11.1.6 Compact Sets in Metric Space
As usual, we are not worrying about empty sets.
Definition 11.1.34 A metric space K is compact if whenever C is an open cover of
there exists a finite subset of C
such that K ⊆∪k=1nUk. In words, every open cover admits a
The above definition is equivalent to the same statement with the provision that each open set in C is
an open ball. See Problem 15 on Page 782.
This is the real definition given above. However, in metric spaces, it is equivalent to another definition
called sequentially compact.
Definition 11.1.35 A metric space K is sequentially compact means that whenever
there exists a subsequence
= x ∈ K for some point x. In words, every
sequence has a subsequence which converges to a point in the set.
Definition 11.1.36 Let X be a metric space. Then a finite set of points
is called an ε net if
If, for every ε > 0 a metric space has an ε net, then we say that the metric space is totally
Lemma 11.1.37 If a metric space
is sequentially compact, then it is separable and totally
Proof: Pick x1 ∈ K. If B
then stop. Otherwise, pick x2
have been chosen, either
in which case, you have
net or this does not happen in which case, you can pick xn+1
The process must terminate since otherwise, the sequence would need to have a convergent
subsequence which is not possible because every pair of terms is farther apart than
. Thus for every
there is an ε
net. Thus the metric space is totally bounded. Let Nε
denote an ε
. Then this is a countable dense set. It is countable because it is the countable union
of finite sets and it is dense because given a point, there is a point of D
Also recall that a complete metric space is one for which every Cauchy sequence converges to a point in
the metric space.
The following is the main theorem which relates these concepts.
Theorem 11.1.38 For
a metric space, the following are equivalent.
is sequentially compact.
is complete and totally bounded.
be a sequence. Suppose it fails to have a convergent subsequence. Then it
follows right away that no value of the sequence is repeated infinitely often. If
has a limit point
, then it follows from Proposition 11.1.18
there would be a convergent subsequence converging to this
limit point. Therefore, assume ∪k=1∞
has no limit point. This is equivalent to saying
has no limit point for each
Thus these are closed sets by Theorem 11.1.9
because they contain all of their limit points due to the fact that they have none. Hence the open
yield an open cover. This is an increasing sequence of open sets and none of them contain all the values of
the sequence because no value is repeated for infinitely many indices. Thus this is an open cover which has
no finite subcover contrary to 1.
is sequentially compact, then by Lemma
, it is totally bounded. If
Cauchy sequence, then there is a subsequence which converges to
x ∈ X
by assumption. However, from
this requires the original Cauchy sequence to converge.
is totally bounded, there must be a countable dense subset of
. Just take the
union of 1∕
nets for each k ∈ ℕ
is completely separable by Theorem
Lindeloff property. Hence, if X
is not compact, there is a countable set of open sets
but no finite subset does. Consider the nonempty closed sets Fn
and pick xn ∈ Fn
be a 1∕
net for X
. We have for some m,B
for infinitely many
values of n
because there are only finitely many balls and infinitely many indices. Then out of the finitely
pick one xmk+1k+1
for infinitely many n
. Then obviously
is a Cauchy sequence
Hence for p < q,
Now take a subsequence xnk ∈ B
and it follows that
However, x ∈ Fn for each n since each Fn is closed and these sets are nested. Thus x ∈∩nFn contrary to
the claim that