Interior in Double Pointed Topology

From ProofWiki
Jump to navigation Jump to search

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 $S$.

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

Define $A \subseteq S$ by:

$A := \left\{{s \in S }\,\middle\vert\,{ \forall d \in D: \left({s, d}\right) \in X}\right\}$


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 $\vartheta$.


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:

$\forall s \in U': \forall d \in D: \left({s, d}\right) \in X$

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 $\vartheta$.

By Set Interior is Largest Open Set, we conclude:

$U = A^\circ$

and so:

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

as desired.

$\blacksquare$


Also see