Theorem
Ifis an open connected subset of
then
is polygonally connected.
Proof
Supposeis open and connected and
Letbe all points of
which can be polygonally connected to
Then
Let
Sinceis open for any
there is an open neighbourhood
of
such that
can be polygonally connected to any point of
hence
hence
is open.
The setconsists of elements which cannot be polygonally connected to
so
is
open.
Hence
whereand
are disjoint open subsets of
Since
is connected, we have either
or
but sinceso
and
is polygonally connected.