İspat: $x\in\overline{A}$ olsun.
$$x\in\overline{A}\Rightarrow (x\in A^{\circ}\vee x\notin A^{\circ})$$
I. Durum: $x\in A^{\circ}$ olsun. Bu durumda $\overline{A}\subseteq A^{\circ}\cup A^s$ koşulu sağlanır.
II. Durum: $x\notin A^{\circ}$ olsun.
$\left.\begin{array}{rr} x\in\overline{A}\Rightarrow (\forall U\in\mathcal{U}(x))(U\cap A\neq \emptyset)\\ x\notin A^{\circ}\Rightarrow (\forall U\in\mathcal{U}(x))(U\nsubseteq A)\Rightarrow (\forall U\in\mathcal{U}(x))(U\cap (\setminus A)=\emptyset)\end{array}\right\}\Rightarrow$
$\Rightarrow (\forall U\in\mathcal{U}(x))(U\cap A\neq \emptyset)(U\cap (\setminus A)=\emptyset)$
$\Rightarrow x\in A^s $
yani
$(x\in \overline{A})(x\notin A^{\circ})\Rightarrow x\in A^s$
yani
$$\overline{A}\subseteq A^{\circ}\cup A^s\ldots (1)$$
$x\in A^s\Rightarrow (\forall U\in\mathcal{U}(x))(U\cap A\neq \emptyset)(U\cap (\setminus A)=\emptyset)$
$\hspace{3.3 cm}\Rightarrow (\forall U\in\mathcal{U}(x))(U\cap A\neq \emptyset)$
$\hspace{3.3 cm}\Rightarrow x\in \overline {A}$
yani
$$A^s\subseteq \overline{A}\ldots (2)$$
öte yandan
$$A^{\circ} \subseteq A\subseteq \overline{A}\ldots (3)$$ Dolayısıyla
$$(2),(3)\Rightarrow A^{\circ} \cup A^s\subseteq \overline{A}\ldots (4)$$
$(1)$ ve $(4)$ bize isteneni verir.