nyuuzyou/wikis huggingface dataset:
Count of a's and b's in Fibonacci String

== Theorem ==
Let $S_n$ denote the $n$th Fibonacci string.
Then for $n \ge 3$, $S_n$ has:
:$F_{n - 2}$ instances of $\text a$
:$F_{n - 1}$ instances of $\text b$.
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:$S_n$ has $F_{n - 2}$ instances of $\text a$ and $F_{n - 1}$ instances of $\text b$.
=== Basis for the Induction ===
$\map P 3$ is the case:
:$S_n = \text {ba}$
It can be seen that $S_n$ has $F_1 = 1$ instance of $\text a$ and $F_2 = 1$ instance of $\text b$.
Thus $\map P 3$ is seen to hold.
$\map P 4$ is the case:
:$S_n = \text {bab}$
It can be seen that $S_n$ has $F_2 = 1$ instance of $\text a$ and $F_3 = 2$ instances of $\text b$.
Thus $\map P 4$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $4 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$S_k$ has $F_{k - 2}$ instances of $\text a$ and $F_{k - 1}$ instances of $\text b$
and:
:$S_{k - 1}$ has $F_{k - 3}$ instances of $\text a$ and $F_{k - 2}$ instances of $\text b$
from which it is to be shown that:
:$S_{k + 1}$ has $F_{k - 1}$ instances of $\text a$ and $F_k$ instances of $\text b$.
=== Induction Step ===
This is the induction step:
By definition of Fibonacci string:
:$S_{k + 1} = S_k S_{k - 1}$
concatenated.
By the induction hypothesis:
:$S_k$ has $F_{k - 2}$ instances of $\text a$ and $F_{k - 1}$ instances of $\text b$
and:
:$S_{k - 1}$ has $F_{k - 3}$ instances of $\text a$ and $F_{k - 2}$ instances of $\text b$
So:
:$S_{k + 1}$ has $F_{k - 2} + F_{k - 3}$ instances of $\text a$
and so by definition of Fibonacci numbers:
:$S_{k + 1}$ has $F_{k - 1}$ instances of $\text a$
and:
:$S_{k + 1}$ has $F_{k - 1} + F_{k - 2}$ instances of $\text b$
and so by definition of Fibonacci numbers:
:$S_{k + 1}$ has $F_k$ instances of $\text b$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:for all $n \in \Z$ such that $n \ge 3$: $S_n$ has $F_{n - 2}$ instances of $\text a$ and $F_{n - 1}$ instances of $\text b$.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $36$
Category:Fibonacci Strings
Countability Axioms Preserved under Open Continuous Surjection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ has one of the following properties, then $T_B$ has the same property:
:First-Countability
:Second-Countability
== Proof ==
=== Proof for First-Countability ===
=== Proof for Second-Countability ===
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Countability Axioms
Category:Open Mappings
Category:Continuous Mappings
Category:Surjections
Countability Properties Preserved under Projection Mapping

== Theorem ==
Let $\sequence {\struct {S_\alpha, \tau_\alpha} }$ be a sequence of topological spaces.
Let $\ds \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\sequence {\struct {S_\alpha, \tau_\alpha} }$.
Let $\pr_\alpha: \struct {S, \tau} \to \struct {S_\alpha, \tau_\alpha}$ denote the projection on the $\alpha$ coordinate.
Then $\pr_\alpha$ preserves the following countability properties.
That is, if $\struct {S, \tau}$ has one of the following properties, then each of $\struct {S_\alpha, \tau_\alpha}$ has the same property.
:Separability
:First-Countability
:Second-Countability
== Proof ==
First note that Projection from Product Topology is Continuous.
It follows from Continuous Image of Separable Space is Separable that separability is preserved under projections.
Next note that Projection from Product Topology is Open.
It follows from Countability Axioms Preserved under Open Continuous Surjection that:
:First-Countability
:Second-Countability
are preserved under projections.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Countability Axioms
Category:Continuous Mappings
Category:Projections
Countable Basis for P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\tau_p$ be the topology induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$.
For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$.
 
Then:
:$\BB_p = \set {\map {B_{p^{-n} } } q : q \in \Q, n \in \Z}$
is a countable basis for $\struct{\Q_p, \tau_p}$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
From Sequence of Powers of Number less than One, $\sequence{p^{-n}}$ is a real null sequence.
From Null Sequence induces Local Basis in Metric Space, for all $a \in \Q_p$ the set $\set {\map {B_{p^{-n} } } a : n \in \Z}$ is a local basis of $a$.
From Union of Local Bases is Basis, the set:
:$\BB' = \ds \bigcup_{a \in \Q_p} \set {\map {B_{p^{-n} } } a : n \in \Z} = \set {\map {B_{p^{-n} } } a : a \in Q_p, n \in \Z}$
is a basis for $\tau_p$.
Let $a \in \Q_p$ and $n \in \Z$.
From Rational Numbers are Dense Subfield of P-adic Numbers:
:$\Q$ is everywhere dense in $\Q_p$.
Hence:
:$\exists q \in \Q: q \in \map {B_{p^{-n} } } a$
From Characterization of Open Ball in P-adic Numbers,
:$\map {B_{p^{-n} } } q = \map {B_{p^{-n} } } a$
Hence:
:$\map {B_{p^{-n} } } a \in \BB_p$
Since $q$ and $n$ were arbitrary, then:
:$\BB' \subseteq \BB_p$
It follows that $\BB_p$ is a basis for $\tau_p$.
It remains to show that $\BB_p$ is countable.
For all $q \in \Q$, let $\map {\BB_p} q = \set {\map {B_{p^{-n} } } q : n \in \Z}$.
Then: 
:$\BB_p = \ds \bigcup_{q \mathop \in \Q} \map {\BB_p} q$
From Integers are Countably Infinite and Surjection from Countably Infinite Set iff Countable:
:$\forall q \in \Q: \map {\BB_p} q$ is a countable set.
From Rational Numbers are Countably Infinite:
:$\BB_p$ is a countable union of countable sets.
From Countable Union of Countable Sets is Countable:
:$\BB_p$ is countable.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.7$
Category:Topology of P-adic Numbers
Countable Basis for P-adic Numbers/Closed Balls

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\tau_p$ be the topology induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$.
For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {B_\epsilon^-} a$ denote the closed $\epsilon$-ball of $a$.
 
Then:
:$\BB_p = \set {\map {B^-_{p^{-n} } } q : q \in \Q, n \in \Z}$
is a countable basis for $\struct{\Q_p, \tau_p}$.
== Proof ==
For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$.
 
From Open Ball in P-adic Numbers is Closed Ball:
:$\BB_p = \set {\map {B^-_{p^{-n} } } q : q \in \Q, n \in \Z} = \set {\map {B_{p^{-n + 1} } } q : q \in \Q, n \in \Z} = \set {\map {B_{p^{-n} } } q : q \in \Q, n \in \Z}$
From Countable Basis for P-adic Numbers:
:$\BB_p$ is a countable basis for $\struct{\Q_p, \tau_p}$.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.7$
Category:Topology of P-adic Numbers
Countable Basis for P-adic Numbers/Cosets

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\tau_p$ be the topology induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$.
Let $\Z_p$ be the $p$-adic integers.
Then:
:$\BB_p = \set {q + p^n \Z_p : q \in \Q, n \in \Z}$
is a countable basis for $\struct{\Q_p, \tau_p}$.
== Proof ==
For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {B_\epsilon^-} a$ denote the closed $\epsilon$-ball of $a$.
 
From Closed Balls of P-adic Number:
:$\BB_p = \set {q + p^n \Z_p : q \in \Q, n \in \Z} = \set {\map {B^-_{p^{-n} } } q : q \in \Q, n \in \Z}$
From Countable Closed Ball Basis for P-adic Numbers:
:$\BB_p$ is a countable basis for $\struct{\Q_p, \tau_p}$.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.7$
Category:Topology of P-adic Numbers
Countable Basis of Real Number Line

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\BB$ be the set of subsets of $\R$ defined as:
:$\BB = \set {\openint a b: a, b \in \Q,\ a < b}$
That is, $\BB$ is the set of open intervals of $\R$ whose endpoints are rational numbers.
Then $\BB$ forms a countable basis of $\struct {\R, \tau_d}$
== Proof ==
Let $U \in \tau_d$.
Let $x \in U$.
Let:
:$\BB' = \set {\openint c d: c, d \in \R,\ a < b}$
By Basis for Euclidean Topology on Real Number Line, $\BB'$ is an analytic basis for the Euclidean topology on $\R$.
Then let $\openint c d \in \BB'$ such that $x \in \openint c d \subseteq U$.
That is, $c < x < d$.
By Between two Real Numbers exists Rational Number, let $a \in \Q$ such that $c < a < x$.
By Between two Real Numbers exists Rational Number, let $b \in \Q$ such that $x < b < d$.
Then $\openint a b \in \BB$ and $x \in \openint a b$.
Let $y \in \openint a b$.
Then $c < a < y < b < d$ implies $y \in \openint c d$.
Thus, $\openint a b \subseteq \openint c d \subseteq U$.
It follows that $\BB$ is a basis of $\struct {\R, \tau_d}$
It remains to be confirmed that $\BB$ is countable.
By Rational Numbers are Countably Infinite, $\Q$ is countable.
By Cartesian Product of Countable Sets is Countable, $\Q \times \Q$ is countable.
Define the mapping $f: \Q \times \Q \to \BB$ by:
:$\map f {q, r} := \begin {cases}
\openint q r & : q < r \\
\openint 0 1 & : q \ge r
\end {cases}$
For each $\openint a b \in \BB$, we have $\map f {a, b} = \openint a b$.
Then $f$ is a surjection.
By Image of Countable Set under Mapping is Countable, $\BB$ is countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $2$
Category:Real Number Line with Euclidean Topology
Category:Topological Bases
Countable Closed Ordinal Space is Metrizable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a metrizable space.
== Proof ==
From Countable Closed Ordinal Space is Second-Countable, $\closedint 0 \Gamma$ has a basis which is $\sigma$-locally finite.
From Ordinal Space is Completely Normal, $\closedint 0 \Gamma$ is a completely normal space.
From Sequence of Implications of Separation Axioms it follows that $\closedint 0 \Gamma$ is a regular space.
The result follows from Metrizable iff Regular and has Sigma-Locally Finite Basis.
== Sources ==
* : Part $\text {II}$: Counterexamples: $41$. Closed Ordinal Space $[0, \Gamma] \ (\Gamma < \Omega)$: $5$
Category:Ordinal Spaces
Category:Metrizable Topologies
Countable Closed Ordinal Space is Second-Countable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a second-countable space.
== Proof ==
From Basis for Open Ordinal Topology, the set $\BB$ of subsets of $\closedint 0 \Gamma$ of the form:
:$\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$
for $\alpha, \beta \in \hointr 0 \Gamma$, forms a basis for $\closedint 0 \Gamma$.
As $\Gamma$ strictly precedes $\Omega$, there are a countable number of points of $\closedint 0 \Gamma$.
Each point in $\closedint 0 \Gamma$ has a countable basis.
From Countable Union of Countable Sets is Countable, it follows that $\closedint 0 \Gamma$ has a countable basis.
== Sources ==
* : Part $\text {II}$: Counterexamples: $41$. Closed Ordinal Space $[0, \Gamma] \ (\Gamma < \Omega)$: $5$
Category:Ordinal Spaces
Category:Second-Countable Spaces
Countable Compactness is Preserved under Continuous Surjection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is countably compact, then $T_B$ is also countably compact.
== Proof ==
Let $T_A$ be countably compact.
Take a countable open cover $\UU$ of $T_B$.
From Preimage of Cover is Cover, $\VV := \set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$.
$\VV$ is a countable cover because there it is bijective with $\UU$.
By hypothesis, $\phi$ is continuous.
For all $U \in \UU$, $U$ is open in $T_B$.
It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$.
So $\VV$ is a countable open cover of $T_A$.
$T_A$ is countably compact, so we take a finite subcover:
:$\set {\phi^{-1} \sqbrk {U_1}, \ldots, \phi^{-1} \sqbrk {U_n} }$
We have that $\phi$ is surjective.
So from Surjection iff Right Inverse:
:$\phi \sqbrk {\phi^{-1} \sqbrk A} = A$
So:
:$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} } } = \set {U_1, \ldots, U_n} \subseteq \UU$
is a finite subcover of $\UU$ on $T_B$.
== Also see ==
* Compactness Properties Preserved under Continuous Surjection
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Countably Compact Spaces
Category:Continuous Mappings
Category:Surjections
Countable Complement Space Satisfies Countable Chain Condition

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ satisfies the countable chain condition.
== Proof ==
We have that a Countable Complement Space is Irreducible.
Therefore there is no disjoint set of open sets of $T$.
So the result holds vacuously.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $3$
Category:Countable Chain Condition
Category:Countable Complement Topology
Countable Complement Space is Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a connected space.
== Proof ==
We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $4$
Category:Connected Spaces
Category:Countable Complement Topology
Countable Complement Space is Irreducible

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is an irreducible space.
== Proof ==
Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$.
We have that both $U_1$ and $U_2$ are both uncountable if $S$ is.
From Uncountable Subset of Countable Complement Space Intersects Open Sets, they intersect each other.
Hence the result from definition of irreducible space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $4$
Category:Irreducible Spaces
Category:Countable Complement Topology
Countable Complement Space is Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a Lindelöf space.
== Proof ==
By definition, $T$ is a Lindelöf space every open cover of $S$ has a countable subcover.
Let $\CC$ be an open cover of $T$.
Let $U \in \CC$ be any set in $C$.
It covers all but a countable number of points of $T$.
So for each of those points we pick an element of $\CC$ which covers each of those points.
Hence we have a countable subcover of $T$.
So by definition $T$ is a Lindelöf space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $2$
Category:Lindelöf Spaces
Category:Countable Complement Topology
Countable Complement Space is Locally Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a locally connected space.
== Proof ==
We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Locally Connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $4$
Category:Locally Connected Spaces
Category:Countable Complement Topology
Countable Complement Space is Pseudocompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is pseudocompact.
== Proof ==
We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Pseudocompact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $4$
Category:Pseudocompact Spaces
Category:Countable Complement Topology
Countable Complement Space is T1

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology.
Then $T$ is a $T_1$ (Fréchet) space.
== Proof ==
We have that the Countable Complement Topology is Expansion of Finite Complement Topology. 
We also have that a Finite Complement Space is $T_1$.
Then from Separation Properties Preserved by Expansion, we have that $T$ is a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $1$
Category:T1 Spaces
Category:Countable Complement Topology
Countable Complement Space is not Countably Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a countably compact space.
== Proof ==
Consider $U \in \tau$.
By definition of countable complement topology, $\relcomp S U$ is countably infinite.
Then for each $x \in \relcomp S U$, $\relcomp S U \setminus \set x$ is countably infinite.
Then for each $x \in \relcomp S U$, $\relcomp S {\relcomp S U \setminus \set x} = U \cup \set x$ is open.
Hence $\set {U \cup \set x: x \in \relcomp S U}$ is an open cover.
This cover is countable because it is equivalent to $\relcomp S U$.
 $T$ is a countably compact space.
Then by definition there exists a finite subcover $\set {U \cup \set {x_1}, \ldots, U \cup \set {x_n} }$ of $T$.
Then:
:$\paren {U \cup \set {x_1} } \cup \dotsb \cup \paren {U \cup \set {x_n} } = U \cup \set {x_1, \ldots, x_n}$
Since $\relcomp S U$ is not finite, it follows that $U \cup \set {x_1, \ldots, x_n} \ne S$.
Thus there is no finite subcover of $T$.
It follows by Proof by Contradiction that $T$ cannot be a countably compact space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $2$
Category:Countably Compact Spaces
Category:Countable Complement Topology
Countable Complement Space is not Countably Metacompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not countably metacompact.
== Proof ==
From Uncountable Subset of Countable Complement Space Intersects Open Sets, the intersection of any open sets is uncountable.
So for any open cover $\CC$ of $T$, every point is in an infinite number of open sets of $T$.
So no refinement of any open cover of $T$ can be point finite.
Hence the result by definition of countably metacompact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $5$
Category:Countably Metacompact Spaces
Category:Countable Complement Topology
Countable Complement Space is not First-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a first-countable space.
== Proof ==
 some $x \in S$ has a countable local basis.
That means:
:there exists a countable set of sets $\BB_x \subseteq \tau$ 
such that:
:$\forall B \in \BB_x: x \in B$
and such that:
:every open neighborhood of $x$ contains some $B \in \BB_x$.
So:
By definition, each of $S \setminus B$ is countable.
From Countable Union of Countable Sets is Countable it follows that $\ds \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is also countable.
So $S \setminus \set x$ and therefore $S$ is also countable.
From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $x \in S$ has a countable local basis must be false.
Hence by definition $T$ can not be first-countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $3$
Category:First-Countable Spaces
Category:Countable Complement Topology
Countable Complement Space is not Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a separable space.
== Proof ==
Let $U$ be a countable subset of $S$.
By the definition of $T$, $U$ is closed.
As $U$ is countable but $S$ is uncountable:
:$U \subsetneq S$
and so:
:$U \ne S$
From Closed Set Equals its Closure:
:$U^- = U$
where $U^-$ is the closure of $U$.
Thus:
:$U^- \ne S$
So by definition, $U$ is not everywhere dense in $T$.
As $U$ is arbitrary, there is no countable set which is everywhere dense in $T$.
The result follows by definition of separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $3$
Category:Separable Spaces
Category:Countable Complement Topology
Countable Complement Space is not Sigma-Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $\sigma$-compact space.
== Proof ==
From Compact Sets in Countable Complement Space, the only compact sets in $T$ are finite.
A countable union of finite sets can not be an uncountable set.
Hence the result by definition of $\sigma$-compact space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $2$
Category:Sigma-Compact Spaces
Category:Countable Complement Topology
Countable Complement Space is not T2

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $T_2$ (Hausdorff) space.
== Proof ==
We have:
:Countable Complement Space is Irreducible
:Irreducible Hausdorff Space is Singleton
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $1$
Category:Hausdorff Spaces
Category:Countable Complement Topology
Countable Complement Space is not T3, T4 or T5

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $T_3$ space, $T_4$ space or $T_5$ space.
== Proof ==
We have that a Countable Complement Space is a $T_1$ space.
From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space.
We then have that a Countable Complement Space is not $T_2$.
From Regular Space is $T_2$ Space, $T$ is not a regular space.
By definition, a regular space is a space that is both a $T_0$ space and a $T_3$ space.
But $T$ is a $T_0$ space and not a regular space.
So it follows that $T$ cannot be a $T_3$ space.
Next we have that a Normal Space is a $T_3$ Space.
But as $T$ is not a $T_3$ space, $T$ cannot be a normal space.
By definition, a normal space is a space that is both a $T_1$ space and a $T_4$ space.
But $T$ is a $T_1$ space and not a normal space.
So it follows that $T$ cannot be a $T_4$ space.
Finally we have that a $T_5$ Space is a $T_4$ Space.
But as $T$ is not a $T_4$ space, $T$ cannot be a $T_5$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $1$
Category:T3 Spaces
Category:T4 Spaces
Category:T5 Spaces
Category:Countable Complement Topology