# Intersection and union of interiors

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.