# Separation Axioms on Double Pointed Topology/T5 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 $T_5$ 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

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous): Notes: Part $1$: Basic Definitions: Section $2$. Separation Axioms: $1$