avewright/proofwiki-math huggingface dataset:
Theorem: Product Space is Connected iff Factors are Connected

Let $I$ be an indexing set.
Let $\family {T_\alpha}_{\alpha \mathop \in I}$ be an indexed family of topological spaces.
Let $T = \ds \prod_{\alpha \mathop \in I} T_\alpha$ be the Cartesian space of $\family {T_\alpha}_{\alpha \mathop \in I}$.
Let:
:$T = \ds \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$
where $\ds \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$ denotes the closure of $\ds \bigcup_{\alpha \mathop \in I} S_\alpha$ in $T$.
Let $\tau$ be a topology on $T$ such that
:${S'}_\alpha \subseteq \ds \prod T_\alpha$
where ${S'}_\alpha = \set {\family {y_\beta} \in T: y_\beta = x_\beta \text { for all } \beta \ge \alpha}$ is homeomorphic to $S_{\alpha - 1} \times T_\alpha$.
Then $T$ is connected {{iff}} each of $T_\alpha: \alpha \in I$ are connected.

Proof:
Let the {{Axiom-link|Choice}} be assumed.
Let $I$ be well-ordered.
Let $x = \family {x_\alpha} \in T$ be some arbitrary fixed element of $T$.
Let $S_\alpha = \set {\family {y_\beta} \in T: y_\beta = x_\beta \text { for all } \beta \ge \alpha}$.
We have that $S_\alpha$ is homeomorphic to $S_{\alpha - 1} \times T_\alpha$.
Then from Finite Product Space is Connected iff Factors are Connected, $S_\alpha$ is connected {{iff}} $S_{\alpha - 1}$ is.
Let $\alpha$ be a limit ordinal.
Then:
:$S_\alpha = \ds \paren {\bigcup_{\beta \mathop < \alpha} S_\beta}^-$
where $X^-$ denotes the closure of $X$.
So if each $S_\beta$ is connected for $\beta < \alpha$, it follows that $S_\alpha$ must likewise be connected, as $\family \gamma$ is nested.
Thus:
:$T = \ds \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$ is connected.
{{qed}}
{{AoC}}
Theorem: Continuous Mapping from Compact Space to Hausdorff Space Preserves Local Connectedness

Let $T_1 = \struct {S_1, \tau_1}$ be a compact topological space.
Let $T_2 = \struct {S_2, \tau_2}$ be a $T_2$ (Hausdorff) space.
Let $f: T_1 \to T_2$ be a continuous mapping.
Let $T_1$ be locally connected.
Then $T_2$ is also locally connected.

Proof:
Let $H$ be a component of an open set $U$ of $T_2$.
By definition of continuous mapping, $f^{-1} \sqbrk U$ is an open set of $T_1$.
Let $G$ be a component of $f^{-1} \sqbrk U$.
Thus by Continuous Image of Connected Space is Connected, $f \sqbrk G$ is connected in $T_2$.
Thus either:
:$f \sqbrk G \subseteq H$
or:
:$f \sqbrk G \cap H = \O$
Thus every component of $f^{-1} \sqbrk H$ is a component of $f^{-1} \sqbrk U$.
From Component of Locally Connected Space is Open, $f^{-1} \sqbrk H$ is open.
Let $S_1 \setminus G$ denote the complement of $f^{-1} \sqbrk H$ relative to $S_1$.
By definition of closed set, $S_1 \setminus f^{-1} \sqbrk H$ is closed in $T_1$.
From Closed Subspace of Compact Space is Compact, $S_1 \setminus f^{-1} \sqbrk H$ is also compact in $T_1$.
From Continuous Image of Compact Space is Compact, $f \sqbrk {S_1 \setminus f^{-1} \sqbrk H} = S_2 \setminus H$ is compact in $T_2$.
From Compact Subspace of Hausdorff Space is Closed, $S_2 \setminus H$ is closed in $T_2$.
Thus by definition of closed set, $H$ is open in $T_2$.
{{Proofread}}
{{qed}}
Theorem: Minimum Degree Bound for Simple Planar Graph

Let $G$ be a simple connected planar graph.
Then: 
:$\map \delta G \le 5$
where $\map \delta G$ denotes the minimum degree of vertices of $G$.

Proof:
{{AimForCont}} $G$ is a simple connected planar graph with $\map \delta G \ge 6$. 
Let $m$ and $n$ denote the number of edges and vertices respectively in $G$. 
Then by the Handshake Lemma: 
{{begin-eqn}}
{{eqn | l = 2 m
      | r = \sum_{i \mathop = 1}^n \map {\deg_G} {v_i}
      | c = where $\map {\deg_G} {v_i}$ is the degree of vertex $v_i$
}}
{{eqn | o = \ge
      | r = \sum_{i \mathop = 1}^n 6
}}
{{eqn | r = 6 n
}}
{{eqn | ll = \leadsto
      | l = m
      | o = \ge
      | r = 3 n
}}
{{end-eqn}}
This contradicts the Linear Bound Lemma:
:$m \le 3 n - 6$
Hence $\map \delta G \le 5$.
{{qed}}
Category:Graph Theory
9oq2zx8vczpp04j1nb6jfozwpi0m0vq
Theorem: Linear Bound Lemma

Let $G_n$ be a simple connected planar graph with $n$ vertices.
Then:
:$m \le 3 n − 6$
where $m$ is the number of edges.

Proof:
Let $f$ denote the number of faces of $G_n$. 
Let $\sequence {s_i}_{i \mathop = 1}^f$ be a sequence of regions of a planar embedding of $G_n$. 
Consider the sequence $\sequence {r_i}_{i \mathop = 1}^f$ where $r_i$ denotes the number of boundary edges of $s_i$. 
Since $G$ is simple, then by the definition of planar embedding: 
:every region has at least $3$ boundary edges
:every edge is a boundary edge of at most two regions in the planar embedding.
{{explain|Every edge of what? The original graph, or of the planar embedding?}}
Using this two facts, we can find the boundary of $\ds \sum_{i \mathop = 1}^f r_i$ as:
{{explain|Boundary of what exactly?}}
:$3 f \le \ds \sum_{i \mathop = 1}^f r_i \le 2m$
Now, as $f \le \dfrac 2 3 m$:
{{begin-eqn}}
{{eqn | l = n - m + f
      | r = 2
      | c = Euler's Theorem for Planar Graphs
}}
{{eqn | l = n - m + \frac 2 3 m
      | o = \ge
      | r = 2
}}
{{eqn | l = \frac 1 3 m
      | o = \le
      | r = n - 2
}}
{{eqn | l = m
      | o = \le
      | r = 3 n - 6
}}
{{end-eqn}}
{{qed}}
Category:Graph Theory
Category:Named Theorems
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Theorem: Existence of Connected Space which is Totally Pathwise Disconnected

There exists at least one example of a topological space which is both connected and totally pathwise disconnected.

Proof:
Let $T$ be Cantor's leaky tent.
From Cantor's Leaky Tent is Connected, $T$ is a connected space.
From Cantor's Leaky Tent is Totally Pathwise Disconnected, $T$ is a totally pathwise disconnected space.
Hence the result.
{{qed}}
Theorem: Zero Dimensional Space is not necessarily T0

Let $T = \struct {S, \tau}$ be a zero dimensional topological space.
Then $T$ is not necessarily a $T_0$ (Kolmogorov) space.

Proof:
Let $T = \struct {S, \tau}$ be a partition space.
From Partition Topology is Zero Dimensional, $T$ is a zero dimensional topological space.
From Partition Topology is not $T_0$, $T$ is not a $T_0$ (Kolmogorov) space.
{{qed}}
Theorem: Scattered Space is not necessarily T1

Let $T = \struct {S, \tau}$ be a scattered topological space.
Then $T$ is not necessarily a $T_1$ space.

Proof:
Let $T = \struct {S, \tau}$ be a non-trivial particular point space.
From Particular Point Space is Scattered, $T$ is a scattered space.
From Non-Trivial Particular Point Space is not $T_1$, $T$ is not a $T_1$ (Fréchet) space.
{{qed}}
Theorem: Existence of Connected Non-T1 Scattered Space

There exists at least one example of a connected topological space which is not a $T_1$ space, but which is also a scattered space.

Proof:
Let $T$ be the divisor space.
From Divisor Space is Connected, $T$ is a connected space.
From Divisor Space is not $T_1$, $T$ is not a $T_1$ space.
From Divisor Space is Scattered, $T$ is a scattered space.
Hence the result.
{{qed}}
Theorem: Sorgenfrey Line is not Second-Countable

Let $T = \struct {\mathbb R, \tau}$ be the Sorgenfrey line.
Then $T$ is not second-countable.

Proof:
{{Recall|Second-Countable Space|second-countable space}}
{{:Definition:Second-Countable Space}}
Suppose $\BB$ is a basis for $\tau$.
{{Recall|Analytic Basis|(analytic) basis}}
{{:Definition:Analytic Basis/Definition 2}}
Let $U \in \tau$ be arbitrary.
Then:
:$U = \hointr x {x + \epsilon}$
for some $x \in \R$ and $\epsilon \in \R_{>0}$.
Now:
:$\forall y \in U: \exists B_y \in \BB: y \in B_y \subseteq \hointr x {x + \epsilon}$
This $B_y$ has an infimum equal to $x$.
So for different $x$, the corresponding $B_y$ is different.
So the cardinality of $\BB$ is at least $\size \R$, which is uncountable.
{{qed}}
Theorem: Biconnected Set does not necessarily have Dispersion Point

A biconnected set does not necessarily have a dispersion point.

Proof:
Let $T$ be a Miller's biconnected set.
From Miller's Biconnected Set is Biconnected, $T$ is a biconnected set.
From Miller's Biconnected Set is has no Dispersion Point, $T$  does not have a dispersion point.
Hence the result.
{{qed}}
Theorem: Existence of Connected Punctiform Space

There exists at least one example of a connected topological space which is also punctiform.

Proof:
Let $T$ be Cantor's leaky tent.
From Cantor's Leaky Tent is Connected, $T$ is a connected topological space.
From Cantor's Leaky Tent is Punctiform, $T$ is a punctiform space.
Hence the result.
{{qed}}
Theorem: Metric Space is Completely Normal

A metric space $M = \struct {A, d}$ is a completely normal space.

Proof:
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
We have that:
{{begin-itemize}}
{{item||a Metric Space is $T_5$}}
{{item||a Metric Space is $T_1$.}}
{{end-itemize}}
Hence the result.
{{qed}}
Theorem: Metric Space is Perfectly Normal

A metric space $M = \struct {A, d}$ is a perfectly normal space.

Proof:
{{Recall|Perfectly Normal Space|perfectly normal space|index = 1}}
{{:Definition:Perfectly Normal Space/Definition 1}}
From Metric Space is Perfectly $T_4$
:$M$ is a perfectly $T_4$ space.
From Metric Space is $T_1$
:$M$ is a $T_1$ space.
The result follows.
{{qed}}
Theorem: Bounded Metric Space is not necessarily Totally Bounded

Let $M = \struct {A, d}$ be a bounded metric space.
Then it is not necessarily the case that $M$ is totally bounded.

Proof:
Let $d$ be a discrete metric on the open unit interval $\Bbb I := \openint 0 1 \subseteq \R$.
We have that for all $x \in \openint 0 1$ and for all $r \in \R_{> 1}$:
:$\map {B_r} x = \openint 0 1$
where $\map {B_r} x$ denotes the open $r$-ball of $x$.
Thus $\struct {\Bbb I, d}$ is bounded.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number such that $\epsilon < 1$.
Let $x \in \openint 0 1$ be arbitrary.
Then $\map {B_\epsilon} x$ contains $x$ alone.
So, there is no finite $\epsilon$-net for $\openint 0 1$.
By definition of totally bounded metric space, $\struct {\Bbb I, d}$ is not totally bounded.
{{qed}}
Theorem: Total Boundedness is not Preserved under Homeomorphism

Let $M = \struct {A, d}$ be a totally bounded metric space.
Let $M' = \struct {A', d'}$ be a metric space.
Let $M$ be homeomorphic to $M'$.
Then it is not necessarily the case that $M'$ is totally bounded.

Proof:
{{ProofWanted|According to S&S (item $134$), this is proved somehow using the metric $\delta {{=}} \dfrac d {1 + d}$, given some metric space $M {{=}} \struct {A, d}$, but the derivation of this is obscure.}}
Theorem: Tutte's Wheel Theorem

Every $3$-connected graph can be obtained by the following procedure:
* Start with $G_0 := K_4$
* Given $G_i$ pick a vertex $v$
* Split into $v'$ and $v' '$ and add edge $\set {v', v' '}$
This procedure directly follows from the theorem: 
:A graph $G$ is $3$-connected ('''A''') {{iff}} there exists a sequence $G_0, G_1, \dotsc, G_n$ of graphs with the following properties ('''B'''): 
:* $G_0 = K_4$ and $G_n = G$
:* $G_{i + 1}$ has an edge $e = x y$ with $\map \deg x, \map \deg y \ge 3$ and $G_i = G_{i + 1} / e$ for every $i < n$.

Proof:
=== Lemma ===
{{:Tutte's Wheel Theorem/Lemma}}{{qed|lemma}}
Theorem: Complete Metrizability is not Hereditary

Let $T = \struct {S, \tau}$ be a topological space which is completely metrizable.
Let $H \subseteq S$ be a subset of $S$.
Let $\struct {H, \tau_H}$ be the topological subspace of $T$ induced by $H$.
Then it is not necessarily the case that $\struct {H, \tau_H}$ is also completely metrizable.
That is, complete metrizability is not a hereditary property.

Proof:
Let $\struct {\R, d}$ denote the real number line under the Euclidean metric.
Let $\struct {\Q, d}$ denote the rational number space under the Euclidean metric.
We have that $\Q \subset \R$ by definition.
Let $T = \struct {\R, \tau_d}$ be the topological space induced on $\R$ by $d$.
By Real Number Line is Complete Metric Space, $\struct {\R, d}$ is a complete metric space.
Thus, by definition, $T$ is completely metrizable.
Let $T' = \struct {\Q, \tau_d}$ be the topological space induced on $\Q$ by $d$.
By Rational Number Space is not Complete Metric Space, $\struct {\Q, d}$ is not a complete metric space.
Thus, by definition, $T'$ is not completely metrizable.
Hence the result.
{{qed}}
Theorem: Completion Theorem (Metric Space)/Lemma 4

:$\tilde M = \struct {\tilde A, \tilde d}$ is unique up to isometry.

Proof:
Let $M_1 = \struct {\tilde{A_1}, \tilde{d_1}, \phi_1}$ and $M_2 = \struct {\tilde{A_2}, \tilde{d_2}, \phi_2}$ be two completions of $\struct {A, d}$.
Here, $\phi_1: A \to A_1$ and $\phi_2: A \to A_2$ are isometries
By Composite of Isometries is Isometry, $\psi = \phi_2 \circ \phi_1^{-1}$ gives an isometry from $\phi_1 \sqbrk A$ to $\phi_2 \sqbrk A$.
The sets $\phi_1 \sqbrk A$ and $\phi_2 \sqbrk A$ are dense in $A_1$ and $A_2$ respectively.
Thus we can extend $\psi$ continuously to a map $\psi: A_1 \to A_2$.
{{explain|That needs a result to be linked to}}
That is, for $x \in A_1$, we can find a Cauchy sequence $\sequence {w_n}$ in $A_1$ with limit $x$.
Then we define:
:$\ds \map \psi x := \lim_{n \mathop \to \infty} \map \psi {w_n}$
which converges as $A_2$ is complete.
By Metric Space is Hausdorff, $A$ is a $T_2$ (Hausdorff) space.
By $T_2$ Property is Preserved under Homeomorphism, $A_1$ and $A_2$ are also $T_2$ spaces.
Therefore, by Convergent Sequence in Hausdorff Space has Unique Limit, $\psi$ is well-defined.
{{qed|lemma}}
It is to be shown that $\psi$ is surjective:
For $y \in \tilde {A_2}$, let $\sequence {w_n'}$ be a Cauchy sequence in $\phi_2 \sqbrk A$ with limit $y$ in $\tilde {A_2}$.
Let $z_n$ be the preimage of the $w_n'$ under $\psi$.
Then, as $A_2$ is a $T_2$ space:
:$\ds \lim_{n \mathop \to \infty} \map \psi {z_n} = y$
as required.
{{qed|lemma}}
injectivity of $\psi$ holds because $A_1$ is a $T_2$ space, as follows:
Suppose that:
:$\ds \lim_{n \mathop \to \infty} \map \psi {w_n} = \lim_{n \mathop \to \infty} \map \psi {w_n'}$
:$\ds \lim_{n \mathop \to \infty} w_n = w$
and:
:$\ds \lim_{n \mathop \to \infty} w_n' = w'$
For $\epsilon > 0$, pick $M \in \N$ such that $\map \psi {w_n}$, $\map \psi {w_n'}$ lie in the open $\epsilon$-ball $\map {B_{\epsilon / 2} } {\map \psi w}$ for all $n \ge M$.
Then we have:
:$\map {\tilde {d_1} } {w_n, w_n'} = \map {\tilde {d_2} } {\map \psi {w_n}, \map \psi {w_n'} } \le \epsilon$
As $A_1$ is a $T_2$ space, we conclude $w = w'$, and the task is complete.
{{qed|lemma}}
Finally, from Distance Function of Metric Space is Continuous, it follows that $\psi$ is an isometry on all of $A_1$, and the proof is complete.
{{qed}}
Category:Completion Theorem (Metric Space)
81jxa81vbnaeulh20upkgdlsn2v7gcy
Theorem: Sorgenfrey Line is First-Countable

Let $\R$ be the set of real numbers.
Let $\BB = \set {\hointr a b: a, b \in \R}$.
Let $\tau$ be the topology generated by $\BB$, that is, the Sorgenfrey line.
Then $\tau$ is first-countable.

Proof:
{{Recall|First-Countable Space|first-countable space}}
{{:Definition:First-Countable Space}}
Let $\BB_x = \set {\hointr x {x + \dfrac 1 n} : n \in \N_{>0} }$.
We will show that:
:$(1): \quad \BB_x$ is countable
:$(2): \quad \BB_x$ is a local basis at $x$
$(1)$ follows from the fact that $\BB_x$ is a bijection from the set of natural numbers.
$(2)$ is demonstrated as follows:
By definition of local basis, it suffices to show that:
:$\forall U \in \tau: x \in U: \exists B \in \BB_x: B \subseteq U$
Let $U \in \tau$ be arbitrary.
By definition of $\tau$, there exists $\hointr x {x + \epsilon} \subseteq U$ for some $\epsilon \in \R_{>0}$.
By the Axiom of Archimedes there exists $n \in \N$ such that $n > \dfrac 1 \epsilon$ (that is, $\dfrac 1 n < \epsilon$).
So:
:$x \in \hointr x {x + \dfrac 1 n} \subseteq \hointr x {x + \epsilon} \subseteq U$
Therefore $\BB_x$ is a local basis at $x$.
We have seen that the Sorgenfrey line has a countable local basis at every point.
By definition of first-countability, $\tau$ is first-countable.
{{qed}}
Theorem: Metrizable Space is not necessarily Second-Countable

Let $T = \struct {S, \tau}$ be a topological space which is metrizable.
Then it is not necessarily the case that $T$ is second-countable.

Proof:
Let $T$ be an uncountable discrete space.
From Standard Discrete Metric induces Discrete Topology, $T$ is metrizable.
From Uncountable Discrete Space is not Second-Countable, $T$ is not second-countable.
{{qed}}
Theorem: Regular Paracompact Space is not necessarily Metrizable

Let $T = \struct {S, \tau}$ be a topological space which is regular and paracompact.
Then it is not necessarily the case that $T$ is metrizable.

Proof:
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line satisfies all Separation Axioms, $T$ is a regular space.
From Sorgenfrey Line is Paracompact, $T$ is a paracompact space.
From Sorgenfrey Line is not Metrizable, $T$ is not a metrizable space.
{{qed}}
Theorem: Regular Paracompact Space is not necessarily Metrizable

Let $T = \struct {S, \tau}$ be a topological space which is regular and paracompact.
Then it is not necessarily the case that $T$ is metrizable.

Proof:
Let $T$ be the radial interval space.
From Radial Interval Space is Completely Normal, $T$ is a completely normal space.
Hence from Sequence of Implications of Separation Axioms, $T$ is a regular space.
From Radial Interval Space is Paracompact, $T$ is a paracompact space.
From Radial Interval Space is not Metrizable, $T$ is not a metrizable space.
{{qed}}
Theorem: Uniform Space whose Topology is Metrizable is not necessarily Metrizable

Let $\UU$ be a uniformity on a set $S$.
Let $\struct {\struct {S, \UU}, \tau}$ be the uniform space generated from $\UU$.
Let $T = \struct {S, \tau}$ be the uniformizable space yielded by $\struct {\struct {S, \UU}, \tau}$.
Let $T$ be a metrizable space.
Then it is not necessarily the case that $\UU$ is itself a metrizable uniformity.

Proof:
Let $T = \struct {S, \tau}$ be an uncountable discrete ordinal space.
From Uncountable Discrete Ordinal Space is Metrizable, $T$ is a metrizable space.
However, from Uncountable Discrete Ordinal Space has Unmetrizable Uniformity, there exists a uniformity $\UU$ which yields the uniformizable space $T = \struct {S, \tau}$ which is not itself a metrizable uniformity.
{{qed}}
Theorem: Open Sets in Indiscrete Topology

$H$ is an open set of $T$ {{iff}} either $H = S$ or $H = \O$.

Proof:
A set $U$ is open in a topology $\tau$ if $U \in \tau$.
In $\tau = \set {\O, S}$, the only open sets are $\O$ and $S$.
{{qed}}
Theorem: Closed Sets in Indiscrete Topology

$H$ is a closed set of $T$ {{iff}} either $H = S$ or $H = \O$.

Proof:
A set $U$ is closed in a topology $\tau$ {{iff}}:
:$\relcomp S U \in \tau$
where $\relcomp S U$ denotes the complement of $U$ in $S$.
That is, the closed sets are the complements of the open sets.
From Open Sets in Indiscrete Topology, in $\tau = \set {\O, S}$, the only open sets are $\O$ and $S$.
Hence the only closed sets in the indiscrete topology on $S$ are:
:$\relcomp S \O = S$ from Relative Complement of Empty Set
and:
:$\relcomp S S = \O$ from Relative Complement with Self is Empty Set
as stated.
{{qed}}
Theorem: F-Sigma Sets in Indiscrete Topology

$H \subseteq S$ is an $F_\sigma$ ($F$-sigma) set of an indiscrete topological space $T = \struct {S, \set {\O, S} }$ {{iff}} either $H = S$ or $H = \O$.

Proof:
An $F_\sigma$ set is a set which can be written as a countable union of closed sets of $S$.
Hence the only $F_\sigma$ sets of $T$ are made from unions of $T$ and $\O$.
So $T$ and $\O$ are the only $F_\sigma$ sets of $T$.
{{qed}}
Theorem: G-Delta Sets in Indiscrete Topology

$H \subseteq S$ is a $G_\delta$ ($G$-delta) set of an indiscrete topological space $T = \struct {S, \set {\O, S} }$ {{iff}} either $H = S$ or $H = \O$.

Proof:
A $G_\delta$ set is a set which can be written as a countable intersection of open sets of $S$.
Hence the only $G_\delta$ sets of $T$ are made from intersections of $T$ and $\O$.
So $T$ and $\O$ are the only $G_\delta$ sets of $T$.
{{qed}}
Theorem: Subset of Indiscrete Space is Compact

$H \subseteq S$ is compact in an indiscrete topological space $T = \struct {S, \set {\O, S} }$.

Proof:
The subspace $T_H = \struct {H, \set {\O, S \cap H} }$ is trivially also an indiscrete space.
The only open cover of $T_H$ is $\set H$ itself.
The only subcover of $H$ is, trivially, also $\set H$, which is finite.
So $H$ is (trivially) compact in $T$.
{{qed}}
Theorem: Subset of Indiscrete Space is Sequentially Compact

$H \subseteq S$ is sequentially compact in an indiscrete topological space $T = \struct {S, \set {\O, S} }$.

Proof:
From Sequence in Indiscrete Space converges to Every Point, every sequence in $T$ converges to every point of $S$.
So every infinite sequence has a subsequence which converges to every point in $S$.
Hence $H$ is (trivially) sequentially compact in $T$.
{{qed}}
Theorem: Limit Points of Sequence in Indiscrete Space on Uncountable Set

Let $S$ be an uncountable set.
Let $T = \struct {S, \set {\O, S} }$ be the indiscrete topological space on $S$.
Let $\sequence {s_n}$ be a sequence in $T$.
Then every sequence in $T$ has an uncountable number of limit points.

Proof:
From Sequence in Indiscrete Space converges to Every Point, $\sequence {s_n}$ converges to every point of $S$.
As $S$ is uncountable, the result follows.
{{qed}}
Theorem: Partition Topology is not T2.5

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 $T_{2 \frac 1 2}$ space.

Proof:
{{AimForCont}} $T$ is a $T_{2 \frac 1 2}$ space.
Then from $T_{2 \frac 1 2}$ Space is $T_2$, $T$ is a $T_2$ space.
This contradicts the result Partition Space is not $T_2$.
Hence the result, by Proof by Contradiction.
{{qed}}
Theorem: Partition Space is not T2

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 $T_2$ (Hausdorff) space.

Proof:
{{AimForCont}} $T$ is a $T_2$ (Hausdorff) space.
Then from $T_2$ Space is $T_1$, $T$ is a $T_1$ space.
This contradicts the result Partition Space is not $T_1$.
Hence the result, by Proof by Contradiction.
{{qed}}
Theorem: Partition Space is not T1

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 $T_1$ space.

Proof:
{{AimForCont}} $T$ is a $T_1$ space.
Then from $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space.
This contradicts the result Partition Topology is not $T_0$.
Hence the result, by Proof by Contradiction.
{{qed}}
Theorem: Odd-Even Topology is Lindelöf

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 Lindelöf.

Proof:
From Odd-Even Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is Lindelöf.
{{qed}}
Theorem: Odd-Even Topology is Separable

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 separable.

Proof:
From Odd-Even Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is Separable.
{{qed}}
Theorem: Odd-Even Topology is First-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 first-countable.

Proof:
From Odd-Even Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is First-Countable Space.
{{qed}}