Theorem
If
is an open connected subset of
then
is polygonally connected.
Proof
Suppose
is open and connected and![]()
Let
be all points of
which can be polygonally connected to![]()
Then![]()
Let![]()
Since
is open for any
there is an open neighbourhood
of
such that
can be polygonally connected to any point of
hence
hence
is open.
The set
consists of elements which cannot be polygonally connected to
so
is
open.

Hence![]()
where
and
are disjoint open subsets of
Since
is connected, we have either
![]()
or
![]()
but since
so
and
is polygonally connected.