Download 1. Compactness for metric spaces For a metric space (X, d) we will

1. Compactness for metric spaces
For a metric space (X, d) we will as usual use the notation Bx (r) to denote the open
ball of radius r centered at x ∈ X:
Bx (r) = {y ∈ X | d(x, y) < r}
Definition 1.1. We say that a metric space (X, d) is totally bounded if for every
ε > 0 there exist finitely many points xi ∈ X, i = 1, ..., N such that
X=
N
[
Bxi (ε)
i=1
Typically the number N of points xi needed to cover X by radius ε balls will increase
as ε becomes small. But now matter how small ε gets, N stays finite.
Proposition 1.2. If a metric space (X, d) is totally bounded then X is separable.
Proof. Let εn = 1/n for n ∈ N. For each such εn there is a finite set of points
An = {xn1 , ..., xn`n } such that
`n
[
X=
Bxi (1/n)
i=1
Let A = ∪n∈N An . Clearly A is countable and we claim that A = X. Let x ∈ X be
an arbitrary point. For any n ∈ N, there is some yn ∈ An so that x ∈ Byn (1/n). This
gives a sequence yn ∈ A with d(x, yn ) < 1/n. Thus lim yn = x. This shows that every
x ∈ X lies in the closure of A and so A = X.
Proposition 1.3. A metric space (X, d) is second countable if and only if it separable.
The proof is left as an exercise. Recall that we already know that a second countable
space is separable. The content of the proposition is that for metric spaces the converse
is also true.
Lemma 1.4. Let (X, d) be a metric space with the property that every sequence xi ∈ X
has a convergent subsequence. Then X is totally bounded.
Proof. If X is not totally bounded, then there is some ε > 0 so that X cannot be covered
by a finite number of balls of radius ε. We use this to construct a sequence xi ∈ X:
let x1 be arbitrary and choose consecutive elements of the sequence so that xi+1 ∈
/
∪ij=1 Bxj (ε). Let yn = xin be some convergent subsequence of xi with limn→∞ yn = y0 .
Let U = By0 (ε/2) and let yk , y` ∈ U be any two elements (which must exist by the
convergence property) with k < `. Then ik < i` and
d(xik , xi` ) ≤ d(xik , y0 ) + d(y0 , xi` ) < ε/2 + ε/2 = ε
This implies that xi` ∈ Bxik (ε) which is a contradiction (to the way we constructed the
sequence xi ). Thus X must be totally bounded.
Theorem 1.5. A metric space (X, d) is compact if and only if every sequence xi ∈ X
has a convergent subsequence.
1
2
Proof. =⇒ Suppose X is compact and let xi ∈ X be an arbitrary sequence. If there
was no convergent subsequence of xi we could for each y ∈ X find an open set Uy which
contains only finitely many elements of the sequence xi . But then F = {Uy | y ∈ X}
is a cover of X implying that there are finitely many points y1 , ..., ym ∈ X so that
X = Uy1 ∪ ... ∪ Uy1 . This would tell us that the set {xi | i ∈ N} is finite. In particular,
some value xi must occur infinitely many times giving rise to a convergent subsequence.
⇐= Suppose that every sequence xi ∈ X has a convergent subsequence and let F
be a cover of X. Find a countable open cover F 0 = {Ui ⊆ X | Ui is open , i ∈ N} with
the property that for every Ui ∈ F 0 there is some V ∈ F with Ui ⊆ V (the proof of the
existence of such a cover F 0 is left as an exercise).
Claim: F 0 has a finite subcover.
To prove
! the claim, suppose the opposite. Then by picking an arbitrary xn ∈ X −
n−1
[
Ui we obtain a sequence. By the assumption, there must be a subsequence
i−1
yi = xni with converges to y0 . Since F 0 is a cover, there is some m ∈ N with y0 ∈ Um .
But then yj ∈
/ Um for all j ≥ m which is a contradiction. This concludes the proof of
the claim.
Let thus {U1 , .., UN } be a finite cover of X and let Vi ∈ F be such that Ui ⊆ Vi for
all i = 1, ...., N . The {V1 , ..., VN } is also a finite cover of X.
Corollary 1.6. Every compact metric space X is totally bounded, separable and second
countable.
Definition 1.7. Let (X, d) be a metric space. A Cauchy sequence xi ∈ X, i ∈ N
is a sequence with the property that for every ε > 0 there is some i0 ∈ N such that
d(xi , xj ) < ε for all i, j ≥ i0 . Every convergent sequence is clearly a Cauchy sequence
but the converse is not generally true.
A metric space (X, d) is called complete if every Cauchy sequence is a convergent
sequence.
Theorem 1.8. Let (X, d) be a metric space. Then X is compact if and only if X is
complete and totally bounded.
Proof. =⇒ Suppose X is compact, we need to show that X is complete and completely
bounded. To see completeness, let xi ∈ X be any Cauchy sequence in X. By theorem
1.5 there is a convergent subsequence of xi with limit x0 . It is then easy to see that xi
converges with limit x0 (left as an exercise). Thus X is complete.
On the other hand, consider the open cover Fε of X defined for any ε > 0 as
Fε = {Bx (ε) | x ∈ X}
By compactness of X there must be a finite subcover of Fε . This shows that X is
totally bounded.
⇐= Let X be complete and totally bounded. We will show that X is compact by
showing that any sequence xi ∈ X has a convergent subsequence (see theorem 1.5).
3
Let εk = 1/k for k ∈ N. For ε1 , there is a finite cover of X with balls of radius ε1 . Since
xi has infinitely many terms, one of the these radius εk balls contains infinitely many
elements of the said sequence. Let B1 be that ball. For ε2 there is again a finite cover
of X by ε2 balls. Since B1 contains infinitely many xi , there is some radius ε2 ball B2
such that B2 ∩ B1 contains infinitely many of the xi . One then continues this process
indefinitely producing a sequence of radius εk balls Bk so that Bk ∩ Bk−1 ∩ ... ∩ B1
contains infinitely many elements of the sequence xi .
Pick now indices n1 < n2 < n3 < ... such that xn` ∈ B` and set yi = xni . It is easy
to see that yi is Cauchy and so by the completeness assumption on X, it must have a
convergent subsequence. This completes the proof.
Corollary 1.9. Consider Rn equipped with the Euclidean topology. Then a subset A
of Rn is compact if and only if it is closed and bounded.
Proof. A subset A of Rn is totally bounded if and only if it is bounded. On the other
hand, any bounded sequence in Rn has a convergent subsequence. If the sequence lies
in a closed set A then so does the limit. Finally, every Cauchy sequence is bounded
(easy exercise).
Corollary 1.10. The closed interval [a, b] ⊆ R is compact. Finite unions of closed
intervals are also compact.
Corollary 1.11. Let f : X → R be a continuous function with X compact (but not
necessarily a metric space). Then f attains a minimum and maximum on X.
Proof. Since X is compact and f continuous then f (X) is a compact subset of R. Since
it is bounded and closed it must have a maximum and minimum.
2. Exercises
(1) Prove proposition 1.3.
(2) Let (X, d) be a second-countable metric space. Then for every open cover F of
X there exists a countable open cover F 0 = {Ui ⊆ X | Ui is open , i ∈ N} with
the property that for every Ui ∈ F 0 there is some V ∈ F with Ui ⊆ V .
(3) Let (X, d) be a metric space and xi ∈ X a Cauchy sequence which has a
convergent subsequence with limit x0 . Show that then xi is also convergent
with limit x0 .
(4) Consider Rn equipped with the Euclidean topology. Show that A ⊆ Rn is
bounded if and only if it is totally bounded.
(5) Let X be a complete topological space and let A be a closed subset of X endowed
with the relative topology. Show that A is also complete.