Download Compactness - GMU Math 631 Spring 2011

Compactness
These notes discuss the same topic as Sections 26, 27, 28, and 37 of Munkres’
book. Some notions (Σ-product, σ-product, sequential compactness, pseudocompactness, ...) are not discussed in Munkres’ book.
1. Definition and basic properties
S
Recall that a family U of subsets of a set X is called a cover of X if U = X.
If (X, T ) is a topological space and all elements of a cover U are open sets, then U
is called an open cover.
Definition 1. A topological space (X, T ) is called compact if every open cover of
X contains a finite subfamily that still covers X.1
Examples 2.
(1) A discrete space is compact iff it is finite.
(2) R with the standard topology is not compact.
(3) Any set with the trivial topology is compact.
(4) Any set with the cofinite topology is compact.
(5) (The one-point compactification of a discrete space) Let D be a set of any
cardinality, p 6∈ D, and X = D ∪ {p}. Consider the topology T on X
generated by the base {{d} : d ∈ D} ∪ {X \ F : F ⊂ D is finite }. Then
(X, T ) is compact.2
Definition 3. A non empty family
of sets H is called centered if for every non
T
empty finite subfamily F ⊂ H, F =
6 ∅.3
Theorem 4. Let (X, T ) be a topological space. The following conditions are equivalent:
(1) (X, T ) is compact;
(2) Every centered family of closed subsets of X has non empty intersection;
(3) For every centered family F of subsets of X there is p ∈ X such that p ∈ A
for every A ∈ F.
Sketch of proof: Consider the compliments. 2
Theorem 5. Every closed subspace of a compact space is compact.
Theorem 6. Every compact subspace of a Hausdorff space is closed.
Sketch of proof: Let X be a compact subspace of a Hausdorff space (Y, T ). To
show that Y \ X is open, let y ∈ Y \ X. For each x ∈ X fix disjoint Ux , Vx ∈ T so
that x ∈ Ux and y ∈ Vx . From the open cover {Ux : x ∈ X} of X extract a finite
subcover, say {Ux1 , ..., Uxn }. Then Vx1 ∩ ... ∩ Vxn is a neighborhood of Y that does
not intersect X. 2
Theorem 7. Every continuous image of a compact space is compact.
Theorem 8. Every continuous mapping from a compact space to a Hausdorff space
is closed.
1To say short: every open cover has a finite subcover
2Compare this example with the one-point Lindelöfication of a discrete space - see notes on
metric and metrizable spaces.
3Another name: H has the finite intersection property
1
2
Exercise 9. Give an example of a continuous mapping of a compact space onto a
Hausdorff space which is not open.
Recall that continuous surjections are called condensations.
Corollary 10. A condensation of a compact space onto a Hausdorff space is an
homeomorphism.4
2. Products. (And some more set theory.)
Theorem
11. (Tychonoff) If for every a ∈ A, the space (Xa , Ta ) is compact, then
Q
P = a∈A Xa with the Tychonoff product topology is compact.
Proof 1 of Theorem 11 - for finite products only. It is enough to prove for the
product of two spaces, and then to proceed by induction. So let (X, TX ) and (Y, TY )
be compact; we are going to prove that X × Y with the Tychonoff product topology
is compact. We need the following notion useful not only in this proof.
Definition 12. A family of sets V is a refinement of a family of sets U if for every
V ∈ V there is U ∈ U such that V ⊂ U .
The following two facts are obvious:
• If U and V are two covers of the same space, V refines U, and V contains a
finite subcover, then so does U.
• Every open cover of a product X × Y has a refinement consisting of rectangular open sets, that is, sets of the form U × V where U is open in X
and V is open in Y .
So given an open cover U of the product X × Y of compact spaces X and Y we can
without loss of generality assume that U consists of rectangular open sets.
For each x ∈ X, put Yx = {x}×Y . Then Yx is a subspace of X×Y homeomorphic
to Y and thus compact. Fix a finite subfamily say Ux = {Ux,1 × Vx1 , ..., Ux,kx ×
Vx,kx } ⊂ U that covers Yx . Put Wx = Ux,1 ∩ ... ∩ Ux,kx . Then Wx is a neighborhood
of x in X. By compactness of X, the open cover {Wx : x ∈ X} has a finite subcover,
say {Wx1 , ..., Wxm }. Then the “tubes” Wx1 × Y ,..., Wxm × Y Scover X × Y , and
m
each tube Wxi × Y is covered by the finite family Uxi . Then i=1 Uxi is a finite
subfamily of U that coveres X × Y . 2
Proof 2 of Theorem 11 - for finite products only.5 is based on the following
proposition:
Proposition 13. Let Y be a topological space. The following conditions are equivalent:
(1) Y is compact
X,Y
(2) For every topological space X, the projection πX
: X × Y → X is a closed
6
mapping.
X,Y1 ,Y2
So let Y1 and Y2 be two compact spaces. Note that the projection πX
:
X,Y1 ,Y2
X × Y1 × Y2 → X is a composition of two projections πX,Y1
: X × Y1 × Y2 →
4In other words, a Hausdorff compact topology on a set X is minimal in the lattice of all
Hausdorff topologies on X.
5This “exotic” argument is optional
6One of the implications is one of the turn-in homework problems.
3
X,Y1
X × Y1 and πX
: X × Y1 → X. Since Y1 and Y2 are compact, by (1)⇒(2) in
X,Y1 ,Y2
X,Y1
Proposition 13, both mappings πX,Y
and πX
are closed. But then (for every
1
X,Y1 ,Y2
X!) πX,Y1 is closed as a composition of two closed mappings and thus by (2)⇒(1)
in Proposition 13 Y1 × Y2 is compact. 2
To prove Theorem 11 in the general case, we need one set-theoretic consequence
of the Axiom of Choice.
Definition 14. Let S be a set7 and P be a property that some subsets of S have
(and may be some have not). Say that P is a property of finite character if:
• ∅ has P;
• A ⊂ S has P iff all finite subsets of A have P.
Examples 15. The following properties are of finite character:
• pairwise disjoint;
• almost disjoint;8
• centered;
• linked; (=every two elements have non-empty intersection),
• ...
The following properties are not of finite character:
• finite;
• countable;
• closed;
• ...
Axiom 16. (The Techmuller-Tukey lemma) If P is a property of finite character (of subsets of a set S), A ⊂ S and A has P then there is B such that A ⊂ B ⊂ S
and B is a maximal element in the partially ordered set of of all subsets of S that
have property P.
Axiom 16 is known to be equivalent to the axiom of choice.
Lemma 17. Let F be a maximal centered family of subsets of some set X. Then:
(1) If n ∈ N and A1 , ..., An ∈ F, then A1 ∩ ... ∩ An ∈ F;
(2) if B ⊂ X and B ∩ A 6= ∅ for every A ∈ F then B ∈ F;
(3) if B ⊂ X then either B ∈ F or X \ B ∈ F.9
Proof 3 of Theorem 11 - in the general
Q case. We will use Theorem 4. Let H
be a centered family of subsets of P = a∈A Xa (where each Xa is compact). By
Axiom 16 H ⊂ F for some maximal centered family of subsets of X. It suffices to
find p ∈ P such that p ∈ F for every F ∈ F.
Let a ∈ A. Then {πa (F ) : F ∈ F} is a centered family of subsets of Xa . By
compactness of Xa and Theorem 4 there is pa ∈ Xa such that pa ∈ πa (F ) for all
F ∈ F (where the closure is taken in Xa ). Then p = hpa : a ∈ Ai ∈ P . We claim
that p is as Q
desired.
Let U = a∈A Ua be a standard neighborhood of p. Then there is a finite A ⊂ A,
say A = {a1 , ..., an } such that Ua = Xa for all a ∈ A \ A. Let i ∈ {1, ..., n}. Then
7In many applications, S is a family of sets, that is the elements of S are sets.
8See the discussion of Ψ-spaces in the notes “Metric and metrizable spaces”
9We will not use property (3) right now, but it is too nice not to be mentioned.
4
(Uai ). Then Vi intersects
Uai intersects every set πai (F ), F ∈ F. Put Vi = πa−1
i
each set F ∈ F. By maximality of F and part 2 of Lemma 17, Vi ∈ F. Then by
part 1 of Lemma 17, V1 ∩ ... ∩ Vn ∈ F. But V1 ∩ ... ∩ Vn = U . So every standard
neighborhood of p intersects every element of F and thus p is in the closure of every
element of F. 2
Example 18. The box product of infinitely many non trivial10 factors is never
compact.
3. Countable compactness. Compactness in metric spaces. Compact
subspaces of Rn
Recall that a space X is Lindelöf if every open cover of X has a countable
subcover.
Definition 19. A space X is called countably compact if every countable open
cover of X has a finite subcover.
Theorem 20. A space X is compact iff X is both Lindelöf and countably compact.
Recall that for metrizable spaces, the three conditions: second countability, Lindelöfness, and separability, are equivalent.
Theorem 21. Every metrizable compact space is second countable.
Exercise 22.
(1) Give an example of a compact space which is not second
countable.
(2) Give an example of a metrizable space which is not second countable.
Theorem 23. Let X be a T1 space. Then the following conditions are equivalent:
(1) X is countably compact;
(2) Every infinite subset of X has a limit point in X;11
(3) X does not contain infinite closed discrete subspaces.
Sketch of proof. Let D be an infinite closed discrete subspace of X. We can
assume without loss of generality that D is countably infinite. Then {X \(D \{d}) :
d ∈ D} is a countable open cover of X that does not contain a finite subcover.
Now let U = {Un : n ∈ N} be a countable open cover of a space X such that U
does not contain a finite subcover. For each n ∈ N, pick pn ∈ X \ (U1 ∩ ... ∩ Un ).
Then {pn : n ∈ N} is an infinite12 closed, discrete subspace of X 2
Theorem 24. A metrizable space X is compact iff X is countably compact.
Sketch of proof. Let ε > 0. Say that a subset E of a metric space (X, d) is a
ε-net in X if:
(1) For every x ∈ X there is e ∈ E with d(x, e) < ε;
(2) For every distinct e, e0 ∈ E, d(e, E 0 ) ≥ ε.
Note that:
10Of course, what is called non trivial needs specification. Perhaps the weakest form of understanding “non trivial” that makes the statement true is this: every point has a neighborhood
that does not equal the entire space.
11In Munkres’ book this property is called the limit point compactness.
12Explain why
5
• In every metric space (and every ε) there is a ε-net. Indeed, just start from
any point, and keep adding new points while possible.
• A ε-net is a closed discrete subspace.
• Therefore, if the space is countably compact, then a ε-net is finite.
So, having a countably compact metric space (X, d), for every n ∈ N, fix a n1 -net
S
En . Then D = n∈N En is a dense13 countable subspace of X. X is separable,
hence (being metrizable) Lindelöf. But Lindelöf + countably compact = compact.
2
Definition 25. A topological space X is called sequentially compact if every sequence of points of X has a convergent subsequence.
Obviously, compact ⇒ countably compact ⇐ sequentially compact.14
Theorem 26. For every Fréchet space15 X the following conditions are equivalent:
• X is countably compact;
• X is sequentially compact.
So for metrizable spaces, the three conditions: compact, countably compact, and
sequentially compact, are equivalent.
Exercise 27. Give an example of a metrizable space which is Lindelöf but not
compact.
Example 28. The unit interval I = [0, 1] (with the topology inherited from the
standard topology of the real line) is compact.
The proof is based on two facts:
Lemma 29. Every infinite sequence of real numbers contains either an infinite non
decreasing subsequence or an infinite non increasing subsequence.
The following we assume to be known from analysis:16
Axiom 30. (The least upper bound principle) Every set of real numbers
bounded from above has the least upper bound (i. e. the minimum in the set of all
upper bounds.
Every set of real numbers bounded from below has the greatest lower bound.
Example 28 together with some theorems above provide the following fundamental fact:
Theorem 31. A subspace K of Rn (where n ∈ N) is compact iff K is closed and
bounded.17
13Explain why
14A little later, we will see that sequentially compact 6⇒ compact. Much later, we will see that
compact 6⇒ sequentially compact.
15
Hence for every metrizable space
16Recall that a subset A ⊂ R is called bounded from above if there is B ∈ R (called an upper
bound for A) such that a < B for every a ∈ A. The notions “bounded from below” and lower
bounds are defined similarly.
17That is, contained in a ball of some radius with center at the origin.
6
4. More on Σ- and σ-products, countable compactness and
pseudocompactness
Recall:
Definition 32. Let p ∈ P =
Q
a∈A
Xa . Then the sets
σn (p) = {x ∈ P : |{a ∈ A : x(a) 6= p(a)}| ≤ n},
(where n ∈ N)
[
σ(p) = {x ∈ P : {a ∈ A : x(a) 6= p(a)} is finite} =
σn (p),
n∈N
Σ(p) = {x ∈ P : {a ∈ A : x(a) 6= p(a)} is countable}
are called, respectively, the σn -product, σ-product, and Σ-product in P with base
point p.
Proposition 33.
(1) p ∈ σ1 (p) ⊂ σ2 (p) ⊂ · · · σ(p) ⊂ Σ(p);
(2) If A is finite then σ(p) = P 18
(3) If A is countable then Σ(p) = P .
(4) If q ∈ Σ(p) then Σ(q) = Σ(p).
(5) If q ∈ σ(p) then σ(q) = σ(p).
Exercise 34. Is the last part of the previous proposition true for σn -products? If
not, how would you correct the statement?
Proposition 35. Let each Xa be a topological space, P bears the Tychonoff product
topology, and p ∈ P . Then σ(p) (and hence Σ(p) is dense in P .
Proposition 36. Let for each a ∈ A, Xa be compact T1 (and P be equipped with
the Tychonoff product topology and thus compact). Then:
(1) For every n ∈ N, σn (p) is closed in P and hence compact.
(2) Therefore σ(p) is σ-compact19 and hence Lindelöf.
(3) Σ(p) is countably compact.
Sketch of proof. (1) It is easier to show that the compliment P \ σn (p) is open.
(3) We need to show that every infinite subset C ⊂ Σ(p) has a limit point in
Σ(p). Without loss of generality, we can assume that C is countably infinite. For
every c ∈ S
C there is a countable Ac ⊂ A such that c(a) = p(a) for every a 6∈ Ac .
Put A = c∈C Ac . Then A is countable. Put KA = {x ∈ P : x(a) = p(a) for all
a 6∈ A}. Then:
• C ⊂ KA ;
• KA ⊂ Σ(p);
• KA is closed in P .
Since P is compact and thus countably compact, C has a limit point, say z, in P .
Then z ∈ C ⊂ KA = KA ⊂ Σ(p). 2
Example 37. A countably compact space which is not compact.
Let D = {0, 1}, and p ∈ DA where A is uncountable. Then X = Σ(p) is
countably compact by Proposition 36, part 3. On the other hand, X is a proper
dense (by Proposition 35) subspace of P and thus can’t be compact (because a
compact subspace of a Hausdorff space must be closed). 2
18This “if” becomes “iff” as soon as all factors X are > 1 point.
a
19If P is some property, then σ-P means being the union of countably many subspaces each
of which has property P.
7
Exercise 38. Show that a Σ-product in DA is sequentially compact.
Recall that a function f : X → R is bounded if there is a constant M such
that |f (x)| ≤ M for all x ∈ X. A topological space X is pseudocompact if every
continuous function f : X → R is bounded.
Proposition 39. Every countably compact space is pseudocompact.
Proof. Let f : X → R be a continuous function which is not bounded. Then
{f −1 ((−n, n)) : n ∈ N} is a countable open cover of X that does not have a finite
subcover. 2
So we have: compact ⇒ (countably compact) ⇒ pseudocompact.
Example 37 shows that the first of this arrows can’t be reversed. The same can
be said about the second one:
Example 40. Let R be an infinite, maximal almost disjoint family of infinite
subsets of a countably infinite set N . Then the Ψ-space20 Ψ(R) = N ∪ R is
pseudocompact but not countably compact.
Exercise 41. What can you say about Ψ(R) if R is finite?
As we know, a closed subspace of a compact space is compact.
Exercise 42.
• Is a closed subspace of a countably compact space necessarily
countably compact?
• Is a closed subspace of a pseudocompact space necessarily pseudocompact?
• Is a closed subspace of a sequentially compact space necessarily sequentially
compact?
• Is a closed subspace of a Lindelöf space necessarily Lindelöf?
20See Section 5 of the notes “Metric and metrizable spaces”