Open Sets of Double Pointed Topology

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Theorem

Let $\left({S, \tau}\right)$ 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 $\tau$ if and only if for some $U \in \tau$:

$X = U \times D$


Corollary

A subset $X \subseteq S \times D$ is closed in $\tau$ if and only if for some closed set $C$ of $\tau$:

$X = C \times D$


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 $U \in \tau$ 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 is Empty iff Factor is Empty, $U \times V = \varnothing$ for all $U \in \tau$.

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

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

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


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

That open sets of this form constitute all of $\tau$ amounts to showing that:

$\tau = \mathcal B$

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


$\blacksquare$