Theorem
If
is a set (open or closed) and
is the boundary of the set then the closure of
labelled
is equal to the union of
and the boundary of
labelled
More concisely,
![]()
Proof
Suppose
then
hence
Since
and![]()
Suppose now that
If
then good. Suppose then that
but
Each neighbourhood of
intersects
at a point distinct from
hence
therefore ![]()
Hence![]()