Need help showing my example does (or doesn't) work. I am trying to find an example of when $(A \bigcup B)^\circ \supset A^\circ \bigcup B^\circ$. Where $^\circ$ denotes the interior of a set. It has been previously shown in the text that:

The interior of the set $\mathbb{Q}$ of rational numbers is empty.

My intuition is telling me that $(\mathbb{R} - \mathbb{Q} \bigcup \mathbb{Q})^\circ \supset (\mathbb{R} - \mathbb{Q})^\circ \bigcup \mathbb{Q}^\circ$. Since, $(\mathbb{R} - \mathbb{Q} \bigcup \mathbb{Q})^\circ = \mathbb{R} = (-\infty, \infty)$, but I don't exactly know what the size of $(\mathbb{R} - \mathbb{Q})^\circ \bigcup \mathbb{Q}^\circ$ is, all I can think of reducing it down to:
$$(\mathbb{R} - \mathbb{Q})^\circ \bigcup \emptyset$$
$$(\mathbb{R} - \mathbb{Q})^\circ$$
Is there a way I can show $\mathbb{R}^\circ \supset (\mathbb{R} - \mathbb{Q})^\circ$? Is it even true?
 A: Take $A=\mathbb R\setminus \mathbb Q$ and $B=\mathbb Q$. Then $A^\circ =B^\circ=\emptyset$, but $A\cup B= \mathbb R$ so $(A\cup B)^\circ =\mathbb R$.
To show that $\mathbb R\setminus \mathbb Q$ has empty interior, assume that it doesn't, so there exists an interval $(a,b)$ contained in $\mathbb R\setminus \mathbb Q$. But every interval contains rationals, which leads to a contradiction.
A: Any nonempty open interval in $\mathbb R$ contains both rational and irrational points. Therefore the set of irrationals has no interior points either.
A: If you're familiar with the Cantor Set, another example that would work would be:

Let $A$ be the Cantor set, and $B = [0, 1] \setminus A$, so that $B$ is the union of the intervals removed from $[0, 1]$ in the construction of $A$.  Then $B$ is open, so $B^\circ = B$.  But $A$ has empty interior, so that $A^\circ \cup B^\circ = B$.  However, $(A\cup B)^\circ = (0, 1) \supset B$.

Letting $A$ be any closed nowhere dense set and $B$ its complement would work similarly.
