Theorem
If
is a continuous function,
is a compact metric space and
is a metric space then
is uniformly continuous.
Proof
Take
Define an open cover of
as![]()
Since
is continuous, each
is open in
and
is an open cover of![]()
is a compact metric space, so the above cover for
has a Lebesgue number![]()
For each
there exists
such that
and
such that![]()
From the definition of Lebesgue number,
and
is uniformly continuous.