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MAT 410 Intro to General Topology
Spring 2015 (2008)
Counterexamples for different notions of compactness
Definitions. A topological space X is
• compact if every open cover has a finite subcover,
• limit point compact if every infinite subset of X has a limit point in X, and
• sequentially compact if every sequence in X has a subsequence that converges in X.
Theorem:
• Every compact space is limit point compact.
• Every sequentially compact space is limit point compact.
• If X is metrizable, then the three notions of compactness are equivalent.
Corollary:
• If a space has one of the compactness properties, but not all three, then it is not metrizable.
Exercise: What other necessary or sufficient conditions for metrizability do you know?
Counterexamples:
• The space I I is compact but not sequentially compact.
• The space SΩ is sequentially compact (and hence limit point compact) but not compact.
Proofs:
• Since I is compact, the space I I , as a product of compact spaces, is again compact by
Tychonoff’s theorem (§ 37, look forward to the last class of the semester).
• For every x ∈ I consider
binary expansion that does not have an infinite tail of
P its unique
−n for uniquely defined functions f : I 7→ {0, 1}, n ∈ Z+ .
f
(x)2
ones. Formally, x = ∞
n
n=1 n
Use the same symbol to denote the function with codomain I, i.e. consider each fn as an
element of I I . Note that the graph of each fn is a piecewise constant function that alternates
between the value 0 and 1 on the intervals [(j − 1)2−n , j2−n ), 1 ≤ j ≤ 2n .
Let {fnk }∞
k=1 be any subsequence. To show that this subsequence does not converge in the
product topology of I I , which is the topology of pointwise convergence, it suffices to find one
point z such
the sequence {fnk (z)}∞
k=1 ⊆ I does not converge in I.
P∞ that
−n
2j−1
. Verify that if k = 2j − 1 ∈ Z+ is odd then fn2j−1 (z) = 1 whereas if
Let z = j=1 2
+
k = 2j ∈ Z is even then fn2j (z) = 0. The sequence {fnk (z)}∞
k=1 alternates between 0 and
∞
1 and hence it does not converge. Consequently {fnk }k=1 does not converge in I I . Since
∞
∞
{fnk }∞
k=1 was an arbitrary subsequence of {fn }n=1 this shows that {fn }n=1 has no converging
I
subsequence and therefore I is not sequentially compact.
+
• Suppose {xn }∞
n=1 ⊆ SΩ is an infinite sequence. Since the set A = {xn : n ∈ Z } of its values
is countable, it has an upper bound α ∈ SΩ , i.e., for every n ∈ Z+ , xn ≤ α.
Write a0 for the smallest element of SΩ . Recall that as a well-ordered set SΩ has the least
upper bound property and hence by theorem 27.1 every closed interval in SΩ is compact. In
particular the interval [a0 , α] ⊆ SΩ is compact, and therefore limit point compact.
Using a powerful machinery of theorems to be proven soon, one can show that Sα = [a0 , α]
is metrizable, and hence also sequentially compact. Therefore the sequence {xn }∞
n=1 has
a subsequence that converges in Sα , and thus also converges in SΩ , showing that SΩ is
sequentially compact.
The theorems that fill the gap establish that as a compact Hausdorff space SΩ is regular,
hence so are its subspaces SΩ and Sα = [a0 , α]. The space Sα = [a0 , α] is countable and
also second countable. By Urysohn’s theorem it follows that Sα is metrizable and hence
compactness of Sα is equivalent to sequential compactness of Sα .
• Using theorem 26.1 the set SΩ is not compact since it is not a closed subset of the compact
Hausdorff space SΩ .
Directly, the nested collection of sections {Sα : α ∈ SΩ } is an
open cover of SΩ that has no finite subcover since SΩ has no largest element.
All rights reserved, Matthias Kawski
http: // math. asu. edu/ ~ kawski
Last updated (minor edits) April 13, 2015
First draft: April 7, 2008.