Consider a topological space \(E\). For subsets \(A, B \subseteq E\) we have the equality \[
A^\circ \cap B^\circ = (A \cap B)^\circ\] and the inclusion \[
A^\circ \cup B^\circ \subseteq (A \cup B)^\circ\] where \(A^\circ\) and \(B^\circ\) denote the interiors of \(A\) and \(B\).

Let’s prove that \(A^\circ \cap B^\circ = (A \cap B)^\circ\).

We have \(A^\circ \subseteq A\) and \(B^\circ \subseteq B\) and therefore \(A^\circ \cap B^\circ \subseteq A \cap B\). As \(A^\circ \cap B^\circ\) is open we then have \(A^\circ \cap B^\circ \subseteq (A \cap B)^\circ\) because \(A^\circ \cap B^\circ\) is open and \((A \cap B)^\circ\) is the largest open subset of \(A \cap B\).

Conversely, \(A \cap B \subseteq A\) implies \((A \cap B)^\circ \subseteq A^\circ\) and similarly \((A \cap B)^\circ \subseteq B^\circ\). Therefore we have \((A \cap B)^\circ \subseteq A^\circ \cap B^\circ\) which concludes the proof of the equality \(A^\circ \cap B^\circ = (A \cap B)^\circ\).

One can also prove the inclusion \(A^\circ \cup B^\circ \subseteq (A \cup B)^\circ\). However, the equality \(A^\circ \cup B^\circ = (A \cup B)^\circ\) doesn’t always hold. Let’s provide a couple of counterexamples.

For the first one, let’s take for \(E\) the plane \(\mathbb R^2\) endowed with usual topology. For \(A\), we take the unit close disk and for \(B\) the plane minus the open unit disk. \(A^\circ\) is the unit open disk and \(B^\circ\) the plane minus the unit closed disk. Therefore \(A^\circ \cup B^\circ = \mathbb R^2 \setminus C\) is equal to the plane minus the unit circle \(C\). While we have \[A \cup B = (A \cup B)^\circ = \mathbb R^2.\]

For our second counterexample, we take \(E=\mathbb R\) endowed with usual topology and \(A = \mathbb R \setminus \mathbb Q\), \(B = \mathbb Q\). Here we have \(A^\circ = B^\circ = \emptyset\) thus \(A^\circ \cup B^\circ = \emptyset\) while \(A \cup B = (A \cup B)^\circ = \mathbb R\).

The union of the interiors of two subsets is not always equal to the interior of the union.