Theorem
Letbe a metric. If
and
are closed subsets of
with
then open sets
and
exist such that
and
Proof
For eachdefine
and for each
Defineand
Both setsand
are open because each is the union of a family of open sets. Since
includes each
Similarly B subset B_1 .
Supposethen for some
and for some
Supposethen
with
but
This is a contradiction hence