Theorem
Let
be the set of components of a space
The components of
form a partition of
i.e.
and
if![]()
Proof
Any element
belongs to at least one connected subspace i.e. at least one of the
All the
are connected. If
for any
then x would have to be in some other connected subspace
But then
- a contradiction. Hence
for some![]()
Take two sets
and
Suppose
and
Then
is a connected subset of
containing both
and
- a contradiction since
if![]()