Proposition 5

Let $A$ be an open set in $X$; then for every subset $B$ of $X$ we have

For suppose $x \in A \cap \overline{B}$; then if $V$ is any neighbourhood of $x$, $V \cap A$ is a neighbourhood of $x$, since $A$ is open; hence $V \cap A \cap B$ is not empty and therefore $x$ lies in the closure of $A \cap B$.

If $x$ lies in the closure of $A$ but not in $A$, then every neighbourhood of $x$ contains a point of $A$ other than $x$; but if $x \in A$ it can happen that $x$ has a neighbourhood which contains no point of $A$ except $x$. We say then that $x$ is an isolated point of $A$. In particular, $x$ is isolated in the whole space $X$ if and only if $\{x\}$ is an open set.

A closed set which has no isolated points is called a perfect set.