Download PDF

closed subsets of a compact set are
compact∗
Wkbj79†
2013-03-21 16:36:19
Theorem 1. Suppose X is a topological space. If K is a compact subset of X,
C is a closed set in X, and C ⊆ K, then C is a compact set in X.
The below proof follows e.g. [?]. A proof based on the finite intersection
property is given in [?].
Proof. Let I be an indexing set and F = {Vα | α ∈ I} be an arbitrary open
cover for C. Since X \C is open, it follows that F together with X \C is an open
cover for K. Thus, K can be covered by a finite number of sets, say, V1 , . . . , VN
from F together with possibly X \ C. Since C ⊂ K, V1 , . . . , VN cover C, and it
follows that C is compact.
The following proof uses the finite intersection property.
Proof. Let I be an indexing set and {Aα }α∈I be \
a collection of X-closed sets
contained in C such that, for any finite J ⊆ I,
Aα is not empty. Recall
α∈J
that, for every α ∈ I, Aα ⊆ C ⊆ K. Thus, for every α ∈ I, Aα = K ∩ Aα .
Therefore, {Aα }α∈I
\ are K-closed subsets of K (see this page)
\ such that, for
any finite J ⊆ I,
Aα is not empty. As K is compact,
Aα is not empty
α∈J
α∈I
(again, by this result). This proves the claim.
References
[1] J.L. Kelley, General Topology, D. van Nostrand Company, Inc., 1955.
[2] S. Lang, Analysis II, Addison-Wesley Publishing Company Inc., 1969.
∗ hClosedSubsetsOfACompactSetAreCompacti created: h2013-03-21i by: hWkbj79i version: h34691i Privacy setting: h1i hTheoremi h54D30i
† This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
1
[3] G.J. Jameson, Topology and Normed Spaces, Chapman and Hall, 1974.
[4] I.M. Singer, J.A. Thorpe, Lecture Notes on Elementary Topology and Geometry, Springer-Verlag, 1967.
2