Definition 11.1.34A metric space K is compact if whenever C is an open cover of
K, (∪C ⊇ K, each set of C is open)
there exists a finitesubset of C
{U1,⋅⋅⋅,Un}
such that K ⊆∪_{k=1}^{n}U_{k}. In words, every open cover admits afinite sub-cover.
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 11.1.35A metric space K is sequentially compactmeans that whenever
{xn}
⊆ K,there exists a subsequence
{xn }
k
such that lim_{k→∞}x_{nk} = x ∈ K for some point x. In words, everysequence has a subsequence which converges to a point in the set.
Definition 11.1.36Let X be a metric space. Then a finite set of points
{x1,⋅⋅⋅,xn}
is called an ε net if
n
X ⊆ ∪ k=1B(xk,ε)
If, for every ε > 0 a metric space has an ε net, then we say that the metric space is totallybounded.
Lemma 11.1.37If a metric space
(K,d)
is sequentially compact, then it is separable and totallybounded.
Proof: Pick x_{1}∈ K. If B
(x1,ε)
⊇ K, then stop. Otherwise, pick x_{2}
∕∈
B
(x1,ε)
. Continue this
way. If
{x1,⋅⋅⋅,xn}
have been chosen, either K ⊆∪_{k=1}^{n}B
(xk,ε)
in which case, you have
found an ε net or this does not happen in which case, you can pick x_{n+1}
∕∈
∪_{k=1}^{n}B
(xk,ε)
.
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 ε. Thus for every
ε > 0, there is an ε net. Thus the metric space is totally bounded. Let N_{ε} denote an ε net. Let
D = ∪_{k=1}^{∞}N_{1∕2k}. 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 within 1∕2^{k} of it.
■
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 11.1.38For
(X, d)
a metric space, thefollowing are equivalent.
(X, d)
is compact.
(X, d)
is sequentially compact.
(X, d)
is complete and totally bounded.
Proof: 1.
=⇒
2. Let
{xn}
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 ∪_{n=1}^{∞}
{xn }
has a limit point
in X, then it follows from Proposition 11.1.18 there would be a convergent subsequence converging to this
limit point. Therefore, assume ∪_{k=1}^{∞}
{xn}
has no limit point. This is equivalent to saying
that ∪_{k=m}^{∞}
{xk}
has no limit point for each m. 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
sets
(∪∞k=m {xn})C
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.
2.
=⇒
3. If
(X, d)
is sequentially compact, then by Lemma 11.1.37, it is totally bounded. If
{x }
n
is a
Cauchy sequence, then there is a subsequence which converges to x ∈ X by assumption. However, from
Theorem 11.1.16 this requires the original Cauchy sequence to converge.
3.
=⇒
1. Since
(X, d)
is totally bounded, there must be a countable dense subset of X. Just take the
union of 1∕2^{k} nets for each k ∈ ℕ. Thus
(X, d)
is completely separable by Theorem 11.1.32 has the
Lindeloff property. Hence, if X is not compact, there is a countable set of open sets
{U }
i
_{i=1}^{∞} which
covers X but no finite subset does. Consider the nonempty closed sets F_{n} and pick x_{n}∈ F_{n}
where
X ∖∪ni=1Ui ≡ X ∩(∪ni=1Ui)C ≡ Fn
Let
{xk }
m
_{m=1}^{Mk} be a 1∕2^{k} net for X. We have for some m,B
(xk ,1∕2k)
mk
contains x_{
n} for infinitely many
values of n because there are only finitely many balls and infinitely many indices. Then out of the finitely
many
{xk+1}
m
for which B
(xk+1,1∕2k+1)
m
intersects B
(xk ,1∕2k)
mk
, pick one x_{mk+1}^{k+1} such that
B
( k+1 k+1)
xmk+1,1∕2
contains x_{n} for infinitely many n. Then obviously
{ k }
xmk
_{k=1}^{∞} is a Cauchy sequence
because
( ) 1 1 1
d xkmk,xkm+k1+1 ≤ 2k + 2k+1 ≤ 2k−-1
Hence for p < q,
( p q ) q−∑ 1 ( k k+1 ) ∑∞ 1 1
d xmp, xmq ≤ d xmk,xmk+1 < 2k−1 = 2p−2
k=p k=p
Now take a subsequence x_{nk}∈ B
( )
xkmk,2−k
and it follows that
k
lk→im∞ xnk = kli→m∞x mk = x ∈ X
However, x ∈ F_{n} for each n since each F_{n} is closed and these sets are nested. Thus x ∈∩_{n}F_{n} contrary to
the claim that