Theorem
If
is a T4 space containing more than one point, then there exists a non constant continuous function![]()
Proof
Suppose
are distinct points. Since
is T4 it must be normal and T1.
For any T1 space
and![]()
so that any singleton set is closed. Since
and
are distinct
and
are disjoint.
Therefore, since
is normal a continuous function
exists such that
and![]()
Hence
is a continuous non constant function from
into![]()