Theorem
A subset
of
is connected if and only if
cannot be expressed as the union
of nonempty , mutually separated subsets
and
of![]()
Proof
If
is not connected then
where
and
and
are both open and closed in![]()
Suppose
then![]()
Since
then
then
- a contradiction.
Similarly![]()
Now suppose
and
then![]()
Similarly![]()
Hence
and
are closed in
and
is disconnected.