Interior of Subset of Double Pointed Topological Space

Theorem
Let $\struct {S, \tau_S}$ be a topological space.

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

Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.

Let $X \subseteq S \times D$ be a subset of $S \times D$.

Define $A \subseteq S$ by:


 * $A := \set {s \in S: \paren {\forall d \in D: \tuple {s, d} \in X} }$

Then the interior of $X$ in $\tau$ is:


 * $X^\circ = A^\circ \times D$

Proof
By Open Sets of Double Pointed Topology, $X^\circ$ must be of the form:


 * $X^\circ = U \times D$

with $U$ open in $\tau_S$.

By Set Interior is Largest Open Set, we have for any open set $U' \times D$ of $\tau$ that:


 * $U' \times D \subseteq X \iff U' \times D \subseteq X^\circ = U \times D$

By Cartesian Product of Subsets, we have for open sets $V, V'$:


 * $V \subseteq V' \iff V \times D \subseteq V' \times D$

since $D$ is non-empty.

Combining these two equivalences, we have:


 * $U' \times D \subseteq X \iff U' \subseteq U$

The condition that $U' \times D \subseteq X$ can be expressed by:


 * $\forall s \in U': \forall d \in D: \tuple {s, d} \in X$

By definition of the set $A$, this is equivalent to:


 * $U' \subseteq A$

Combining this with the above, it follows that:


 * $U' \subseteq A \iff U' \subseteq U$

Recall that $U$ is open in $\tau_S$.

By Set Interior is Largest Open Set, we conclude:


 * $U = A^\circ$

and so:


 * $X^\circ = A^\circ \times D$

as desired.

Also see

 * Closure of Subset of Double Pointed Topological Space