Theorem
The union of any family of connected subsets of any space
having at least one point
in common is also connected.
Proof
Let
be any family of connected subsets of
such that for each
there exists
with
for each
so that
and![]()
Let
be the space
with the discreet topology and let
be continuous.
Since each
is connected and
is continuous on![]()
Since
for each
for all![]()
Hence
is not onto
and
is connected.