Theorem
Any open subspace of a separable space is separable.
Proof
Let
be a separable space and let
be a countable dense subset of![]()
If
is an open subset of
define ![]()
Let
be an open subset of
hence
is open in
hence
contains at least one![]()
A subset of a topological space
is dense in
if and only if every open subset of
contains a point of the subset. Therefore
is dense in![]()
contains a countable dense subset
hence
is separable.