Theorem
Suppose
is onto where
is a
space. A necessary and sufficient condition for
to be a homeomorphism is
1.
for every![]()
or
2.![]()
Proof
A topological space
is a
space if each singleton set is closed so that
for each
With this definition each metric space is a
space.
Since
is a homeomorphism it is one to one and
or![]()
Now we show that
is one to one. Suppose
then![]()
Since
is a
space,
and
is one to one. Hence
for every
and
is a homeomorphism.