Theorem
Ifis an open connected subset ofthenis polygonally connected.
Proof
Supposeis open and connected and
Letbe all points ofwhich can be polygonally connected to
Then
Let
Sinceis open for anythere is an open neighbourhoodofsuch thatcan be polygonally connected to any point ofhencehenceis open.
The setconsists of elements which cannot be polygonally connected tosois
open.
Hence
whereandare disjoint open subsets ofSinceis connected, we have either
or
but sincesoandis polygonally connected.