Separation Axioms on Double Pointed Topology/T5 Axiom

From ProofWiki
Jump to navigation Jump to search

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 $T_5$ 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:

\(\displaystyle \paren {A \times D}^- \cap \paren {B \times D}\) \(=\) \(\displaystyle \paren {A^- \times D} \cap \paren {B \times D}\) Closure of Subset of Double Pointed Topological Space
\(\displaystyle \) \(=\) \(\displaystyle \paren {A^- \cap B} \times D\) Cartesian Product Distributes over Intersection
\(\displaystyle \) \(=\) \(\displaystyle \O \times D\) $A$ and $B$ are separated
\(\displaystyle \) \(=\) \(\displaystyle \varnothing\) Cartesian Product is Empty iff Factor is Empty

Mutatis mutandis, also:

$\paren {B \times D}^- \cap \paren {A \times D} = \O$

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:

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

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:

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

From Cartesian Product is Empty iff Factor is Empty, $U' \cap V' = \O$ 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 of Subset of Double Pointed Topological Space, we know that:

$X^- = {\pr_1 \sqbrk X}^- \times D$
$Y^- = {\pr_1 \sqbrk Y}^- \times D$

where $\pr_1$ is the first projection.


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

$s \in {\pr_1 \sqbrk X}^- \cap \pr_1 \sqbrk Y$

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

$\tuple {s, d} \in Y$

However, by the expression for $X^-$ above, also $\tuple {s, d} \in X^-$.

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

Thus:

${\pr_1 \sqbrk X}^- \cap \pr_1 \sqbrk Y = \O$

and analogously, we derive:

$\pr_1 \sqbrk X \cap {\pr_1 \sqbrk Y}^- = \O$

that is, $\pr_1 \sqbrk X$ and $\pr_1 \sqbrk Y$ are separated.

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

$U \cap V = \varnothing$
${\pr_1 \sqbrk X}^- \subseteq U$
${\pr_1 \sqbrk Y}^- \subseteq V$

By Cartesian Product of Subsets, we have:

$X^- = {\pr_1 \sqbrk X}^- \times D \subseteq U \times D$
$Y^- = {\pr_1 \sqbrk Y}^- \times D \subseteq V \times D$

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:

\(\displaystyle \paren {U \times D} \cap \paren {V \times D}\) \(=\) \(\displaystyle \paren {U \cap V} \times D\) Cartesian Product Distributes over Intersection
\(\displaystyle \) \(=\) \(\displaystyle \O \times D\) by choice of $U$ and $V$
\(\displaystyle \) \(=\) \(\displaystyle \O\) Cartesian Product is Empty iff Factor is Empty

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

$\blacksquare$


Sources