Interior Equals Closure of Subset of Discrete Space

Theorem
Let $S$ be a set.

Let $\vartheta$ be the discrete topology on $S$.

Let $A \subseteq S$.

Then:
 * $A = A^\circ = A^-$

where:
 * $A^\circ$ is the interior of $A$
 * $A^-$ is the closure of $A$.

Proof
Let $T = \left({S, \vartheta}\right)$ be the discrete space on $S$.

Then by definition $\vartheta = \mathcal P \left({S}\right)$, that is, is the power set of $S$.

Let $A \subseteq S$.

Then from All Sets in Discrete Topology are Clopen it follows that $A$ is both open and closed in $T$.

From Closed Set Equals its Closure we have that $A = A^-$.

From Set Interior is Largest Open Set, we have that $A^\circ = A$.