Theorem
Suppose
Then
is continuous if and only if
there exists an open set
with
The converse is also true.
Proof
Supposeis continuous. Take any
and any open neighbourhood
of
Then
and
Sinceis continuous
Since
we have
Let
represent an open set of
Suppose
then
sois an open neighbourhood ofAn open set
exists such that
and
But
hence for each
where
Hence
is the union of open subsets of
and so is an open subset.