Definition 2.5.1A metric space K is compact if whenever C is an open coverof K, there exists a finitesubset of C
{U1,⋅⋅⋅,Un }
such that K ⊆∪_{k=1}^{n}U_{k}. In words,every open cover admits a finite sub-cover.
This is the real definition given above. However, in metric spaces, it is equivalent to
another definition called sequentially compact.
Definition 2.5.2A metric space K is sequentially compactmeans thatwhenever
{xn}
⊆ K, there exists a subsequence
{xn }
k
such that lim_{k→∞}x_{nk} = x ∈ Kfor some point x. In words, every sequence has a subsequence which converges to a pointin the set.
Definition 2.5.3Let X be a metric space. Then a finite set of points
{x1,⋅⋅⋅,xn}
iscalled an ε net if
X ⊆ ∪n B (xk,ε)
k=1
If, for every ε > 0 a metric space has an ε net, then we say that the metric space is totallybounded.
Lemma 2.5.4If a metric space
(K,d)
is sequentially compact, then it is separableand totally bounded.
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 2.5.5For
(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 2.2.4 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 2.1.9 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 2.5.4, it is totally bounded. If
{xn}
is a Cauchy sequence, then there is a subsequence which converges to x ∈ X by
assumption. However, from Theorem 2.2.2 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 2.4.5 has the Lindeloff property. Hence, if X is not compact, there is a countable set
of open sets
{Ui}
_{i=1}^{∞} which covers X but no finite subset does. Consider the nonempty
closed sets F_{n} and pick x_{n}∈ F_{n} where
n n C
X ∖∪i=1Ui ≡ X ∩ (∪i=1Ui) ≡ Fn
Let
{ }
xkm
_{m=1}^{Mk} be a 1∕2^{k} net for X. We have for some m,B
( )
xkmk,1∕2k
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
{ }
xkm+1
for which B
( )
xkm+1,1∕2k+1
intersects
B
(xk ,1∕2k)
mk
, pick one x_{mk+1}^{k+1} such that B
(xk+1 ,1∕2k+1)
mk+1
contains x_{n} for infinitely
many n. Then obviously
( ) q−1 ( ) ∞
d xp ,xq ≤ ∑ d xk ,xk+1 < ∑ --1- = -1--
mp mq k=p mk mk+1 k=p2k−1 2p−2
Now take a subsequence x_{nk}∈ B
(xk ,2−k)
mk
and it follows that
lim xn = lim xk = x ∈ X
k→ ∞ k k→∞ mk
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
{Ui}
_{i=1}^{∞} covers X. ■
Lemma 2.5.6The closed interval
[a,b]
in ℝ is compact.
To show this, suppose it is not. Then there is an open cover C which admits no finite
subcover for
[a,b]
. Let I_{0} =
[a,b]
. Now consider the two intervals
[a, a+2b]
,
[a+2b,b]
.
One of these, maybe both cannot be covered with finitely many sets of C since
otherwise, there would be a finite collection of sets from C covering
[a,b]
. Let I_{1} be the
interval which has no finite subcover. Now do for it what was done for I_{0}. Split it in
half and pick the half which has no finite covering of sets of C. Thus there is a
“nested” sequence of closed intervals I_{0}⊇ I_{1}⊇ I_{2}
⋅⋅⋅
, each being half of the preceding
interval. Say I_{n} =
[an,bn]
. Note that the b_{n} are decreasing or staying the same as n
increases and the a_{n} are increasing or staying the same as n increases. Then if
k,m ∈ ℕ,k ≤ m
a ≤ a < b
k m m
and if k > m,
ak < bk ≤ bm
From completeness of ℝ, (remember calculus), for given m ∈ ℕ, a_{k}≤ b_{m} for all k and so the
least upper bound of the set of left endpoints a must be no larger than all the b_{m}.
Therefore, a is in each of the intervals I_{m}. In particular it is contained in some
set O ∈C. Thus for some δ > 0,
[a− δ,a+ δ]
, having length 2δ, is contained in
O. For k large enough, the interval
[ak,bk]
has length less than δ but contains a.
Therefore, it is contained in
[a− δ,a+ δ]
and so must be contained in a single set of C
contrary to the construction. This contradiction shows that in fact
[a,b]
is compact.
■
This can be used along with the above big theorem to show that ℝ is complete in the
sense that every Cauchy sequence converges. See Problem 4 on Page 100. In fact, the least
upper bound property used in the above and presented in Caclulus is equivalent to saying
that every Cauchy sequence converges.