$A^\mathrm{o}$, the interior of $A$, is the union of all open subsets of $A$.
If $H=\Bbb Q$ and $K=\Bbb R \backslash \Bbb Q$ then $H^\mathrm{o}=K^\mathrm{o}=\emptyset$ but $(H\cup K)^\mathrm{o}=\Bbb R^\mathrm{o}=\Bbb R$
I don't understand why $H^\mathrm{o}=K^\mathrm{o}=\emptyset$
Could somebody explain why this is the case?
Suppose $x \in H^\mathrm{o}$. Then there is an open interval $(a,b)$ such that $a < x < b$ and $(a,b) \subset H$. But you can show that each open interval contains irrational numbers: for example, take $\sqrt{2}$ and translate it by some fraction $q$ so that $a < q + \sqrt{2} < b$. So no open interval is a subset of $\mathbb{Q}$. (Another way to see this is that every open interval is uncountable while $\mathbb{Q}$ is countable; so it cannot possibly contain an interval.)
The argument for $K$ is similar, but it uses that $\mathbb{Q}$ is dense in $\mathbb{R}$. In other words, for any irrational and any open interval surrounding it, you can find a rational close to it. So every open interval contains a rational, which implies $K^\mathrm{o}$ is empty.
