nyuuzyou/wikis huggingface dataset:
Countable Complement Space is not Weakly Countably Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not weakly countably compact.
== Proof ==
By definition, $T$ is weakly countably compact every infinite subset of $S$ has a limit point in $S$.
Let $H \subseteq S$ be a countable set.
From Limit Points of Countable Complement Space, it contains all its limit points.
Let $x \in H$.
$\relcomp S {\relcomp S H \cup \set x} = H \setminus \set x$ is countable.
Then $\relcomp S H \cup \set x$ is an open neighborhood of $x$. 
However $H \cap \paren {\paren {\relcomp S H \cup \set x} \setminus \set x} = H \cap \relcomp S H = \O$.
So $x$ is not a limit point of $H$.
Since $x$ is arbitrary, there are no limit points in $H$. 
This shows that $T$ is not weakly countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $7$
Category:Weakly Countably Compact Spaces
Category:Countable Complement Topology
Countable Complement Topology is Expansion of Finite Complement Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be the countable complement topology on an infinite set $S$.
Let $T' = \struct {S, \tau'}$ be the finite complement topology on the same infinite set $S$.
Then $\tau$ is an expansion of $\tau'$.
== Proof ==
Let $U \in \tau', U \ne \O$.
Then $\relcomp S U$ is finite by definition of finite complement topology.
Then by definition of countable complement topology, we have that $U \in \tau$.
So $\tau' \subseteq \tau$ and hence the result by definition of expansion.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $1$
Category:Countable Complement Topology
Category:Finite Complement Topology
Countable Complement Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement space.
Then $\tau$ is a topology on $T$.
== Proof ==
By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially countable.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both countable.
Hence from Countable Union of Countable Sets is Countable their union is also countable and so $\relcomp S H$ is countable.
So $H = A \cap B \in \tau$ as its complement is countable.
Now let $\UU \subseteq \tau$.
Then:
:$\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
by De Morgan's Laws: Complement of Union.
But as:
:$\forall U \in \UU: \relcomp S U \in \tau$
each of the $\relcomp S U$ is countable.
Hence so is their intersection.
So $\ds \relcomp S {\bigcup \UU}$ is countable which means:
:$\ds \bigcup \UU \in \tau$
So $\tau$ is a topology on $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology
Category:Countable Complement Topology
Countable Discrete Space is Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is a Lindelöf space.
== Proof ==
We have:
:Countable Discrete Space is $\sigma$-Compact
:$\sigma$-Compact Space is Lindelöf
So if $S$ is countable, $T$ is a Lindelöf space.
== Also see ==
* Uncountable Discrete Space is not Lindelöf
== Sources ==
* : Part $\text {II}$: Counterexamples: $2$. Countable Discrete Topology: $8$
Category:Lindelöf Spaces
Category:Discrete Topology
Countable Discrete Space is Second-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is second-countable.
== Proof ==
From Basis for Discrete Topology, the set:
:$\BB := \set {\set x: x \in S}$
is a basis for $T$.
There is a trivial one-to-one correspondence $\phi: S \leftrightarrow \BB$ between $S$ and $\BB$:
:$\forall x \in S: \map \phi x = \set x$
Let $S$ be countable.
Then $\BB$ is also countable by definition of countability.
So we have that $T$ has a countable basis, and so is second-countable by definition.
== Also see ==
* Uncountable Discrete Space is not Second-Countable
== Sources ==
* : Part $\text {II}$: Counterexamples: $2$. Countable Discrete Topology: $8$
Category:Second-Countable Spaces
Category:Discrete Topology
Countable Discrete Space is Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is separable.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Uncountable Discrete Space is not Separable
== Sources ==
* : Part $\text {II}$: Counterexamples: $2$. Countable Discrete Topology: $8$
Category:Separable Spaces
Category:Discrete Topology
Category:Countable Discrete Space is Separable
Countable Discrete Space is Separable/Proof 1

== Theorem ==
== Proof ==
Let $T = \left({S, \tau}\right)$ be a countable discrete topological space.
From Countable Discrete Space is Second-Countable:
:$T$ is second-countable.
From Second-Countable Space is Separable:
:$T$ is separable.
Category:Countable Discrete Space is Separable
Countable Discrete Space is Sigma-Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable discrete topological space.
Then $T$ is $\sigma$-compact.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Uncountable Discrete Space is not Sigma-Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $2$. Countable Discrete Topology: $8$
Category:Sigma-Compact Spaces
Category:Discrete Topology
Category:Countable Discrete Space is Sigma-Compact
Countable Discrete Space is Sigma-Compact/Proof 1

== Theorem ==
== Proof ==
We have that Singleton Set in Discrete Space is Compact.
We also have that $S$ is the union of all its singleton sets:
:$\ds S = \bigcup_{x \mathop \in S} \set x$
As $S$ is countable, it is the union of countably many compact sets.
Hence the result, by definition of $\sigma$-compact.
Category:Countable Discrete Space is Sigma-Compact
Countable Discrete Space is not Weakly Countably Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable discrete space.
Then $T$ is not weakly countably compact.
== Proof ==
Let $A \subseteq S$ be an infinite subset of $S$.
Then as Set in Discrete Topology is Clopen, $A$ is closed in $T$.
From Closed Set Equals its Closure, $A = A^-$ where $A^-$ is the closure of $A$.
From the definition, $x$ is a limit point of $A$ if it belongs to the closure of $A$ but is not an isolated point of $A$.
But then we have Topological Space is Discrete iff All Points are Isolated.
So $A$ has no limit points in $T$.
So $T$ is not weakly countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $5 \text { - } 7$. Partition Topology: $4$
Category:Discrete Topology
Category:Weakly Countably Compact Spaces
Countable Excluded Point Space is Second-Countable

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be a countable excluded point space.
Then $T$ is a second-countable space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $14$. Countable Excluded Point Topology: $6$
Category:Countable Excluded Point Space is Second-Countable
Category:Second-Countable Spaces
Category:Excluded Point Topology
Countable Excluded Point Space is Second-Countable/Proof 1

== Theorem ==
== Proof ==
Consider the set $\BB$ defined as:
:$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$
From Basis for Excluded Point Space, $\BB$ is a basis for $T$, and trivially has the same cardinality as $S$.
So by definition, if $S$ is countable, then $T$ is second-countable.
Category:Countable Excluded Point Space is Second-Countable
Countable Excluded Point Space is Separable

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be a countable excluded point space.
Then $T$ is a separable space.
== Proof ==
The closure of the set $S$ (trivially) equals $S$.
That is, $S$ is everywhere dense in $T$.
But as $S$ is countable it follows by definition that $T$ is separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $14$. Countable Excluded Point Topology: $6$
Category:Separable Spaces
Category:Excluded Point Topology
Countable Finite Complement Space is not Locally Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable finite complement topology.
Then $T$ is not locally path-connected.
== Proof ==
Let $\BB$ be a basis for $T$.
Let $B \in \BB$.
 $f: \closedint 0 1 \to B$ is a path on $T$.
Then $f$ is by definition continuous.
Now consider the set:
:$F = \set {\map {f^{-1} } x: x \in S}$
From Continuity Defined from Closed Sets, each of the elements of $F$ is closed.
Also, from Mapping Induces Partition on Domain, the elements of $F$ are pairwise disjoint.
From Basis of Countable Finite Complement Topology consists of Countably Infinite Sets, $B$ is a countably infinite set.
Hence $F$ is also countably infinite by nature of $f$ being a mapping.
Furthermore, we have $\ds \bigcup F = \closedint 0 1$.
Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\closedint 0 1$.
From Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets, this is impossible.
From this contradiction, $f$ cannot be continuous, and so cannot be a path on $B$.
So $B$ cannot be path-connected.
So, by definition, $T$ is not locally path-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18$. Finite Complement Topology on a Countable Space: $10$
Category:Locally Path-Connected Spaces
Category:Finite Complement Topology
Countable Finite Complement Space is not Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable finite complement topology.
Then $T$ is not path-connected.
== Proof ==
 $f: \closedint 0 1 \to S$ is a path on $T$.
Then $f$ is by definition continuous.
By definition of the finite complement topology, for all $x \in S$, the set $\set x$ is closed.
Now consider the set:
:$F = \set {\map {f^{-1} } x: x \in S}$
From Continuity Defined from Closed Sets, each of the elements of $F$ is closed.
Also, from Mapping Induces Partition on Domain, the elements of $F$ are pairwise disjoint.
As $T$ is a countable finite complement topology, its underlying set $S$ is countably infinite by definition.
Hence $F$ is also countably infinite by nature of $f$ being a mapping.
Furthermore, we have $\ds \bigcup F = \closedint 0 1$.
Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\closedint 0 1$.
From Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets, this is impossible.
From this contradiction, $f$ cannot be continuous, and so cannot be a path on $T$.
So $T$ cannot be path-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18$. Finite Complement Topology on a Countable Space: $10$
Category:Path-Connected Spaces
Category:Finite Complement Topology
Countable Fort Space is Metrizable

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Then $T$ is a metrizable space.
== Proof ==
We have:
:Fort Space is Regular
:Countable Fort Space is Second-Countable.
The result follows from Urysohn's Metrization Theorem.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23$. Countable Fort Space: $6$
Category:Metrizable Topologies
Category:Fort Spaces
Countable Fort Space is Perfectly Normal

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Then $T$ is a perfectly normal space.
== Proof ==
We have from Closed Set of Countable Fort Space is $G_\delta$ that every closed set in $T$ is a $G_\delta$ set.
From Fort Space is $T_5$ and $T_5$ Space is $T_4$ Space, we have that a Fort space is a $T_4$ space.
From Fort Space is $T_1$ it follows by definition that $T$ is a perfectly normal space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23$. Countable Fort Space: $3$
Category:Perfectly Normal Spaces
Category:Fort Spaces
Countable Fort Space is Second-Countable

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Then $T$ is a second-countable space.
== Proof ==
From Subset of Countably Infinite Set is Countable, $S \setminus \set p$ is countable.
Let $f: \N \to S$ be an enumeration of $S \setminus \set p$.
For brevity, let us write $s_n$ for $\map f n$.
Now define, for $n \in \N$, $S_n = S \setminus \set {s_0, \ldots, s_{n - 1} }$.
Note that $s_n \ne p$ for all $n \in \N$, so that $p \in S_n$ for all $n$.
By Relative Complement of Relative Complement, $S \setminus S_n = \set {s_0, \ldots, s_{n - 1} }$ and so $S_n \in \tau_p$.
Also, define $S'_n = \set {s_n}$.
Since $s_n \ne p$, $S'_n$ is open in $\tau_p$ as well.
From Finite Union of Countable Sets is Countable, $\BB := \set {S_n : n \in \N} \cup \set {S'_n : n \in \N}$ is countable.
Let us verify that $\BB$ forms a basis for $\tau_p$.
So let $U \subseteq S$ be open in $\tau_p$.
If $p \notin U$ then $\ds U = \bigcup \set {S'_n: s_n \in U}$ because the $s_n$ form an enumeration of $S \setminus \set p$.
Now if $p \in U$, then by definition of the Fort topology, $S \setminus U$ must be finite.
It follows that $A = \set {n \in \N: s_n \in S \setminus U}$ is also finite.
From Subset of Naturals is Finite iff Bounded, there exists an $N \in \N$ such that $N - 1 > n$ for all $n \in A$.
Therefore, $A \subseteq \set {0, \ldots, N - 1}$ and hence:
:$S \setminus U \subseteq \set {s_0, \ldots, s_{N - 1} }$
From Set Complement inverts Subsets, also:
:$S_N = S \setminus \set {s_0, \ldots, s_{N - 1} } \subseteq U$
It follows that $\BB$ is a basis for $\tau_p$.
Since $\BB$ is countable, we conclude $T$ is second-countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23$. Countable Fort Space: $6$
Category:Second-Countable Spaces
Category:Fort Spaces
Countable Fort Space is Separable

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Then $T$ is a separable space.
== Proof ==
This result follows trivially from Countable Space is Separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23$. Countable Fort Space: $6$
Category:Separable Spaces
Category:Fort Spaces
Countable Function on Power Set of Sample Space is Discrete Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space such that $\Sigma$ is the power set of $\Omega$.
Let $f: \Omega \to \R$ be a function such that $\Img f$ is countable.
Then $f$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.
== Proof ==
By definition, $\map {f^{-1} } x \subseteq \Omega$.
But then $\map {f^{-1} } x \in \powerset \Omega$.
Hence the result.
== Sources ==
* : $\S 2.1$: Probability mass functions: Exercise $2$
Category:Probability Theory
Countable Hypothesis Set has Finished Tableau

== Lemma ==
Let $\mathbf H$ be a countable set of WFFs of propositional logic.
Then there exists a finished tableau whose root node is $\mathbf H$.
== Proof ==
The Tableau Extension Lemma shows that each finite hypothesis set $\mathbf H$ is the root of some finished tableau.
It remains, then, to show that the result still applies when $\mathbf H = \set {\mathbf A_1, \ldots, \mathbf A_n, \ldots}$ is countably infinite.
Let $\mathbf H_n = \set {\mathbf A_1, \ldots, \mathbf A_n} \subset \mathbf H$.
We will say that a finite tableau $T_n$ with root $\mathbf H$ is finished for $\mathbf H_n$ if the tableau $T'_n$ is finished where:
:$T'_n$ is the same as $T_n$ except:
:$T'_n$ has the root $\mathbf H_n$ instead of $\mathbf H$.
We can use the Tableau Extension Lemma countably many times, and get a sequence of finite tableaux $T_0, T_1, \ldots, T_n, \ldots$ such that:
:$T_0$ has only a root node;
:For each $n > 0$, $T_n$ is an extension of $T_{n-1}$ such that $T_n$ is finished for $\mathbf H_n$;
:For each $n > 0$, $T_n$ has the property that no contradictory branch $\Gamma$ of $T_{n-1}$ gets extended when forming $T_n$.
Now, let $T$ be the union $\ds T = \bigcup_{k \mathop = 0}^\infty T_k$.
Let $\Gamma$ be a branch of $T$.
Suppose $\Gamma$ is contradictory, with complementary pair $\mathbf A, \neg \mathbf A$.
Then $\exists n \in \N$ such that both $\mathbf A$ and $\neg \mathbf A$ are in $T_n$.
Then $\Gamma \cap T_n$ is already a contradictory branch of $T_n$.
So, by our method of construction of $T$, this branch $\Gamma \cap T_n$ is never extended past stage $n$, so $\Gamma = \Gamma \cap T_n$ and $\Gamma$ is finite.
On the other hand, suppose $\Gamma$ is not contradictory.
Then the construction ensures that $\Gamma$ is a finished branch.
So $T$ is a finished tableau whose root node is $\mathbf H$.
== Sources ==
* : $\S 1.11$: Compactness: Lemma $1.11.2$
Category:Propositional Tableaux
Countable Infinite Product of Real Number Spaces is Homeomorphic to Fréchet Metric Space

== Theorem ==
Let $\struct {\R, \tau_d}$ denote the real number line under the Euclidean topology.
Let $T = \struct {\R^\omega, \tau} = \ds \prod_{i \mathop \in \N} \struct {\R, \tau_d}$ denote the countable-dimensional real Cartesian space under the product topology $\tau$.
Let $\struct {\R^\omega, d}$ be the Fréchet space on $\R^\omega$, where:
:$\map d {x, y} = \ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} }$
Then the topology induced by $d$ is exactly the product topology $\tau$.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $37$. Fréchet Space: $7$
Category:Fréchet Product Metric
Countable Intersection of Events is Event

== Theorem ==
Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
The event space $\Sigma$ of $\EE$ has the property that:
:$\quad A_1, A_2, \ldots \in \Sigma \implies \ds \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$
That is, the countable intersection of events is also an event.
== Proof ==
By definition, a probability space $\struct {\Omega, \Sigma, \Pr}$ is a measure space.
So, again by definition, an event space $\Sigma$ is a $\sigma$-algebra on $\Omega$.
From Sigma-Algebra is Delta-Algebra:
:$\ds A_1, A_2, \ldots \in \Sigma \implies \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$
by definition of $\delta$-algebra.
== Also see ==
* Elementary Properties of Event Space
== Sources ==
* : $1$: Events and probabilities: $1.4$: Probability spaces: $(13)$
Category:Event Spaces
Countable Local Basis in Compact Complement Topology

== Theorem ==
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Let $p \in \R$.
Then sets of the form:
:$\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$
form a countable local basis for $p$.
== Proof ==
Let $p \in \R$.
Let:
:$\BB_p = \set {\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to: n \in \N}$
Let $n \in \N$ and $P_n \in \BB$, so that:
:$P_n = \openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$
Then:
:$\R \setminus P_n = \closedint {-n} {p - \dfrac 1 n} \cup \closedint {p + \dfrac 1 n} n$
which is the union of two compact sets in $\R$ and therefore itself compact.
Clearly $p \in P_n$.
So $\BB$ is a set of open neighborhoods of $p$ in $T$.
Now let $U \in \tau$ such that $p \in U$.
Then $V := \R \setminus U$ is compact and is therefore bounded.
Suppose $\sup V = M, \inf V = m$.
Then $\exists n_B \in \N: n_T > M, -n_B > m$.
But as $p \notin V$ we have:
:$\exists \epsilon \in \R: p + \epsilon \notin V, p - \epsilon \notin V$
So $\exists n \in \N: \epsilon > \dfrac 1 n$
and so by making $n$ large enough you can fix it so that:
:$\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to \subseteq U$
so fulfilling the conditions for $\BB$ to be a local basis for $p$ which is countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $5$
Category:Compact Complement Topology
Countable Open Covers Condition for Separated Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B \subseteq S$
For all $X \subseteq S$, let $X^-$ denote the closure of $X$ in $T$.
Let:
:$\UU = \set {U_n : n \in \N}$ be a countable open cover of $A : \forall n \in \N : {U_n}^- \cap B = \O$
Let:
:$\VV = \set {V_n : n \in \N}$ be a countable open cover of $B : \forall n \in \N : {V_n}^- \cap A = \O$
Then:
:$A$ and $B$ can be separated in $T$
== Proof ==
By definition of cover:
:$\ds A \subseteq \bigcup_{n \mathop \in \N} U_n$
We have:
:$\ds A \cap \paren{\bigcup_{n \mathop \in \N} {V_n}^-} = \O$
From Subset of Set Difference iff Disjoint Set:
:$(1): \quad \ds A \subseteq \paren {\bigcup_{n \mathop \in \N} U_n} \setminus \paren {\bigcup_{n \mathop \in \N} {V_n}^-}$
Similarly:
:$(2): \quad \ds B \subseteq \paren {\bigcup_{n \mathop \in \N} V_n} \setminus \paren {\bigcup_{n \mathop \in \N} {U_n}^-}$
For each $n \in \N$, let:
:${U_n}' = U_n \setminus \paren {\ds \bigcup_{p \mathop \le n} {V_p}^-}$
For each $n \in \N$, let:
:${V_n}' = V_n \setminus \paren {\ds \bigcup_{p \mathop \le n} {U_p}^-}$
=== Lemma 1 ===
Let:
:$U = \ds \bigcup_{n \mathop \in \N} {U_n}'$
and
:$V = \ds \bigcup_{n \mathop \in \N} {V_n}'$
From Lemma 1:
:$U \cap V = \O$
=== Lemma 2 ===
We have:
Similarly, from $(2)$ above:
It has been shown:
:there exists $U, V \in \tau$ such that $A \subseteq U, B \subseteq V$ and $U \cap V = \O$.
Hence, by definition, $A$ and $B$ can be separated in $T$.
Category:Separated Sets
Category:Countable Open Covers Condition for Separated Sets
Countable Open Covers Condition for Separated Sets/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {U_n}_{n \mathop \in \N}$ be a family of subsets of $S$.
Let $\family {V_n}_{n \mathop \in \N}$ be a family of subsets of $S$.
For each $n \in \N$, let:
:${U_n}' = U_n \setminus \paren {\ds \bigcup_{p \mathop \le n} {V_p}^-}$
For each $n \in \N$, let:
:${V_n}' = V_n \setminus \paren {\ds \bigcup_{p \mathop \le n} {U_p}^-}$
Then:
:$\forall n, m \in \N : {U_n}' \cap {V_m}' = \O$ 
== Proof ==
Let $n, m \in \N$.
, let $m \le n$.
We have:
As $m \le n$, then:
:${U_m}^- \in \set { {U_p}^- : p \le n}$
We have:
Also:
Finally:
From Empty Set is Subset of All Sets:
:$\O \subseteq {U_m}' \cap {V_n}'$
By definition of set equality:
:${U_m}' \cap {V_n}' = \O$
Similarly:
:${U_n}' \cap {V_m}' = \O$
The result follows.
Category:Countable Open Covers Condition for Separated Sets
Countable Open Covers Condition for Separated Sets/Lemma 2

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {U_n}_{n \mathop \in \N}$ be a countable family of open sets.
Let $\family {V_n}_{n \mathop \in \N}$ be a countable family of open sets.
For each $n \in \N$, let:
:${U_n}' = U_n \setminus \paren {\ds \bigcup_{p \mathop \le n} {V_p}^-}$
For each $n \in \N$, let:
:${V_n}' = V_n \setminus \paren {\ds \bigcup_{p \mathop \le n} {U_p}^-}$
Let:
:$U = \ds \bigcup_{n \mathop \in \N} {U_n}'$
and
:$V = \ds \bigcup_{n \mathop \in \N} {V_n}'$
Then:
:$U$ and $V$ are open in $T$.
== Proof ==
By , it is sufficient to show that:
:$\forall n \in \N : {U_n}', {V_n}' \in \tau$
Let $n \in \N$.
We have:
From Topological Closure is Closed:
:${V_p}^-$ is a closed set.
By definition of closed set:
:$\relcomp S {V_p^-}$ is an open set.
Hence:
:${U_n}'$ is the finite intersection of open sets.
By :
:${U_n}'$ is an open set.
Similarly:
:${V_n}'$ is an open set.
The result follows from .
Category:Countable Open Covers Condition for Separated Sets
Countable Open Ordinal Space is Metrizable

== 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 metrizable space.
== Proof ==
From Countable Open Ordinal Space is Second-Countable, $\hointr 0 \Gamma$ has a basis which is $\sigma$-locally finite.
From Ordinal Space is Completely Normal, $\hointr 0 \Gamma$ is a completely normal space.
From Sequence of Implications of Separation Axioms it follows that $\hointr 0 \Gamma$ is a regular space.
The result follows from Metrizable iff Regular and has Sigma-Locally Finite Basis.
== Sources ==
* : Part $\text {II}$: Counterexamples: $40$. Open Ordinal Space $[0, \Gamma) \ (\Gamma < \Omega)$: $5$
Category:Ordinal Spaces
Category:Metrizable Topologies