Theorem
Let
be a metric. If
and
are closed subsets of
with
then open sets
and
exist such that
and
Proof
For each
define
and for each
Define
and
Both setsandare open because each is the union of a family of open sets. Since includes each
Similarly B subset B_1 .
Suppose
then for some
and for some
Suppose
then
with
but
This is a contradiction hence