Let
be a topological space, and let
be a subset of
First we assume that
is closed and show that
contains all of its limit points. Let
be a limit point of
We proceed by contradiction and assume that
Then
which is an open set. This means that there is a neighborhood (namely
) of
that contains no points in
which contradicts the fact that
is a limit point of
We conclude that
and
contains all of its limit points.
For the converse, assume that
contains all of its limit points. To show that
is closed, we prove that
is open. We again proceed by contradiction and assume that
is not open. Then there exists
such that no neighborhood of
is entirely contained in
This implies that every neighborhood of
contains a point in
and so
is a limit point of
It follows that
contradicting the fact that
We conclude that
is open and
is closed.