Inequality involving closure and intersection Let the closure of a set $A$ be $\bar A$. On Page 62, Introduction to Boolean Algebras,Steven Givant,Paul Halmos(2000), an exercise goes like,

Show that $P \cap \bar Q \subseteq \overline{(P \cap Q)}$, whenever $P$ is open.

I felt muddled in face of this sort of exercises. Is there some way to deal with these problems and be assured about the result?

Let $x$ in $P\cap\bar Q$. Since $x$ is in $\bar Q$, there exists a sequence $(x_n)_n$ in $Q$ such that $x_n\to x$. Since $x$ is in $P$ and $P$ is open, $x_n$ is in $P$ for every $n$ large enough, say, $n\geqslant n_0$. Hence, for every $n\geqslant n_0$, $x_n$ is in $P\cap Q$. Thus, $x$ is in $\overline{P\cap Q}$ as limit of $(x_n)_{n\geqslant n_0}$.
