Open Sets of Double Pointed Topology

Theorem
Let $\left({S, \vartheta}\right)$ be a topological space.

Let $D$ be a doubleton endowed with the indiscrete topology.

Let $\left({S \times D, \tau}\right)$ be the double pointed topology on $S$.

Then $X \subseteq S \times D$ is open in $\tau$ iff for some $U \in \vartheta$:


 * $X = U \times D$

Proof
By definition, $\tau$ is the product topology on $X \times D$.

That is, $\tau$ has as a basis sets of the form:


 * $U \times V$

with $U \in \vartheta$ and $V$ open in $D$.

Since $D$ is endowed with the indiscrete topology, either $V = \varnothing$ or $V = D$.

In the former case, by Cartesian Product Empty iff Factors are Empty, $U \times V = \varnothing$ for all $U \in \vartheta$.

In particular, then, it follows that $\tau$ has the basis:


 * $\mathcal B = \left\{{U \times D: U \in \vartheta}\right\}$

since $U = \varnothing \in \vartheta$, and so the basis elements with $V = \varnothing$ are also accounted for.

It is therefore clear that all $U \times D$ with $U \in \vartheta$ are open in $\tau$.

That opens of these form constitute all of $\tau$ amounts to showing that:


 * $\tau = \mathcal B$

i.e., that $\mathcal B$ is already a topology on $S \times D$.