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COMPACTNESS VS. SEQUENTIAL COMPACTNESS
PMAMC&OC SASMS FALL 2004
CHRIS ALMOST
Definitions
Definition. A topological space is a pair (X, τ ) where X is a set and τ is a collection
of subsets of X that satisfies the following properties:
(1)
(2)
(3)
(4)
X∈τ
∅∈τ
S
If {Uα }α∈A is a collection of sets from τ then α∈A Uα ∈ τ .
If U, V ∈ τ then U ∩ V ∈ τ .
The sets in τ are refered to as the open sets.
We often decline to explicitly name the collection of open sets when working
with topological spaces.
Definition. A topological space X is compact if whenever {Uα }α∈A is a collection
of open sets and
[
X=
Uα
α∈A
then there exist α1 , . . . , αk ∈ A such that
X=
k
[
Uαi
i=1
This definition is usually stated succinctly as “X is compact if every open cover
of X admits a finite subcover”.
Definition. A topological space X is sequentially compact if whenever (xn )∞
n=1 is a
sequence in X then (xn )∞
has
a
convergent
subsequence.
(Recall
that
a
sequence
n=1
(xn )∞
n=1 converges to a point x ∈ X if for any open set U , x ∈ U implies that there
exists NU such that xn ∈ U for all n ≥ NU .)
Sequentially Compact but not Compact
Definition. Given a linearly ordered set (P, <), define the order topology on P to
be the topology generated by the collection of sets of the form
{x ∈ P | x < a} and {x ∈ P | x > a}
where a ∈ P . It follows that the open sets of this topology are unions of open
intervals.
Date: November 17, 2004.
1
2
CHRIS ALMOST
The order topology on (R, <) corresponds with the usual metric topology on the
real line. This is the topology studied in elementary calculus courses.
Let Ω denote the first uncountable ordinal, so
Ω = {α | α is a countable ordinal}
Imbue Ω with the order topology. Consider the collection of open sets
{{x ∈ Ω | x < α}}α∈Ω
Then clearly
[
Ω=
{x ∈ Ω | x < α}
α∈Ω
Given any countable collection of ordinals {α1 , α2 , . . . } ⊆ Ω, the union
∞
[
{x ∈ Ω | x < αn }
n=1
is not equal to Ω since it is countable (each of the sets in the union is countable).
It follows that no finite collection of these sets will cover Ω, so Ω is not compact.
Definition. Let (xn )∞
n=1 be a sequence in a linearly ordered set (P, <). An index
n ≥ 1 is a peak point of (xn )∞
n=1 if xn ≥ xk for all k ≥ n.
Lemma. (Peak Point Lemma) Let (xn )∞
n=1 be a sequence in a linearly ordered set
(P, <). Then (xn )∞
n=1 has a monotone subsequence.
See Spivak’s Calculus.
∞
Proof. If (xn )∞
n=1 has infinitely many peak points n1 , n2 , . . . then (xnk )k=1 is a
∞
monotone non-increasing subsequence. Otherwise suppose that (xn )n=1 has finitely
many peak points n1 < · · · < nk . Let m1 > nk . Then m1 is not a peak point,
so there is m2 > m1 such that xm1 ≤ xm2 . Similarily, m2 is not a peak point, so
there is m3 > m2 such that xm2 ≤ xm3 . Continuing this this process we obtain a
monotone non-decreasing subsequence (xmk )∞
k=1 .
∞
Let (αn )∞
n=1 be a sequence in Ω. I will show that (αn )n=1 has a convergent
subsequence, and thus Ω is sequentially compact. By the Peak Point lemma (αn )∞
n=1
has a monotone subsequence (αnk )∞
k=1 . If it is non-decreasing then let
α=
∞
[
αnk
k=1
otherwise let α be the smallest element of the set {αn1 , αn2 , . . .} (there is a smallest
element because Ω is well-ordered). Then (xnk )∞
k=1 converges to α, since for any
open interval around α, say {x ∈ Ω | γ1 < x < γ2 }, by defintion of α and because
{xnk }∞
k=1 is monotone, the sequence will eventually be inside of the interval. Thus
Ω is sequentially compact.
Compact but not Sequentially Compact
Definition. Given a collection {Xα }α∈A of topological spaces, the product space is
Y
Xα = {(xα )α∈A | α ∈ A and α ∈ Xα }
α∈A
PMAMC&OC
3
The product topology on the product space is the topology generated by the sets
{p−1
α (U ) | α ∈ A and U ⊆ Xα is open}
The only fact about the product topology that we will need for this talk is that
a sequence (x(n) )∞
n=1 converges in the product topology if and only if the sequence
(n)
(xα )∞
converges
for each α ∈ A.
n=1
Theorem. (Tychonov) Let {Xα }α∈A be a collection of compact topological spaces.
Then
Y
X :=
Xα
α∈A
is compact in the product topology.
Let X = [0, 1][0,1] , the set of all funtions from [0, 1] to [0, 1] (with the usual
topology on [0, 1] inherited from R). [0, 1] is compact by the Heine-Borel Theorem,
so X is compact in the product topology by Tychonov’s Theorem.
Define a sequence (fn )∞
n=1 in X by
fn (x) = the nth digit in the binary expansion of x
(If a number could have 2 different binary expansions, favor the one that ends in
an infinite string of zeroes over the one that ends in an infinite string of one’s.)
∞
∞
Suppose that (fnk )∞
k=1 is a convergent subsequence of (fn )n=1 . Then (fnk (x))k=1
converges for each x ∈ [0, 1]. Define y ∈ [0, 1] by setting
(
1 if n = nk and k is even
th
n digit of y =
0 otherwise
Then (fnk (y))∞
k=1 does not converge. This contradiction shows that X is not sequentially compact.