nyuuzyou/wikis huggingface dataset:
Countable Open Ordinal Space is Second-Countable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Then $\hointr 0 \Gamma$ is a second-countable space.
== Proof ==
From Basis for Open Ordinal Topology, the set $\BB$ of subsets of $\hointr 0 \Gamma$ 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 $\hointr 0 \Gamma$.
As $\Gamma$ strictly precedes $\Omega$, there are a countable number of points of $\hointr 0 \Gamma$.
Each point in $\hointr 0 \Gamma$ has a countable basis.
From Countable Union of Countable Sets is Countable, it follows that $\hointr 0 \Gamma$ has a countable basis.
== Sources ==
* : Part $\text {II}$: Counterexamples: $40$. Open Ordinal Space $[0, \Gamma) \ (\Gamma < \Omega)$: $5$
Category:Ordinal Spaces
Category:Second-Countable Spaces
Countable Particular Point Space is Lindelöf

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a countable particular point space.
Then $T$ is a Lindelöf apace.
== Proof ==
Consider the open cover of $T$:
:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
As $S$ is countable, then so is $\CC$, as we can set up a bijection from $\phi: S \setminus \set p \leftrightarrow \CC$:
:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
Hence $\CC$ is its own countable subcover.
The result follows by definition of Lindelöf apace.
== Sources ==
* : Part $\text {II}$: Counterexamples: $9$. Countable Particular Point Topology: $5$
Category:Lindelöf Spaces
Category:Particular Point Topology
Countable Product of First-Countable Spaces is First-Countable

== Theorem ==
Let $I$ be an indexing set with countable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be first-countable.
Then $\struct {S, \tau}$ is also first-countable.
== Proof ==
Let $x = \family {x_\alpha} \in S$.
Let $\BB_\alpha = \set {N^\alpha_i : i \in \N}$ be a countable local basis for $x_\alpha$ in $\struct {S_\alpha, \tau_\alpha}$ for each $\alpha \in I$.
Let $\pr_\alpha$ denote the projection of $S$ onto $S_\alpha$.
Let $\LL_\alpha = \set {\map {\pr_\alpha^{-1} } {N^\alpha_i}: N^\alpha_i \in B_\alpha}$.
Let $\KK_J = \ds \set {\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size J < \infty$.
From Product Space Local Basis Induced from Factor Spaces Local Bases:
:$\ds \BB_x = \bigcup_{J \mathop \subset I \mathop , \size J \mathop < \infty} \KK_J$ forms a local basis of $x$ in the product space $\struct {S_\alpha, \tau_\alpha}$
Now, each of the $\LL_\alpha$'s is countable. 
Since the $\KK_J$'s can be identified with a finite product of countable sets, they are each countable
From Countable Union of Countable Sets is Countable, $\BB_x$ forms a countable local basis of $x$ in the product space $\struct {S_\alpha, \tau_\alpha}$.
The result follows.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:First-Countable Spaces
Category:Product Spaces
Countable Product of Second-Countable Spaces is Second-Countable

== Theorem ==
Let $I$ be an indexing set with countable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be second-countable.
Then $\struct {S, \tau}$ is also second-countable.
== Proof ==
Let $\BB_\alpha = \set {N^\alpha_i : i \in \N}$ be a countable basis for $\tau_\alpha$ for each $\alpha \in I$.
Let $\pr_\alpha$ denote the projection of $S$ onto $S_\alpha$.
Let $\LL_\alpha = \set {\map {\pr_\alpha^{-1} } {N^\alpha_i}: N^\alpha_i \in B_\alpha}$.
Let $\ds \KK_J = \set {\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size J < \infty$.
From Product Space Basis Induced from Factor Space Bases:
:$\ds \BB = \bigcup_{J \mathop \subset I \mathop , \size J \mathop < \infty} \KK_J$ forms a basis of the product space.
Now, each of the $\LL_\alpha$'s is countable. 
Since the $\KK_J$'s can be identified with a finite product of countable sets, they are each countable
From Countable Union of Countable Sets is Countable, $\BB$ forms a countable basis of $S$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Second-Countable Spaces
Category:Product Spaces
Countable Product of Separable Spaces is Separable

== Theorem ==
Let $I$ be an indexing set with countable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be separable.
Then $\struct {S, \tau}$ is also separable.
== Proof ==
For each $\alpha \in I$, let $D_\alpha$ be a countable everywhere dense subset of $\struct {S_\alpha, \tau_\alpha}$.
Let $D = \ds \prod_{\alpha \mathop \in I} D_\alpha$.
From Countable Union of Countable Sets is Countable, $D$ is a countable set of $S$.
From Natural Basis of Product Topology, the set $\BB$ of cartesian products of the form $\ds \prod_{\alpha \mathop \in I} U_\alpha$ where:
:for all $\alpha \in I : U_\alpha \in \tau_\alpha$
:for all but finitely many indices $\alpha : U_\alpha = X_\alpha$
is a basis for $\tau$.
From Analytic Basis Characterization of Denseness, to show that $D$ is an everywhere dense subset in $\struct {S, \tau}$ it is sufficient to show that:
:$\forall U \in \BB : U \cap D \ne \O$
Let $U = \ds \prod_{\alpha \mathop \in I} U_\alpha \in \BB$.
Then:
:$\forall \alpha \in I : U_\alpha \cap D_\alpha \ne \O$
For all $\alpha \in I$, let $x_\alpha \in U_\alpha \cap D_\alpha$.
Then:
Thus:
:$U \cap D \ne \O$
The result follows.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Separable Spaces
Category:Product Topology
Countable Set has Choice Function

== Theorem ==
Let $S$ be a countable set.
Let $\mathbb S = \powerset S \setminus \set \O$ be the power set of $S$ excluding the empty set $\O$.
Then there exists a choice function $C$ for $S$:
:$\forall x \in \mathbb S: \map C x \in x$
== Proof ==
From Countable Set is Well-Orderable, we have that $S$ is a well-orderable set.
The result follows from Well-Orderable Set has Choice Function.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice
Category:Countable Sets
Category:Choice Functions
Countable Set has Measure Zero

== Theorem ==
Let $S$ be a countable set.
Then the measure of $S$ is $\map m S = 0$.
== Proof ==
Let $\ds \set {x_i}_{i \mathop = 1}^\infty$ be an enumeration of the elements of $S$.
For any (strictly) positive real number $\epsilon$, define $A_i$ as an open cover for $S$:
:$A_i = \paren {x_i - 2^{-i} \epsilon, x_i + 2^{-i} \epsilon}$
Then:
:$\ds S \subseteq \bigcup_{i \mathop = 1}^\infty A_i$
and:
:$\ds \map m {\bigcup A_i} \le \sum_{i \mathop = 1}^\infty 2^{1 - i} \epsilon = 2 \epsilon$ From the infinite geometric series
Since our choice of $\epsilon$ was arbitrary, for any positive real $z$ we can construct a set $T$ such that $S \subseteq T$ and $\map m T \le z$.
Hence $X$ has zero measure.
Category:Analysis
Countable Set is Null Set under Lebesgue Measure

== Theorem ==
Let $S \subseteq \R$ be a countable set.
Then $\map \lambda S = 0$, where $\lambda$ is Lebesgue measure.
That is, $S$ is a $\lambda$-null set.
== Proof ==
By Surjection from Natural Numbers iff Countable, there exists a surjection $f: \N \to S$.
It follows that:
:$S = \ds \bigcup_{n \mathop \in \N} \set{\map f n}$
As Lebesgue Measure is Diffuse, it holds that:
:$\forall n \in \N: \map \lambda {\set{\map f n}} = 0$
Thus, by Null Sets Closed under Countable Union, it follows that:
:$\map \lambda S = 0$
== Sources ==
* : $\S 10$: Problem $10.4$
Category:Lebesgue Measure
Countable Set is Well-Orderable

== Theorem ==
Let $S$ be a countable set.
Then $S$ is well-orderable.
== Proof ==
By the Well-Ordering Principle, the set of natural numbers $\N$ under the usual ordering $\le$ forms a well-ordered set.
By definition of countable set, there exists an injection:
:$f: S \to \N$
Let $V$ be a basic universe.
By definition of basic universe:
:$S \in V$
and:
:$\N \in V$
By the , both $S$ and $\N$ are classes.
From Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable, it follows that $S$ is well-orderable.
Hence the result.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Discussion
Category:Countable Sets
Category:Well-Orderings
Countable Set may have Uncountable Limit Points

== Theorem ==
Let $S$ be an uncountable set.
Let $H \subseteq S$ be a countable subset of $S$.
Let $T = \struct {H, \tau}$ be a topological space on the underlying set $H$.
Then despite the fact that $H$ is countable, the set of limit points of $T$ may be uncountable.
== Proof ==
Let $\R$ be the set of real numbers.
Let $\Q$ be the set of rational numbers.
Let $x \in \R$.
Then from Real Number is Limit Point of Rational Numbers in Real Numbers, $x$ is a limit point of $T$.
As $x$ is arbitrary, it follows that every element of $\R$ is a limit point of $T$.
From Rational Numbers are Countably Infinite, $\Q$ is a countable set.
From Real Numbers are Uncountable, $\R$ is an uncountable set.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Definition $3.7.10 \ \text {(c)}$
Category:Limit Points
Countable Space is Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $S$ is a countable set.
Then $T$ is a separable space.
== Proof ==
By definition, a topological space $T = \struct {S, \tau}$ is separable if there exists a countable subset of $S$ which is everywhere dense in $T$.
The closure of $S$ in $S$ is trivially $S$.
So, by definition, $S$ is everywhere dense in $S$.
As $S$ is countable by definition, the result follows.
Category:Separable Spaces
Category:Countable Sets
Countable Space is Sigma-Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space, where $S$ is a countable set.
Then $T$ is $\sigma$-compact.
== Proof ==
From Finite Space Satisfies All Compactness Properties, for every $p \in S$, $\set p$ is compact.
Then $\ds S = \bigcup_{p \mathop \in S} \set p$ is a countable union of compact sets.
Thus by definition $T$ is $\sigma$-compact.
Category:Sigma-Compact Spaces
Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound

== Theorem ==
Let $\Omega$ denote the minimal uncountable well-ordered set.
Let $\omega$ be a countable subset of $\Omega$.
Then $\omega$ has an upper bound in $\Omega$.
== Proof ==
By the minimal uncountable well-ordered set, $\Omega$ is uncountable
Then $\omega \ne \Omega$, for the former is countable and the latter is not.
Consider the union:
:$\ds \bigcup_{x \mathop \in \omega} S_x$
of initial segments $S_x$ in $\omega$.
By the definition of $\Omega$, any $S_x$ is countable.
Thus $\ds \bigcup_{x \mathop \in \omega} S_x$ is a countable union of countable sets.
This union is also countable, and so also cannot be all of $\Omega$, as $\Omega$ is uncountable.
From Union of Initial Segments is Initial Segment or All of Woset, there exists some $y \in \Omega$ such that:
:$\ds \bigcup_{x \mathop \in \omega} S_x = S_y$
Such a $y$ is an upper bound for $\omega$ by the definition of an initial segment.
== Sources ==
* $\S P.19$
Category:Order Theory
Countable Union of Countable Sets is Countable

== Theorem ==
Let the be accepted.
Then it can be proved that a countable union of countable sets is countable.
== Informal Proof ==
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $1.$ Introduction: $1.4$ Problems: $6$
* : Chapter $1$: Preliminaries: Exercises: $1.12$
Category:Set Union
Category:Countable Sets
Category:Countable Union of Countable Sets is Countable
Countable Union of Countable Sets is Countable/Informal Proof

== Theorem ==
== Proof ==
Consider the countable sets $S_0, S_1, S_2, \ldots$ where $\ds S = \bigcup_{i \mathop \in \N} {S_i}$.
Assume that none of these sets have any elements in common.
Otherwise, we can consider the sets $S_0' = S_0, S_1' = S_1 \setminus S_0, S_2' = S_2 \setminus \paren {S_0 \cup S_1}, \ldots$
All of these are countable by Subset of Countable Set‎ is Countable, and they have the same union $\ds S = \bigcup_{i \mathop \in \N} {S_i'}$.
Now we write the elements of $S_0', S_1', S_2', \ldots$ in the form of a (possibly infinite) table:
:$\begin{array} {*{4}c}
 {a_{00}} & {a_{01}} & {a_{02}} & \cdots \\
 {a_{10}} & {a_{11}} & {a_{12}} & \cdots \\
 {a_{20}} & {a_{21}} & {a_{22}} & \cdots \\
 \vdots & \vdots & \vdots & \ddots \\
\end{array}$
where $a_{ij}$ is the $j$th element of set $S_i$.
This table clearly contains all the elements of $\ds S = \bigcup_{i \mathop \in \N} {S_i}$.
Furthermore, we have that $\phi: S \to \N \times \N, a_{ij} \mapsto \tuple {i, j}$ is an injection.
By Cartesian Product of Countable Sets is Countable, there exists an injection $\psi: \N \times \N \to \N$.
Then $\psi \circ \phi: S \to \N$ is also an injection by Composite of Injections is Injection.
That is to say, $S$ is countable.
== Sources ==
* : $\S 2.2$: Countable sets: Theorem $2$
Category:Countable Union of Countable Sets is Countable
Countable Union of Countable Sets is Countable/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of countable sets.
Define:
:$\ds S = \bigcup_{n \mathop \in \N} S_n$
For all $n \in \N$, let $\FF_n$ denote the set of all injections from $S_n$ to $\N$.
Since $S_n$ is countable, $\FF_n$ is non-empty.
Using the , there exists a sequence $\sequence {f_n}_{n \mathop \in \N}$ such that $f_n \in \FF_n$ for all $n \in \N$.
Let $\phi: S \to \N \times \N$, where $\times$ denotes the cartesian product, be the mapping defined by:
:$\map \phi x = \tuple {n, \map {f_n} x}$
where $n$ is the (unique) smallest natural number such that $x \in S_n$.
From the Well-Ordering Principle, such an $n$ exists; hence, the mapping $\phi$ exists.
Since each $f_n$ is an injection, it (trivially) follows that $\phi$ is an injection.
Since $\N \times \N$ is countable, there exists an injection $\alpha: \N \times \N \to \N$.
From Composite of Injections is Injection, the mapping $\alpha \circ \phi: S \to \N$ is an injection.
Hence, $S$ is countable.
== Sources ==
* : $2.6$
* : Appendix $\text{A}$: Set Theory: Countable Sets
Category:Countable Union of Countable Sets is Countable
Countable Union of Countable Sets is Countable/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of countable sets.
Define:
:$\ds S = \bigcup_{n \mathop \in \N} S_n$.
For all $n \in \N$, let $\FF_n$ be the set of all surjections from $\N$ to $S_n$.
Since $S_n$ is countable, it follows by Surjection from Natural Numbers iff Countable that $\FF_n$ is non-empty.
Using the , there exists a sequence $\sequence {f_n}_{n \mathop \in \N}$ such that $f_n \in \FF_n$ for all $n \in \N$.
Let $\phi: \N \times \N \to S$, where $\times$ denotes the cartesian product, be the surjection defined by:
:$\map \phi {m, n} = \map {f_m} n$
Since $\N \times \N$ is countable, it follows by Surjection from Natural Numbers iff Countable that there exists a surjection $\alpha: \N \to \N \times \N$.
Since the composition of surjections is a surjection, the mapping $\phi \circ \alpha: \N \to S$ is a surjection.
By Surjection from Natural Numbers iff Countable, it follows that $S$ is countable.
== Sources ==
* : $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Theorem $7.5$
Category:Countable Union of Countable Sets is Countable
Countable Union of Meager Sets is Meager

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $\family {A_n}_{n \in \N}$ be a countable set of meager subsets of $X$.
Let:
:$\ds A = \bigcup_{n \mathop = 1}^\infty A_n$ 
Then $A$ is meager in $X$. 
== Proof ==
For each $n \in \N$, $A_n$ is meager and hence there exists a countable set $\family {A_{n, m} }_{m \in \N}$ of nowhere dense sets in $X$ such that:
:$\ds A_n = \bigcup_{m \mathop = 1}^\infty A_{n, m}$
Hence:
:$\ds A = \bigcup_{n \mathop = 1}^\infty \bigcup_{m \mathop = 1}^\infty A_{n, m} = \bigcup_{\tuple {n, m} \in \N^2} A_{n, m}$
From Cartesian Product of Countable Sets is Countable, $A$ is therefore the countable union of nowhere dense sets in $X$.
So $A$ is meager in $X$.
== Sources ==
* : $1.$ Introduction: $1.4$ Problems: $7$
Category:Meager Spaces
Category:Set Union
Countable iff Cardinality not greater than Aleph Zero

== Theorem ==
Let $X$ be set.
$X$ is countable :
$\card X \le \aleph_0$
where:
:$\card X$ denotes the cardinality of $X$
:$\aleph_0 = \card \N$ by Aleph Zero equals Cardinality of Naturals.
== Proof ==
:$X$ is countable 
:
:there exists an injection $f: X \to \N$ by definition of countable set
:
:$\card X \le \card \N$ by Injection iff Cardinal Inequality
:
:$\card X \le \aleph_0$
== Sources ==
* 
Category:Countable Sets
Category:Aleph Mapping
Countably Additive Function Dichotomy by Empty Set

== Theorem ==
Let $\AA$ be a $\sigma$-algebra.
Let $\overline \R$ denote the extended set of real numbers.
Let $f: \AA \to \overline \R$ be a function be a countably additive function.
Then the exactly one of the following is true:
:$\paren 1$: $\map f \O = 0$
:$\paren 2$: $\map f \O = + \infty$. Moreover, $\map f A = + \infty$ for all $A \in \AA$.
:$\paren 3$: $\map f \O = - \infty$. Moreover, $\map f A = - \infty$ for all $A \in \AA$.
== Proof ==
Suppose $\map f \O \ne 0$.
Then:
Furthermore, for each $A \in \AA$ we have:
Category:Countably Additive Functions
Category:Empty Set
Countably Additive Function also Finitely Additive

== Theorem ==
Let $\AA$ be a $\sigma$-algebra.
Let $\overline \R$ denote the extended set of real numbers.
Let $f: \AA \to \overline \R$ be a countably additive function.
Then $f$ is a finitely additive function.
== Proof ==
We have that $f$ is defined as countably additive :
:$\ds \map f {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map f {A_i}$
where $\sequence {A_i}$ is any sequence of pairwise disjoint elements of $\AA$.
We need to show that:
:$\ds \forall n \in \N: \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$
The claim is trivial if $f=+\infty$ or $f=-\infty$, constantly.
Thus we assume $\map f \O = 0$ by Countably Additive Function Dichotomy by Empty Set.
Let $n \in \N$ be any arbitrary natural number.
Let $\sequence {B_i}$ be the sequence of pairwise disjoint elements of $\AA$ defined as:
:$B_i = \begin{cases}
A_i & : i \le n \\
\O & : i > n
\end{cases}$
It follows that:
:$\ds \bigcup_{i \mathop \ge 1} B_i = \bigcup_{i \mathop = 1}^n A_i$
Thus:
Hence the result.
Category:Set Systems
Category:Measure Theory
Countably Additive Function of Null Set

== Theorem ==
Let $\AA$ be a $\sigma$-algebra.
Let $\overline \R$ denote the extended set of real numbers.
Let $f: \AA \to \overline \R$ be a function be a countably additive function.
Suppose that there exists at least one $A \in \AA$ where $\map f A$ is a finite number.
Then:
: $\map f \O = 0$
== Proof ==
Suppose that $A \in \AA$ such that $\map f A$ is a finite number.
So, let $\map f A = x$.
Consider the sequence $\sequence {S_i} \subseteq \AA$ defined as:
:$\forall i \in \N: S_i = \begin{cases} A & : i = 0 \\ \O & : i > 0 \end{cases}$
Then:
:$\ds \bigcup_{i \mathop \ge 0} S_i = A$
Hence:
It follows directly that:
: $\map f \O = 0$
== Proof 2 ==
This follows immediately from Countably Additive Function Dichotomy by Empty Set.
Category:Set Systems
Category:Countably Additive Functions
Countably Compact First-Countable Space is Sequentially Compact

== Theorem ==
A countably compact first-countable topological space is also sequentially compact.
== Proof 1 ==
== Proof 2 ==
Category:Sequentially Compact Spaces
Category:Countably Compact Spaces
Category:First-Countable Spaces
Category:Countably Compact First-Countable Space is Sequentially Compact
Countably Compact First-Countable Space is Sequentially Compact/Proof 2

== Theorem ==
== Proof ==
Let $T = \struct {S, \tau}$ be a countably compact first-countable topological space.
By definition of sequentially compact, it is sufficient to show that every infinite sequence in $S$ has a convergent subsequence.
Let $\sequence {s_n}$ be any sequence in $S$.
By Infinite Sequence in Countably Compact Space has Accumulation Point, $\sequence {s_n}$ has an accumulation point $p \in S$.
As $T$ is first-countable, $p$ has a countable local basis, say:
:$\set {V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}$
Then a subsequence $\sequence {s_{n_i} }$, where $s_{n_i} \in V_i$, converges to $p$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Countably Compact First-Countable Space is Sequentially Compact
Countably Compact Lindelöf Space is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a Lindelöf space which is also countably compact.
Then $T$ is compact.
== Proof ==
By the definitions:
:If $T = \struct {S, \tau}$ is a Lindelöf space then every open cover of $S$ has a countable subcover.
:If $T = \struct {S, \tau}$ is a countably compact space then every countable open cover of $S$ has a finite subcover.
It follows trivially that every open cover of $S$ has a finite subcover.
Hence the result by definition of compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Countably Compact Spaces
Category:Compact Spaces
Category:Lindelöf Spaces
Countably Compact Metric Space is Compact

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $M$ be countably compact.
Then $M$ is compact.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Metric Space is Compact iff Countably Compact
Category:Metric Spaces
Category:Countably Compact Spaces
Category:Compact Spaces
Category:Countably Compact Metric Space is Compact
Countably Compact Metric Space is Compact/Proof 1

== Theorem ==
== Proof ==
This follows directly from:
: Metric Space is First-Countable
: Countably Compact First-Countable Space is Sequentially Compact
: Sequentially Compact Metric Space is Second-Countable
: Second-Countable Space is Compact iff Countably Compact
Category:Countably Compact Metric Space is Compact
Countably Compact Metric Space is Compact/Proof 2

== Theorem ==
== Proof ==
We have that a Metric Space is Countably Compact iff Sequentially Compact.
Then we have that a Sequentially Compact Metric Space is Separable.
For each $n$, a metric space which is countably compact can be covered by finitely many open $\paren {1/n}$-balls: $\map {B_{1/n} } {x_i}$.
So $\set {x_i}$ is a dense subset of $A$ which is countable.
So if a metric space is countably compact it is by definition second-countable.
The result follows from Second-Countable Space is Compact iff Countably Compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Countably Compact Metric Space is Compact