nyuuzyou/wikis huggingface dataset:
Separable C*-Algebra has Sequential Approximate Identity

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a separable $\text C^\ast$ algebra.
Then there exists a sequence $\family {e_n}_{n \mathop \in \N}$ forming an approximate identity of $A$.
== Proof ==
Let $\set {a_n : n \in \N}$ be a countable everywhere dense subset of $A$.
From Existence of Approximate Identity of C*-Algebra arising from Dense Ideal, there exists a directed set $\struct {\Lambda, \preceq}$ and net $\family {e_\lambda}_{\lambda \mathop \in \Lambda}$ forming an approximate identity of $A$.
The issue is that, at this point, $\Lambda$ may not be countable.
Hence we desire to extract an $\tuple {\le, \preceq}$-increasing sequence $\sequence {\lambda_n}_{n \mathop \in \N}$ in $\Lambda$ such that $\family {e_{\lambda_n} }_{n \mathop \in \N}$ forms an approximate identity of $A$.
Take $n \in \N$. 
For $1 \le j \le n$, we have that $\family {a_j e_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $a_j$.
Hence for each $j$ we can find $\mu_{j, n} \in \Lambda$ such that:
:$\norm {a_j - a_j e_\lambda} < 2^{-n}$ for each $\lambda \succeq \mu_{j, n}$.
Hence $\Lambda$ is directed, there exists $\lambda_n \in \Lambda$ such that:
:$\lambda_n \succeq \mu_{j, n}$ for each $1 \le j \le n$.
Then:
:$\norm {a_j - a_j e_\lambda} < 2^{-n}$ for each $\lambda \succeq \lambda_n$.
By construction, for $i \le j$ we have $\lambda_i \preceq \lambda_j$ and hence $e_{\lambda_i} \le_A e_{\lambda_j}$ where $\le_A$ is the canonical preordering.
Hence from Approximate Identity for Everywhere Dense Subset is Approximate Identity of C*-Algebra, it will suffice to show that:
:$\sequence {a_k e_{\lambda_n} }_{n \mathop \in \N}$ converges to $a_k$ for each $k \in \N$.
Now suppose that $\epsilon > 0$ and $k \in \N$. 
Take $N > k$ such that $2^{-N} < \epsilon$.
Then for $n \ge N$ we have $\lambda_n \succeq \lambda_N$ and hence:
:$\norm {a_k - a_k e_{\lambda_n} } < 2^{-N} < \epsilon$ 
Hence the sequence $\sequence {a_k e_{\lambda_n} }_{n \in \N}$ converges to $a_k$ for each $k \in \N$.
Hence by Approximate Identity for Everywhere Dense Subset is Approximate Identity of C*-Algebra, $\sequence {e_{\lambda_n} }_{n \in \N}$ is an approximate identity of $A$.
== Sources ==
* : Corollary $4.4$
Category:Approximate Identities of C*-Algebras
Separable Degree is At Most Equal To Degree

== Theorem ==
Let $E / F$ be a finite field extension.
Then $\index E F_s$ is finite, and:
:$\index E F_s \le \index E F$
where:
:$\index E F$ denotes the degree of $E / F$
:$\index E F_s$ denotes the separable degree of $E / F$.
== Proof ==
== Sources ==
* : Chapter $\text V$: $\S 4$: Separable Extensions: Theorem $4.1$
Category:Separable Field Extensions
Separable Degree of Field Extensions is Multiplicative

== Theorem ==
Let $E / F / k$ be a tower of fields. Then
:$\index E k_s = \index E F_s \index F k_s$
where $\index E F_s$ denotes the separable degree of $E / F$.
== Proof ==
If $E / k$ is infinite, both sides are infinite.
Now assume $E / k$ is finite.
Let $L$ be the algebraic closure of $k$.
Let $\set{\sigma_i}$ be the family of distinct embedding of $F$ to $L$ fixing $k$,
By definition 2 of separable degree, the set $\set{\sigma_i}$ has $\index F k_s$ elements.
For each $i$, let $\set{\tau_{i j}}$ be the family of distinct extensions of $\sigma_i$ to $E$.
We can view $L$ as an algebraic closure of $F$. By definition 2 of separable degree, each $\sigma_i$ has precisely
:$\index E F_s$
extensions to embeddings of $E$ in $L$.
The set of embeddings $\set{\tau_{i j}}$ contains precisely
:$\index E F_s \index F k_s$
elements.
Any embedding of $E$ into $L$ fixing $k$ must be one of the $\tau_{i j}$, and thus we have
:$\index E k_s = \index E F_s \index F k_s$
== Sources ==
* : Chapter $\text V$: $\S4$: Separable Extensions: Theorem $4.1$
Category:Separable Field Extensions
Separable Discrete Space is Countable/Proof 1

== Theorem ==
== Proof ==
Let $T$ be separable.
By Space is Separable iff Density not greater than Aleph Zero:
:$\map d T \le \aleph_0$
where:
:$\map d T$ denotes the density of $T$
:$\aleph$ denotes the aleph mapping.
By definition of density:
:$\exists A \subseteq S: A$ is everywhere dense and $\map d T = \card A$
where $\card A$ denotes the cardinality of $A$.
By definition of everywhere dense set:
:$A^- = S$
where $A^-$ denotes the closure of $A$.
By Set in Discrete Topology is Clopen:
:$A$ is closed
Then by Set is Closed iff Equals Topological Closure:
:$A^- = A$
Thus by Countable iff Cardinality not greater than Aleph Zero:
:$S$ is countable.
== Sources ==
* 
Category:Separable Discrete Space is Countable
Separable Elements Form Field

== Theorem ==
Let $E / F$ be an algebraic field extension.
Let $K$ be the relative separable closure of $F$ in $E$.
Then $K$ is an intermediate field of $E / F$.
== Proof ==
We need to show that $K$ is a field.
By Transitivity of Separable Field Extensions, an algebraic extension generated by a family of separable elements is separable.
== Source ==
* : Chapter $1$: $\S4$: Separable and Inseparable Extensions: Proposition $4.20$
* : Chapter $\text V$: $\S4$: Separable Extensions: Theorem $4.5$
Category:Separable Field Extensions
Separable Metacompact Space is Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be a separable topological space which is also metacompact.
Then $T$ is a Lindelöf space.
== Proof 1 ==
== Proof 2 ==
Category:Separable Spaces
Category:Metacompact Spaces
Category:Lindelöf Spaces
Category:Separable Metacompact Space is Lindelöf
Separable Metacompact Space is Lindelöf/Proof 1

== Theorem ==
== Proof ==
$T$ is separable there exists a countable subset of $S$ which is everywhere dense.
$T$ is metacompact every open cover of $S$ has an open refinement which is point finite.
$T$ is a Lindelöf space if every open cover of $S$ has a countable subcover.
Having established the definitions, we proceed.
Let $T = \left({S, \tau}\right)$ be separable and metacompact.
 there exists an open cover $\UU$ of $S$ which has no countable subcover.
As $T$ is metacompact, $\UU$ has an open refinement $\VV$ which is point finite.
By nature of $\UU$, which has no countable subcover, $\VV$ is uncountable.
By hypothesis, $T$ is separable.
Let $\SS$ be a countable subset of $S$ which is everywhere dense.
Then each $V \in \VV$ contains some $s \in \SS$.
So some $s \in \SS$ is contained in an uncountable number of elements of $\VV$.
Thus, by definition, $\VV$ is not point finite.
Thus no uncountable open refinement $\VV$ of $\UU$ exists which is point finite.
It follows that $\VV$ must be countable.
Thus $\UU$ has a countable subcover.
That is, by definition, $T$ is a Lindelöf space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Separable Metacompact Space is Lindelöf
Separable Metacompact Space is Lindelöf/Proof 2

== Theorem ==
== Proof ==
$T$ is metacompact every open cover of $S$ has an open refinement which is point finite.
$T$ is a Lindelöf space if every open cover of $S$ has a countable subcover.
Having established the definitions, we proceed.
Let $\UU$ be an open cover of $S$.
Let $\VV$ be a point finite open refinement of $\UU$.
By Point Finite Set of Open Sets in Separable Space is Countable, $\VV$ is countable.
Define a mapping $H$ on $\VV$ thus:
:$\forall V \in \VV: \map H V = \set {U \in \UU: V \subseteq U}$
By Image of Countable Set under Mapping is Countable, the image of $H$ is countable.
Call this image $I$.
Since $\VV$ is a refinement of $\UU$, $\O \notin I$.
By the Axiom of Countable Choice, $I$ has a choice function $c$.
Then $G = c \circ H: \VV \to \UU$ is a mapping such that:
:$\forall V \in \VV: V \subseteq \map G V$
Then $\QQ = \map G \VV$ is countable by Image of Countable Set under Mapping is Countable.
Each element of $\QQ$ is an element of $\UU$ by the definition of $G$.
Let $x \in S$.
Then since $\VV$ is a cover for $S$:
: $\exists V \in \VV: x \in V$
Then $x \in V \subseteq \map G V \in \QQ$.
Thus $\QQ$ is a countable subcover of $\UU$.
Thus each open cover of $S$ has a countable subcover, so $T$ is a Lindelöf space.
Category:Separable Metacompact Space is Lindelöf
Separable Metric Space is Homeomorphic to Subspace of Fréchet Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space whose induced topology is separable.
Then $M$ is homeomorphic to a subspace of the Fréchet space $\struct {\R^\omega, d}$ on the countable-dimensional real Cartesian space $\R^\omega$.
== Proof ==
Let $f: M \to \R^\omega$ be the mapping defined as:
:$\forall x \in M: \map f x = \sequence {\map d {x, x_i} }$
where $\set {x_i}$ is a countable dense subset of $A$.
It remains to be shown that $f$ is a homeomorphism.
== Sources ==
* : Part $\text {II}$: Counterexamples: $37$. Fréchet Space: $8$
Category:Fréchet Product Metric
Category:Separable Spaces
Separable Metric Space is Second-Countable

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $M$ be separable.
Then $M$ is second-countable.
== Proof ==
By the definition of separability, we can choose a subset $S \subseteq X$ that is countable and everywhere dense.
Define:
:$\BB = \set {\map {B_{1/n} } x: x \in S, \, n \in \N_{>0} }$
where $\map {B_\epsilon } x$ denotes the open $\epsilon$-ball of $x$ in $M$.
We have that Cartesian Product of Countable Sets is Countable.
Hence, by Image of Countable Set under Mapping is Countable, it follows that $\BB$ is countable.
Let $\tau$ denote the topology on $X$ induced by the metric $d$.
It suffices to show that $\BB$ is an analytic basis for $\tau$.
From Open Ball of Metric Space is Open Set, we have that $\BB \subseteq \tau$.
We use Equivalence of Definitions of Analytic Basis.
Let $y \in U \in \tau$.
By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} y \subseteq U$.
By the Axiom of Archimedes, there exists a natural number $n > \dfrac 2 \epsilon$.
That is:
:$\dfrac 2 n < \epsilon$
and so:
:$\map {B_{2/n} } y \subseteq \map {B_\epsilon} y$.
From Subset Relation is Transitive, we have $\map {B_{2/n} } y \subseteq U$.
By the definition of everywhere denseness, and by Equivalence of Definitions of Adherent Point, there exists an $x \in S \cap \map {B_{1/n} } y$.
By , it follows that $y \in \map {B_{1/n} } x$.
For all $z \in \map {B_{1/n} } x$, we have:
That is:
:$\map {B_{1/n} } x \subseteq \map {B_{2/n} } y$
From Subset Relation is Transitive, we have:
:$y \in \map {B_{1/n} } x \subseteq U$
Hence the result.
== Sources ==
* : Exercise $2.23$
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 37$
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Second-Countable Spaces
Category:Separable Spaces
Separable Normed Vector Space Isometrically Isomorphic to Linear Subspace of Space of Bounded Sequence

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a separable normed vector space over $\Bbb F$.
Let $\struct {\map {\ell^\infty} {\Bbb F}, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences.
Then there exists a linear subspace $Y$ of $\map {\ell^\infty} {\Bbb F}$ such that: 
:$X$ is isometrically isomorphic to $Y$.
== Proof ==
Let $\mathcal S = \set {x_n : n \in \N}$ be a countable everywhere dense subset of $X$. 
By Existence of Support Functional, for each $n \in \N$ there exists $f_n \in X^\ast$ such that $\norm {f_n}_{X^\ast} = 1$ and $\map {f_n} {x_n} = \norm {x_n}$.
Then, for each $x \in X$, we have: 
:$\cmod {\map {f_n} x} \le \norm x$ 
from Supremum Operator Norm as Universal Upper Bound.
So, we have: 
:$\ds \sup_{n \in \N} \cmod {\map {f_n} x} \le \norm x$ 
So we can define $T : X \to \map {\ell^\infty} {\Bbb F}$ by: 
:$T x = \sequence {\map {f_n} x}_{n \mathop \in \N} = \tuple {\map {f_1} x, \map {f_2} x, \ldots}$
From Image of Vector Subspace under Linear Transformation is Vector Subspace, $T \sqbrk U$ is a vector subspace of $\map {\ell^\infty} {\Bbb F}$.
It remains to show that $T$ is a linear isometry. 
Let $\alpha, \beta \in \Bbb F$ and $x, y \in X$. 
We have:
so $T$ is a linear transformation.
Also, for $x \in X$:
So $T$ is a bounded linear transformation.
So by Continuity of Linear Transformations, $T$ is continuous. 
Let $x_k \in \mathcal S$.
Note that while: 
:$\ds \sup_{n \mathop \in \N} \cmod {\map {f_n} {x_k} } \le \norm {x_k}$
we have: 
:$\map {f_k} {x_k} = \norm {x_k}$
So we actually have: 
:$\ds \norm {T x_k}_\infty = \sup_{n \mathop \in \N} \cmod {\map {f_n} {x_k} } = \norm {x_k}$
for each $k \in \N$.
Now take general $x \in X$.
Then there exists a sequence $\sequence {n_k}_{k \mathop \in \N}$ such that: 
:$x_{n_k} \to x$ 
as $k \to \infty$.
Since $T$ is continuous , we have: 
:$T x_{n_k} \to T x$
as $k \to \infty$, by Sequential Continuity is Equivalent to Continuity in Metric Space.
So taking $k \to \infty$ in: 
:$\norm {T x_{n_k} }_\infty = \norm {x_{n_k} }$
we have: 
:$\norm {T x}_\infty = \norm x$
for each $x \in X$.
So $T : X \to Y$ is a linear isometry.
Hence $X$ and $Y$ are isometrically isomorphic.
Category:Isometric Isomorphisms (Normed Vector Spaces)
Category:Separable Spaces
Separable Space need not be First-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is separable.
Then $T$ does not necessarily have to be first-countable.
== Proof ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
We have that a Finite Complement Topology is Separable.
But we also have that an Uncountable Finite Complement Space is not First-Countable.
Hence the result, by Proof by Counterexample.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Countability Properties
* : Part $\text {II}$: Counterexamples: $19$. Finite Complement Topology on an Uncountable Space: $4$
Category:First-Countable Spaces
Category:Separable Spaces
Separable Space satisfies Countable Chain Condition

== Theorem ==
Let $T = \struct {S, \tau}$ be a separable topological space.
Then $T$ satisfies the countable chain condition.
== Proof ==
In order to demonstrate that $T$ satisfies the countable chain condition, it is sufficient to demonstrate that every disjoint set of open sets of $T$ is countable.
Because $T$ is separable, there exists a subset $\set {y_n : n \in \N}$ of $S$ which is everywhere dense in $S$.
Now consider an indexed family $\family {U_j}_{j \mathop \in J}$ of non-empty open sets of $T$ such that:
:$\forall i, j \in J, i \ne j: U_i \cap U_j = \O$
Using Equivalence of Definitions of Everywhere Dense this implies that for every $j \in J$ there has to exist $n_j \in \N$ such that $y_{n_j} \in U_j$.
This gives rise to a well-defined mapping $f: J \to \N$ via $\map f j := n_j$.
In particular $f$ is injective:
 there were to exist $i, j \in J$, $i \ne j$ such that $n_i = n_j$.
Then:
:$y_{n_i} \in U_i \cap U_j$
But the latter is the empty set by assumption.
From this contradiction it follows that $J$ is countable by definition.
This concludes the proof. 
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Separable Spaces
Separable Topological Space remains Separable in Coarser Topology

== Theorem ==
Let $\struct {X, \tau_2}$ be a separable topological space. 
Let $\tau_1$ be a topology on $X$ such that:
:$\tau_1 \subseteq \tau_2$
That is, such that $\tau_1$ is coarser than $\tau_2$. 
Then $\struct {X, \tau_1}$ is separable.
== Proof ==
Let $\cl_1$ and $\cl_2$ denote the topological closure in $\struct {X, \tau_1}$ and $\struct {X, \tau_2}$ respectively.
Let $D$ be a countable everywhere dense subset of $\struct {X, \tau_2}$.
From Topological Closure in Coarser Topology is Larger, we have: 
:$\map {\cl_2} D \subseteq \map {\cl_1} D$
Since $D$ is everywhere dense in $\struct {X, \tau_2}$, we have $\map {\cl_2} D = X$.
So:
:$\map {\cl_1} D = X$
So $D$ is everywhere dense in $\struct {X, \tau_1}$.
Category:Separable Spaces
Separated Morphism is Quasi-Separated

== Theorem ==
Let $f$ be a separated morphism of schemes.
Then $f$ is quasi-separated.
== Proof ==
Let $f$ be a separated morphism of schemes.
By definition, the diagonal morphism $\Delta_f$ is a closed immersion.
By Closed Immersion is Quasi-Compact $\Delta_f$ is quasi-compact.
Thus, by definition, $f$ is quasi-separated.
Category:Algebraic Geometry
Category:Schemes
Separated Sets are Clopen in Union

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $A$ and $B$ be separated sets in $T$.
Let $H = A \cup B$ be given the subspace topology.
Then $A$ and $B$ are each both open and closed in $H$.
== Proof ==
By hypothesis, $A$ and $B$ are separated:
:$A \cap B^- = A^- \cap B = \O$
Then:
Since the intersection of a closed set with a subspace is closed in the subspace, $B$ is closed in $H$.
Since $A = H \setminus B$ and $B$ is closed in $H$, $A$ is open in $H$.
By the same argument with the roles of $A$ and $B$ reversed, $A$ is closed in $H$ and $B$ is open in $H$.
Hence the result.
Category:Separated Sets
Separated Sets are Disjoint

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B \subseteq S$ such that $A$ and $B$ are separated in $T$.
Then $A$ and $B$ are disjoint:
:$A \cap B = \O$
== Proof ==
Let $A$ and $B$ be separated in $T$.
Then:
== Sources ==
* 
Category:Separated Sets
Separated Subsets of Linearly Ordered Space under Order Topology

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
:$A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$
:$B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Then $A^*$ and $B^*$ are themselves separated sets of $T$.
== Proof ==
From the lemma:
:$A \subseteq A^*$
:$B \subseteq B^*$
:$A^* \cap B^* = \O$
Let $p \notin A^* \cup A^-$.
Thus $p \notin A^*$ and $p \notin A^-$.
Then there exists an open interval $\openint s t$ which is disjoint from $A$ such that $p \in \openint s t$.
Now $\openint s t$ can only intersect $A^*$ only if it intersects some $\closedint a b \subseteq A^*$ where $a, b \in A$.
But we have:
:$\openint s t \cap A = \O$
and as $a, b \in A$ it follows that:
:$\openint s t \subseteq \openint a b$
That means $p \in A^*$.
But we have $p \notin A^*$.
Therefore:
:$\openint s t \cap A^* = \O$
Thus:
:$p \notin \paren {A^*}^-$
Hence:
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $4$
Category:Linearly Ordered Spaces
Separated Subsets of Linearly Ordered Space under Order Topology/Lemma

== Lemma ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
:$A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$
:$B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Then:
:$(1): \quad A \subseteq A^*$
:$(2): \quad B \subseteq B^*$
:$(3): \quad A^* \cap B^* = \O$
== Proof ==
Let $a \in A$.
Then:
Similarly, $B \subseteq B^-$.
 $A^* \cap B^* \ne \O$.
Then:
:$\exists p: p \in A^* \cap B^*$
Hence:
:$\exists a, b \in A, c, d \in B: p \in \closedint a b \cap \closedint c d$
But because $A$ and $B$ are separated sets:
:$c, d \notin A$
and:
:$a, b \notin B$
and so:
:$\closedint a b \cap \closedint c d = \O$
Thus $p \notin \closedint a b \cap \closedint c d$
It follows by Proof by Contradiction that $A^* \cap B^* = \O$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $3$
Category:Linearly Ordered Spaces
Separation Axioms Preserved under Homeomorphism

== Theorem ==
The separation axioms are preserved under homeomorphism.
Let $T_A = \left({S_A, \tau_A}\right), T_B = \left({S_B, \tau_B}\right)$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
=== $T_0$ (Kolmogorov) Space is Preserved under Homeomorphism ===
=== $T_1$ (Fréchet) Space is Preserved under Homeomorphism ===
=== Hausdorff Condition is Preserved under Homeomorphism ===
=== $T_{2 \frac 1 2}$ (Completely Hausdorff) Space is Preserved under Homeomorphism ===
=== $T_3$ Space is Preserved under Homeomorphism ===
=== Regular Space is Preserved under Homeomorphism ===
=== $T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism ===
=== Tychonoff (Completely Regular) Space is Preserved under Homeomorphism ===
=== $T_4$ Space is Preserved under Homeomorphism ===
=== Normal Space is Preserved under Homeomorphism ===
=== $T_5$ Space is Preserved under Homeomorphism ===
=== Completely Normal Space is Preserved under Homeomorphism ===
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Separation Axioms
Separation Axioms on Double Pointed Topology

== Theorem ==
Let $T = \struct {S, \tau_S}$ be a topological space.
Let $D = \struct {\set {a, b}, \tau_D}$ be the indiscrete topology on two points.
Let $T \times D$ be the double pointed topology on $T$.
Then:
:$T \times D$ is not a $T_0$ (Kolmogorov) space, a $T_1$ (Fréchet) space, a $T_2$ (Hausdorff) space or a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
:$T \times D$ is a $T_3$ space, a $T_{3 \frac 1 2}$ space, a $T_4$ space or a $T_5$ space $T$ is.
== Proof ==
=== $T_0$, $T_1$, $T_2$ and $T_{2 \frac 1 2}$ Axioms ===
From:
:Double Pointed Topology is not $T_0$
:$T_1$ Space is $T_0$ Space
:$T_2$ Space is $T_1$ Space
:Completely Hausdorff Space is $T_2$ Space
we have that $T \times D$ is not $T_0$, $T_1$ or $T_2$ or $T_{2 \frac 1 2}$.
From:
:Separation Axioms on Double Pointed Topology: $T_3$ Axiom
$T \times D$ is a $T_3$ space $T$ is.
From:
:Separation Axioms on Double Pointed Topology: $T_{3 \frac 1 2}$ Axiom
$T \times D$ is a $T_{3 \frac 1 2}$ space $T$ is.
From:
:Separation Axioms on Double Pointed Topology: $T_4$ Axiom
$T \times D$ is a $T_4$ space $T$ is.
From:
:Separation Axioms on Double Pointed Topology: $T_5$ Axiom
$T \times D$ is a $T_5$ space $T$ is.
== Sources ==
* : Notes: Part $1$: Basic Definitions: Section $2$. Separation Axioms: $1$
Category:Double Pointed Topologies
Category:Separation Axioms
Category:Separation Axioms on Double Pointed Topology
Separation Axioms on Double Pointed Topology/T3.5 Axiom

== Theorem ==
Let $T_1 = \struct {S, \tau_S}$ be a topological space.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete topology on an arbitrary doubleton $A = \set {a, b}$.
Let $T = \struct {T_1 \times D, \tau}$ be the double pointed topological space on $T_1$.
Then $T \times D$ is a $T_{3 \frac 1 2}$ space $T$ is also a $T_{3 \frac 1 2}$ space.
== Proof ==
== Sources ==
* : Notes: Part $1$: Basic Definitions: Section $2$. Separation Axioms: $1$
Category:Separation Axioms on Double Pointed Topology
Category:T3 1/2 Spaces