[proofplan] Both directions follow from the definition $A^\circ = \bigcup \{ U \in \tau : U \subset A \}$. If $A$ is open, then $A$ itself appears in this union, forcing $A^\circ = A$. Conversely, if $A = A^\circ$, then $A$ equals a union of open sets, so $A$ is open. [/proofplan] [step:Show that $A$ open implies $A = A^\circ$] Suppose $A \in \tau$. By property (1) of the [Properties of the Interior Operator](/theorems/1013), $A^\circ \subset A$. For the reverse inclusion: since $A$ is open and $A \subset A$, the set $A$ belongs to the family $\{ U \in \tau : U \subset A \}$. Therefore \begin{align*} A \subset \bigcup \{ U \in \tau : U \subset A \} = A^\circ. \end{align*} Combining both inclusions gives $A = A^\circ$. [/step] [step:Show that $A = A^\circ$ implies $A$ is open] Suppose $A = A^\circ$. By definition, $A^\circ = \bigcup \{ U \in \tau : U \subset A \}$, which is a union of members of $\tau$. Since any union of open sets is open (by the axioms of a topology), $A^\circ \in \tau$. Hence $A = A^\circ \in \tau$, so $A$ is open. [/step]