Separation Axioms on Double Pointed Topology/T5 Axiom

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $D = \left({\left\{{a, b}\right\}, \vartheta}\right)$ be the indiscrete topology on two points.

Let $T \times D$ be the double pointed topology on $T$.


Then $T \times D$ is a $T_5$ space if and only if $T$ is also a $T_5$ space.


Proof

Necessary Condition

Suppose that $T \times D$ is a $T_5$ space.

Let $A, B \subseteq T$ be two separated sets.

First, we will show that $A \times D$ and $B \times D$ are also separated.


To this end, observe that:

\(\displaystyle \left({A \times D}\right)^- \cap \left({B \times D}\right)\) \(=\) \(\displaystyle \left({A^- \times D}\right) \cap \left({B \times D}\right)\) Closure in Double Pointed Topology
\(\displaystyle \) \(=\) \(\displaystyle \left({A^- \cap B}\right) \times D\) Cartesian Product Distributes over Intersection
\(\displaystyle \) \(=\) \(\displaystyle \varnothing \times D\) $A$ and $B$ are separated
\(\displaystyle \) \(=\) \(\displaystyle \varnothing\) Cartesian Product is Empty iff Factor is Empty

Mutatis mutandis, also:

$\left({B \times D}\right)^- \cap \left({A \times D}\right) = \varnothing$

Thus $A \times D$ and $B \times D$ are separated.


Since $T \times D$ is assumed to be a $T_5$ space, find open sets $U, V$ such that:

$U \cap V = \varnothing$
$A^- \times D \subseteq U$
$B^- \times D \subseteq V$

By Open Sets of Double Pointed Topology, there are open sets $U', V'$ of $\tau$ such that:

$U = U' \times D$
$V = V' \times D$


We have, furthermore:

\(\displaystyle \varnothing\) \(=\) \(\displaystyle U \cap V\)
\(\displaystyle \) \(=\) \(\displaystyle \left({U' \times D}\right) \cap \left({V' \times D}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({U' \cap V'}\right) \times D\) Cartesian Product Distributes over Intersection

From Cartesian Product is Empty iff Factor is Empty, $U' \cap V' = \varnothing$ since $D$ is non-empty.

By Cartesian Product of Subsets, we also infer:

$A^- \subseteq U'$
$B^- \subseteq V'$

It follows that $T$ is a $T_5$ space.

$\Box$


Sufficient Condition

Suppose that $T$ is a $T_5$ space.

Let $X, Y \subseteq T \times D$ be separated sets.

By Closure in Double Pointed Topology, we know that:

$X^- = {\operatorname{pr}_1 \left({X}\right)}^- \times D$
$Y^- = {\operatorname{pr}_1 \left({Y}\right)}^- \times D$

where $\operatorname{pr}_1$ is the first projection.


Suppose now that for some $s \in S$, one has:

$s \in {\operatorname{pr}_1 \left({X}\right)}^- \cap \operatorname{pr}_1 \left({Y}\right)$

Then it would be that for some $d \in D$:

$\left({s, d}\right) \in Y$

However, by the expression for $X^-$ above, also $\left({s, d}\right) \in X^-$.

This is a contradiction, for $X$ and $Y$ were assumed separated.

Thus:

${\operatorname{pr}_1 \left({X}\right)}^- \cap \operatorname{pr}_1 \left({Y}\right) = \varnothing$

and analogously, we derive:

$\operatorname{pr}_1 \left({X}\right) \cap {\operatorname{pr}_1 \left({Y}\right)}^- = \varnothing$

i.e., $\operatorname{pr}_1 \left({X}\right)$ and $\operatorname{pr}_1 \left({Y}\right)$ are separated.

Since $T$ is a $T_5$ space, we find open sets $U, V$ of $\tau$ such that:

$U \cap V = \varnothing$
${\operatorname{pr}_1 \left({X}\right)}^- \subseteq U$
${\operatorname{pr}_1 \left({Y}\right)}^- \subseteq V$

By Cartesian Product of Subsets, we have:

$X^- = {\operatorname{pr}_1 \left({X}\right)}^- \times D \subseteq U \times D$
$Y^- = {\operatorname{pr}_1 \left({Y}\right)}^- \times D \subseteq V \times D$

By Open Sets of Double Pointed Topology, $U \times D$ and $V \times D$ are also open sets in $S \times D$.

Finally, we compute:

\(\displaystyle \left({U \times D}\right) \cap \left({V \times D}\right)\) \(=\) \(\displaystyle \left({U \cap V}\right) \times D\) Cartesian Product Distributes over Intersection
\(\displaystyle \) \(=\) \(\displaystyle \varnothing \times D\) By choice of $U$ and $V$
\(\displaystyle \) \(=\) \(\displaystyle \varnothing\) Cartesian Product is Empty iff Factor is Empty

That is, $T \times D$ is a $T_5$ space.

$\blacksquare$


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