Theorem
Suppose
is a map from a metric space
to a metric space![]()

The statements '
is continuous'
and
'for all open sets
is open in![]()
Proof
Suppose
is continuous and
is an open subset of
Let
then![]()
Since
is open there exists
such that![]()
Since
is continuous there exists
such that![]()
Hence
(1)
For
exists such that (1) is true hence
is open in![]()
Conversely suppose that
is an open subset of
then
is an open subset of
If
then there exists
is an open subset of
Hence
is an open subset of
Then there exists
such that
or![]()