theorem ==
== Proof ==
From the Euler Formula for Sine Function:
:$\ds \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$
Taking the logarithm of both sides:
Differentiating $x$:
Hence the result.
== Sources ==
* : Chapter $\text {B}.20$: The Bernoulli Numbers and some Wonderful Discoveries of Euler
Category:Partial Fractions Expansion of Cotangent
Partial Fractions Expansion of Cotangent/Proof 3

== Theorem ==
== Proof ==
From Euler's Reflection Formula:
:$\forall x \notin \Z: \map \Gamma x \map \Gamma {1 - x} = \dfrac \pi {\map \sin {\pi x} }$
Taking the logarithm of both sides:
Taking the derivative of both sides:
We now have:
== Sources ==
* : Chapter $\text {8}$. Analogues of the Gamma Function
Category:Partial Fractions Expansion of Cotangent
Partial Gamma Function expressed as Integral

== Theorem ==
Let $m \in \Z_{\ge 1}$.
Let $\map {\Gamma_m} x$ denote the partial Gamma function, defined as:
:$\map {\Gamma_m} x := \dfrac {m^x m!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} }$
Then:
:$\ds \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$
for $x > 0$.
== Proof ==
=== Lemma ===
First we establish:
The proof continues by induction on $m$.
For all $m \in \Z_{\ge 1}$, let $\map P m$ be the proposition:
:$\ds \map {\Gamma_m} x = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \map {\Gamma_k} x = k^x \int_0^1 \paren {1 - t}^k t^{x - 1} \rd t$
from which it is to be shown that:
:$\ds \map {\Gamma_{k + 1} } x = \paren {k + 1}^x \int_0^1 \paren {1 - t}^{k + 1} t^{x - 1} \rd t$
=== Induction Step ===
This is the induction step:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$
let:
and let:
Then:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m \in \Z_{\ge 1}: \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$
for $x > 0$.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $19$
Category:Gamma Function
Partial Gamma Function expressed as Integral/Lemma

== Theorem ==
Let $m \in \Z_{\ge 1}$.
Then:
:$(1): \quad \ds \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$
for $x > 0$.
== Proof ==
Let:
Recalculating the limits:
Hence:
Thus $(1)$ can be written:
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $19$
Category:Gamma Function
Partial Ordering/Examples/Ancestry

== Example of Partial Ordering ==
Let $P$ denote the set of all people who have ever lived.
Let $\DD$ denote the relation on $P$ defined as:
:$a \mathrel \DD b$ $a$ is a descendant of or the same person as $b$.
Its dual $\DD^{-1}$ is defined as:
:$a \mathrel {\DD^{-1} } b$ $a$ is an ancestor of or the same person as $b$.
Then $\DD$ and $\DD^{-1}$ are partial orderings on $P$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Examples of Partial Orderings
Partial Ordering/Examples/Arbitrary Example 1

== Example of Partial Ordering ==
Let $X = \set {x, y, z}$.
Let $\RR = \set {\tuple {x, x}, \tuple {x, y}, \tuple {x, z}, \tuple {y, y}, \tuple {z, z} }$.
Then $\RR$ is a partial ordering on $X$.
The strict partial ordering on $X$ corresponding to $\RR$ is its reflexive reduction:
:$\RR^{\ne} = \set {\tuple {x, y}, \tuple {x, z} }$
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: $(0)$
Category:Examples of Partial Orderings
Partial Ordering/Examples/Parallel Lines

== Example of Partial Ordering ==
Let $S$ denote the set of all infinite straight lines embedded in a cartesian plane.
Let $\LL$ denote the relation on $S$ defined as:
:$a \mathrel \LL b$ :
::$a$ is parallel $b$
::if $a$ is not parallel to the $y$-axis, then coincides with or lies below $b$
::but if $b$ is parallel to the $y$-axis, then $a$ coincides with or lies to the right of $b$
Its dual $\LL^{-1}$ is defined as:
:$a \mathrel {\LL^{-1} } b$ :
::$a$ is parallel $b$
::if $a$ is not parallel to the $y$-axis, then coincides with or lies above $b$
::but if $b$ is parallel to the $y$-axis, then $a$ coincides with or lies to the left of $b$.
Then $\LL$ and $\LL^{-1}$ are partial orderings on $S$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Examples of Partial Orderings
Partial Ordering can be Expanded to compare Additional Pair

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $a$ and $b$ be non-comparable elements of $S$.
That is, let:
:$a \not\preceq b$
and:
:$b \not\preceq a$
Let ${\preceq'} = {\preceq} \cup \set {\tuple {a, b} }$.
Let $\preceq'^+$ be the transitive closure of $\preceq'$.
Then:
:$\preceq'^+$ is an ordering.
$\preceq'^+$ can be defined by letting $p \preceq'^+ q$ :
:$p \preceq q$ or
:$p \preceq a$ and $b \preceq q$.
== Proof 1 ==
== Proof 2 ==
Category:Order Theory
Category:Partial Ordering can be Expanded to compare Additional Pair
Partial Ordering can be Expanded to compare Additional Pair/Proof 2

== Theorem ==
== Proof ==
Let $\prec$ be the reflexive reduction of $\preceq$.
Let $\prec' = {\prec} \cup \set {\tuple {a, b} }$.
By Reflexive Reduction of Ordering is Strict Ordering, $\prec$ is a strict ordering.
Define a relation $\prec'_2$ by letting $p \prec'_2 q$ :
:$p \prec q$ or
:$p \preceq a$ and $b \preceq q$
By Strict Ordering can be Expanded to Compare Additional Pair, $\prec'_2$ is a strict ordering and is the transitive closure of $\prec'$.
Then the reflexive closure of $\prec'_2$, ${\prec'_2}^=$ is the transitive reflexive closure of $\prec'$.
From Equivalence of Definitions of Reflexive Transitive Closure, ${\prec'_2}^=$ is the transitive closure of the reflexive closure of $\prec'$.
The reflexive closure of $\prec'$ is $\preceq'$, so ${\prec'_2}^=$ is the transitive closure of $\preceq'$.
But ${\prec'_2}^=$ is clearly the relation defined by the two conditions in the theorem statement.
Category:Partial Ordering can be Expanded to compare Additional Pair
Partial Products of Uniformly Convergent Product Converge Uniformly

== Definition ==
Let $X$ be a set.
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\sequence {f_n}$ be a sequence of bounded mappings $f_n: X \to \mathbb K$.
Let the product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly.
Then the sequence of partial products converges uniformly.
== Proof ==
Let $n_0 \in \N$ be such that the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty f_n$ converges uniformly.
By Product of Bounded Functions is Bounded, $\ds \prod_{n \mathop = 1}^{n_0 - 1} f_n$ is bounded.
By Uniformly Convergent Sequence Multiplied with Function, the sequence of partial products converges uniformly.
== Also see ==
* Uniform Product of Continuous Functions is Continuous
* Infinite Product of Analytic Functions is Analytic
Category:Uniform Convergence
Category:Infinite Products
Partial Sum Congruent to P-adic Integer Modulo Power of p

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $a \in \Z_p$.
Let $a = \ds \sum_{j=0}^\infty d_jp^j$ be the $p$-adic expansion of $a$
For all $n \in \N$, let $a_n = \ds \sum_{j=0}^n d_jp^j$ be the n-th partial sum of the $p$-adic expansion of $a$
Then:
:$\forall n \in \N : a_n \equiv a \pmod{p^{n+1}\Z_p}$
where $a_n \equiv a \pmod{p^{n+1}\Z_p}$ denotes congruence modulo the ideal $\Z_p$.
== Proof ==
We have:
Category:P-adic Integers
Partial Sums of P-adic Expansion forms Coherent Sequence

== Theorem ==
Let $p$ be a prime number.
Let $\ds \sum_{n \mathop = 0}^\infty d_n p^n$ be a $p$-adic expansion.
Let $\sequence {\alpha_n}$ be the sequence of partial sums; that is:
:$\forall n \in \N :\alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$.
Then $\sequence {\alpha_n}$ is a coherent sequence.
== Proof ==
From the definition of a coherent sequence, it needs to be shown that $\sequence {\alpha_n}$ is a sequence of integers such that:
:$(1): \quad \forall n \in \N: 0 \le \alpha_n < p^{n + 1}$
:$(2): \quad \forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$
That the sequence $\sequence {\alpha_n}$ is a sequence of integers follows immediately from the assumption that the series begins at $n = 0$ and so the terms of each summation are integers. 
First note:
=== Partial Sums satisfy (1) ===
This is proved by induction on $n$:
==== Basis for the Induction ====
$n = 0$ 
By definition:
$\alpha_0 = d_0$
By definition of a $p$-adic expansion:
:$0 \le d_0 < p$
So shown for basis for the induction.
==== Induction Hypothesis ====
This is our induction hypothesis:
:$0 \le \alpha_k < p^{k + 1}$
Now we need to show true for $n=k+1$:
:$0 \le \alpha_{k + 1} < p^{k + 2}$
==== Induction Step ====
This is our induction step:
Since:
:$\alpha_{k + 1} = \alpha_k + d_{k + 1} p^{k + 1}$
then:
:$0 \le \alpha_{k + 1} < p^{k + 2}$
By induction:
:$\forall n \in \N: 0 \le \alpha_n < p^{n + 1}$
=== Partial Sums satisfy (2) ===
Let $n \in \N$.
 
Since:
:$\alpha_{n + 1} - \alpha_n = d_{n + 1}p^{n + 1}$
Then: 
:$p^{n + 1} \divides \alpha_{n+1} - \alpha_n$
So:
:$\alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1} }$
The result follows.
== Note ==
The theorem is stated for the special set of $p$-adic expansions where the series index begins at $0$ and not the more general $m \in \Z_{\le 0}$.
From P-adic Integer has Unique P-adic Expansion Representative, it is seen that this set of $p$-adic expansions is indeed the $p$-adic integers.
== Also see ==
* Coherent Sequence is Partial Sum of P-adic Expansion
Category:P-adic Number Theory
Partial Sums of Power Series with Fibonacci Coefficients

== Theorem ==
:$\ds \sum_{k \mathop = 0}^n F_k x^k = \begin {cases}
\dfrac {x^{n + 1} F_{n + 1} + x^{n + 2} F_n - x} {x^2 + x - 1} & : x^2 + x - 1 \ne 0 \\
\dfrac {\paren {n + 1} x^n F_{n + 1} + \paren {n + 2} x^{n + 1} F_n - 1} {2 x + 1} & : x^2 + x - 1 = 0 \end {cases}$
where $F_n$ denotes the $n$th Fibonacci number.
== Proof ==
Multiplying the summation by $x^2 + x - 1$:
If the denominator is $0$, then $x = \dfrac 1 \phi$ or $x = \dfrac 1 {\hat \phi}$ and the numerator is $0$ also.
Thus we can differentiate the numerator and denominator $x$ and use L'Hôpital's Rule:
Hence the result.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $21$
Category:Fibonacci Numbers
Category:Real Power Series
Particle on Curved Surface under no Force moves along Geodesic

== Theorem ==
Consider a particle $P$ which is constrained to move on a curved surface $C$.
Let $P$ be such that no force acts upon it.
Then $P$ moves along a geodesic.
== Proof ==
== Sources ==
* : Chapter $\text {A}.32$: Riemann ()
Category:Calculus of Variations
Particular Affirmative and Particular Negative are Subcontrary iff First Predicate is not Vacuous

== Theorem ==
Consider the categorical statements:
Then:
:$\mathbf I$ and $\mathbf O$ are subcontrary
:
:$\exists x: \map S x$
Using the symbology of predicate logic:
:$\exists x: \map S x \iff \paren {\paren {\exists x: \map S x \land \map P x} \lor \paren {\exists x: \map S x \land \neg \map P x} }$
== Proof ==
=== Sufficient Condition ===
Let $\exists x: \map S x$.
Suppose $\mathbf I$ and $\mathbf O$ are both false.
As $\mathbf I$ is false, then by the Rule of Conjunction:
:$\neg \map P x$
As $\mathbf O$ is false, then by the Rule of Conjunction:
:$\neg \neg \map P x$
and so by Double Negation:
:$\map P x$
It follows by Proof by Contradiction that $\mathbf I$ and $\mathbf O$ are not both false.
Thus, by definition, $\mathbf I$ and $\mathbf O$ are subcontrary statements.
=== Necessary Condition ===
Let $\mathbf I$ and $\mathbf O$ be subcontrary statements
Suppose $\neg \exists x: \map S x$.
From the definition of logical conjunction, it follows that:
:$\neg \paren {\exists x: \map S x \land \map P x}$
Similarly from the definition of logical conjunction, it follows that:
:$\neg \paren {\exists x: \map S x \land \neg \map P x}$
That is, both $\mathbf I$ and $\mathbf O$ are false.
So, by definition, $\mathbf I$ and $\mathbf O$ are not subcontrary.
It follows by Proof by Contradiction that $\exists x: \map S x$.
== Also defined as ==
Some sources gloss over the possibility of $\map S x$ being vacuous and merely report that the particular affirmative and particular negative are subcontrary.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
* : $4$: Propositional Functions and Quantifiers: $4.1$: Singular Propositions and General Propositions
Category:Categorical Statements
Particular Affirmative and Universal Negative are Contradictory

== Theorem ==
Consider the categorical statements:
Then $\mathbf I$ and $\mathbf E$ are contradictory.
Using the symbology of predicate logic:
:$\neg \paren {\paren {\exists x: \map S x \land \map P x} \iff \paren {\forall x: \map S x \implies \neg \map P x} }$
== Proof ==
The argument reverses:
The result follows by definition of contradictory.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
* : $4$: Propositional Functions and Quantifiers: $4.1$: Singular Propositions and General Propositions
Category:Categorical Statements
Particular Point Space is First-Countable

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is first-countable.
== Proof ==
Let $x \in S: x \ne p$.
Consider the set $U_x = \set {x, p} \subseteq S$.
Now let $V \in \tau_p$ be an open set in $S$ such that $x \in V$.
So $x \in V$, by definition of $V$, and $p \in V$ as $V$ is open.
It follows directly that $U_x \subseteq V$.
So $\set {U_x}$ is a local basis at $x$ which is (trivially) countable.
If $x = p$ then $U_x = \set p$ and the same argument still applies.
So every element of $T$ has a countable local basis.
Hence $T$ is first-countable by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $7$
Category:First-Countable Spaces
Category:Particular Point Topology
Particular Point Space is Irreducible/Proof 1

== Theorem ==
== Proof ==
By definition, $T = \struct {S, \tau_p}$ is irreducible every two non-empty open sets of $T$ have non-empty intersection.
Let $U_1$ and $U_2$ be non-empty open sets of $T$.
By definition of particular point space, $p \in U_1$ and $p \in U_2$.
Thus:
:$p \in U_1 \cap U_2$
and so:
:$U_1 \cap U_2 \ne \O$
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $10$
Category:Particular Point Space is Irreducible
Particular Point Space is Locally Compact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is locally compact.
== Proof ==
Let $x \in S$.
Consider the set $\set {p, x}$.
From the definition of particular point topology, $\set {p, x}$ is open in $T$.
By Finite Topological Space is Compact, $\set {p, x}$ is compact.
Let $N$ be a neighborhood of $x$.
Then:
:$\exists U \in \tau_p: x \in U \subseteq N$.
From the definition of particular point topology, since $U \ne \O$, we must have $p \in U$.
Therefore $\set {p, x} \subseteq U \subseteq N$.
Since $N$ is arbitrary, $\set {\set {p, x}}$ is a neighborhood basis for $x$.
The result follows from definition of a locally compact space.
== Also see ==
* Particular Point Space is Weakly Locally Compact
Category:Locally Compact Spaces
Category:Particular Point Topology
Particular Point Space is Locally Path-Connected

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is locally path-connected.
== Proof ==
Consider the set $\BB$ defined as:
:$\BB = \set {\set {x, p}: x \in S}$
Then $\BB$ is a basis for $T$.
Now consider the open set $\set {p, q} \in \BB$.
Let $\mathbb I$ be the closed unit interval in $\R$.
Let $f: \mathbb I \to S$ be the mapping defined as:
:$\forall x \in \mathbb I: \map f x = \begin{cases}
p & : x \in \hointr 0 1 \\
q & : x = 1
\end{cases}$
Let $U \in \tau_p$.
Then either $q \in U$ or $q \notin U$.
If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is half open in $\R$ but open in $\mathbb I$.
So $f: \mathbb I \to S$ is a continuous mapping and so a path from $p$ to $q$.
As $\set {p, q}$ is any element of $\BB$ it follows that $T$ is locally path-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $13$
Category:Locally Path-Connected Spaces
Category:Particular Point Topology
Particular Point Space is Non-Meager

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is non-meager.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Non-Meager Spaces
Category:Particular Point Topology
Category:Particular Point Space is Non-Meager
Particular Point Space is Non-Meager/Proof 1

== Theorem ==
== Proof ==
Suppose $T$ were meager.
Then it would be a countable union of subsets which are nowhere dense in $T$.
Let $H \subseteq S$.
From Closure of Open Set of Particular Point Space, the closure of $H$ is $S$.
From the definition of interior, the interior of $S$ is $S$.
So the interior of the closure of $H$ is not empty.
So $T$ can not be the union of a countable set of subsets which are nowhere dense in $T$.
Hence $T$ is not meager and so by definition must be non-meager.
Category:Particular Point Space is Non-Meager
Particular Point Space is Non-Meager/Proof 2

== Theorem ==
== Proof ==
By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$.
So $\left\{{p}\right\}$ itself is open in $T$.
That is, $p$ is an open point.
The result follows from Space with Open Point is Non-Meager.
Category:Particular Point Space is Non-Meager
Particular Point Space is Non-Meager/Proof 3

== Theorem ==
== Proof ==
By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$.
 $T$ is meager.
By definition, $T$ is meager it is a countable union of subsets of $S$ which are nowhere dense in $T$.
At least one such nowhere dense subset $U$ of $S$ must contain $p$.
By definition, $U$ is nowhere dense in $T$ :
:$U^-$ contains no open set of $T$ which is non-empty
where $U^-$ denotes the closure of $U$.
By definition of particular point space, $U$ is open in $T$.
By Closure of Open Set of Particular Point Space, $U^- = S$.
But $S$ is itself open in $T$ and non-empty, and so $U$ is not nowhere dense.
From this contradiction it follows that $T$ is non-meager
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $14$
Category:Particular Point Space is Non-Meager
Particular Point Space is Path-Connected

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is path-connected.
== Proof ==
Let $q \in S$.
Let $\mathbb I$ be the closed unit interval in $\R$.
Let $f: \mathbb I \to S$ be the mapping defined as:
:$\forall x \in \mathbb I: \map f x = \begin{cases}
p & : x \in \hointr 0 1 \\
q & : x = 1
\end{cases}$
Let $U \in \tau_p$.
Then by definition of particular point space, $p \in U$
Either $q \in U$ or $q \notin U$.
If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is half open in $\R$ but open in $\mathbb I$.
So $f: \mathbb I \to S$ is a continuous mapping and so a path from $p$ to $q$.
As $q$ is any point in $S$, it follows from Path-Connected iff Path-Connected to Point that $T$ is path-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $13$
Category:Path-Connected Spaces
Category:Particular Point Topology
Particular Point Space is Pseudocompact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is pseudocompact.
== Proof ==
We have that:
:a Particular Point Space is Irreducible
:an Irreducible Space is Pseudocompact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $12$
Category:Pseudocompact Spaces
Category:Particular Point Topology
Particular Point Space is Scattered

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is scattered.
== Proof ==
Let $H \subseteq S$ such that $p \notin H$.
From Subset of Particular Point Space is either Open or Closed, $H$ is closed in $T$.
We have that Closed Set in Particular Point Space has no Limit Points.
So if $p \notin H$ then $H$ has no limit points.
So by definition, every element of such a set $H$ is an isolated point.
So if $p \notin H$ then $H$ is by definition not dense-in-itself.
Now let $H \subseteq S$ such that $p \in H$.
We have that $\set p \in \tau_p$.
That is, $p$ is an open point of $T$.
From Point in Topological Space is Open iff Isolated, $p$ is an isolated point.
So if $p \in H$ then $H$ is by definition not dense-in-itself.
Hence the result, by definition of a scattered space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $9$
Category:Scattered Spaces
Category:Particular Point Topology
Particular Point Space is Separable

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is separable.
== Proof ==
By definition, $T$ is separable there exists a countable subset of $S$ which is everywhere dense in $T$.
Consider $U := \set p \subseteq S$.
By definition, $U$ is open in $T$.
From Closure of Open Set of Particular Point Space we have that $U^- = S$, where $U^-$ is the closure of $U$.
By definition, $U$ is everywhere dense in $T$.
$U$ is (trivially) countable.
Hence the result, by definition of a separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $6$
Category:Separable Spaces
Category:Particular Point Topology
Particular Point Space is T0

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is a $T_0$ (Kolmogorov) space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $4$
Category:T0 Spaces
Category:Particular Point Topology
Category:Particular Point Space is T0
Particular Point Space is T0/Proof 1

== Theorem ==
== 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.
Category:Particular Point Space is T0
Particular Point Space is T0/Proof 2

== Theorem ==
== 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.
Category:Particular Point Space is T0