Theorem
Let
and let
be a basis for![]()
is continuous if and only if for each![]()
is an open subset of![]()
Proof
Suppose
is continuous. Since each
is an open subset of
then![]()
Suppose for each![]()
is an open set in
Let
be an open subset of
Since
is a basis for![]()
![]()
Then![]()
Each
is an open set hence
is a union of open sets and
is continuous.