Theorem
Let
be the space of real numbers in the metric space
with the absolute value metric
and let![]()
The closure of
is
and the closure of any subset
of
is a closed subset of![]()
Proof
Let
The closure of
written
is defined as
where
Since![]()
For each![]()
Hence
and![]()
Choose
and
then
hence![]()
To prove the closure of any subset
of
is a closed subset of
suppose
is not closed then
is not open and there is an element
such that for every![]()

Choose![]()
Since
is open there exists
such that![]()
Since
then
so
exists such that![]()
Hence for
there exists
such that
Hence
contradicting![]()