2.7 Continuity And Compactness
How does compactness relate to continuity? It turns out that the continuous image of a
compact set is always compact. This is an easy consequence of the above major
Theorem 2.7.1 Let f : X → Y where
are metric spaces and
f is continuous on X. Then if K ⊆ X is compact, it follows that f
is compact in
Proof: Let C be an open cover of f
Denote by f−1
is an open cover of
. It follows there are finitely many,
which covers K. It follows that
is an open cover for
The following is the important extreme values theorem for a real valued function defined
on a compact set.
Theorem 2.7.2 Let K be a compact metric space and suppose f : K → ℝ
is a continuous function. That is, ℝ is the metric space where the metric is given by
. Then f achieves its maximum and minimum values on K.
Proof: Let λ = sup
Then from the definition of sup,
the existence of a sequence
such that limn→∞f
There is a
subsequence still called
which converges to some
x ∈ K.
Therefore, from continuity,
achieves its maximum value at x
. Similar reasoning
shows that it achieves its minimum value on K
Definition 2.7.3 Let f :
be a function. Then it is said to be
uniformly continuous on X if for every ε >
0 there exists a δ >
0 such that whenever
are two points of X with d
< δ, it follows that ρ
Note the difference between this and continuity. With continuity, the δ could depend on x
but here it works for any pair of points in X.
There is a remarkable result concerning compactness and uniform continuity.
Theorem 2.7.4 Let f :
be a continuous function and let K be
a compact subset of X. Then the restriction of f to K is uniformly continuous.
Proof: First of all, K is a metric space and f restricted to K is continuous. Now suppose
it fails to be uniformly continuous. Then there exists ε > 0 and pairs of points xn,
. Since K
is compact, it is sequentially
compact and so there exists a subsequence, still denoted as
xn → x ∈ K.
n → x
also and so
which is a contradiction. Note the use of Lemma 2.2.5 in the equal sign. ■