[proofplan] We use the definition of the interior as the union of all open subsets of $X$ contained in $A$. Since a topology is closed under arbitrary unions, this union is open. Since every set being unioned is contained in $A$, the union is contained in $A$. Finally, any open subset $U \subset A$ is one of the sets appearing in the defining union, so it is contained in $A^\circ$. [/proofplan] [step:Express the interior as a union over all open subsets contained in $A$] Define the indexing collection \begin{align*} \mathcal{U}_A := \{V \subset X : V \in \tau \text{ and } V \subset A\}. \end{align*} By the definition of the [interior](/page/Interior), \begin{align*} A^\circ = \bigcup_{V \in \mathcal{U}_A} V. \end{align*} [/step] [step:Use closure of the topology under arbitrary unions to show that $A^\circ$ is open] For every $V \in \mathcal{U}_A$, the definition of $\mathcal{U}_A$ gives $V \in \tau$. Since $\tau$ is a [topology](/page/Topology), it is closed under arbitrary unions. Therefore \begin{align*} A^\circ = \bigcup_{V \in \mathcal{U}_A} V \in \tau. \end{align*} [guided] We must prove that $A^\circ$ is an [open set](/page/Open%20Set), meaning that $A^\circ \in \tau$. The definition of the interior writes $A^\circ$ as \begin{align*} A^\circ = \bigcup_{V \in \mathcal{U}_A} V, \end{align*} where \begin{align*} \mathcal{U}_A := \{V \subset X : V \in \tau \text{ and } V \subset A\}. \end{align*} Thus every set $V$ appearing in the union is open in $X$, because $V \in \tau$ by the definition of $\mathcal{U}_A$. The defining property of a [topology](/page/Topology) is that arbitrary unions of members of $\tau$ again belong to $\tau$. Applying this closure property to the family $(V)_{V \in \mathcal{U}_A}$ gives \begin{align*} \bigcup_{V \in \mathcal{U}_A} V \in \tau. \end{align*} Since this union is exactly $A^\circ$, we conclude that \begin{align*} A^\circ \in \tau. \end{align*} So $A^\circ$ is open. [/guided] [/step] [step:Use containment of each member of the union to prove $A^\circ \subset A$] Let $x \in A^\circ$. Since \begin{align*} A^\circ = \bigcup_{V \in \mathcal{U}_A} V, \end{align*} there exists $V_x \in \mathcal{U}_A$ such that $x \in V_x$. By the definition of $\mathcal{U}_A$, one has $V_x \subset A$. Hence $x \in A$. Since $x \in A^\circ$ was arbitrary, $A^\circ \subset A$. [/step] [step:Recognize any open subset of $A$ as one of the sets in the defining union] Let $U \in \tau$ satisfy $U \subset A$. Then $U \subset X$, $U \in \tau$, and $U \subset A$, so $U \in \mathcal{U}_A$. Therefore $U$ is one of the sets in the union defining $A^\circ$, and hence \begin{align*} U \subset \bigcup_{V \in \mathcal{U}_A} V = A^\circ. \end{align*} This proves that every open subset of $A$ is contained in $A^\circ$. Together with the previous steps, this proves that $A^\circ$ is open, $A^\circ \subset A$, and $A^\circ$ is the largest open subset of $A$. [/step]