[proofplan] Each property follows directly from the definition of the interior as the union of all open subsets of $A$. For inclusion, every open subset of $A$ is contained in $A$, so their union is as well. Idempotence follows because $A^\circ$ is itself open, so it is the largest open subset of itself. Preservation of $X$ holds because $X \in \tau$. The finite intersection property reduces to showing that $A^\circ \cap B^\circ$ is the largest open set inside $A \cap B$. Monotonicity follows because every open subset of $A$ is also a subset of $B$ when $A \subset B$. [/proofplan] [step:Establish inclusion by observing that each open subset of $A$ is contained in $A$] By definition, $A^\circ = \bigcup \{ U \in \tau : U \subset A \}$. Each set in this union satisfies $U \subset A$, so the union itself satisfies $A^\circ \subset A$. [/step] [step:Prove idempotence by showing $A^\circ$ is its own interior] Since $A^\circ$ is a union of open sets, $A^\circ \in \tau$. The interior $(A^\circ)^\circ$ is the largest open subset of $A^\circ$. Since $A^\circ$ is open and $A^\circ \subset A^\circ$, it follows that $A^\circ \subset (A^\circ)^\circ$. Conversely, by property (1) applied to $A^\circ$, $(A^\circ)^\circ \subset A^\circ$. Combining the two inclusions gives $(A^\circ)^\circ = A^\circ$. [/step] [step:Verify $X^\circ = X$ using the fact that $X$ is open in every topology] By the axioms of a topological space, $X \in \tau$. Since $X$ is open and $X \subset X$, it appears in the defining union $X^\circ = \bigcup \{ U \in \tau : U \subset X \}$, so $X \subset X^\circ$. The reverse inclusion $X^\circ \subset X$ holds by property (1). Therefore $X^\circ = X$. [/step] [step:Prove the intersection identity $(A \cap B)^\circ = A^\circ \cap B^\circ$] **($\subset$ direction.)** Let $x \in (A \cap B)^\circ$. Then there exists $U \in \tau$ with $x \in U \subset A \cap B$. In particular $U \subset A$, so $x \in A^\circ$, and $U \subset B$, so $x \in B^\circ$. Hence $x \in A^\circ \cap B^\circ$. **($\supset$ direction.)** Let $x \in A^\circ \cap B^\circ$. Then there exist $U_1, U_2 \in \tau$ with $x \in U_1 \subset A$ and $x \in U_2 \subset B$. The set $U_1 \cap U_2$ is open (since $\tau$ is closed under finite intersections), satisfies $x \in U_1 \cap U_2$, and $U_1 \cap U_2 \subset A \cap B$. Therefore $x \in (A \cap B)^\circ$. [/step] [step:Derive monotonicity from the definition of interior] Suppose $A \subset B$. Every open set $U \in \tau$ satisfying $U \subset A$ also satisfies $U \subset B$ (since $A \subset B$). Therefore \begin{align*} A^\circ = \bigcup \{ U \in \tau : U \subset A \} \subset \bigcup \{ U \in \tau : U \subset B \} = B^\circ. \end{align*} [/step]