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Math 525 Some notes on compactness (sec 26)
Compactness
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Let X be a set, A ⊆ X, and U ⊆ 2X . U is a cover of A iff A ⊆ U. If U ′ ⊆ U and
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A ⊆ U ′ , then U ′ is called a subcover of A (relative to U); if U ′ consisits of only finitely
many sets, it is called a finite subcover of A. If the members of U are open relative to a
given topology on X, then U is called an open cover of A. Obviously, any subcover of an
open cover would also be an open cover.
Def 1 Let (X, τ ) be a topological space, A ⊆ X. Then A is τ -compact iff every open cover
U of A has a finite subcover. The space (X, τ ) is compact iff X is τ -compact.
One may check that subset A of (X, τ ) is compact iff subspace (A, τA ) is compact.
Prop A51 The continuous image of a compact set is compact.
Proof : Let A ⊆ (X, τ ) be compact, with f : (X, τ ) → (Y, µ) continuous. Let W be an
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open cover of f (A). Then U = f −1 (W) = {f −1 (W ) ¯ W ∈ W} is an open cover of A,
and by the compactness of A, U must have a finite subcover U ′ . Finally, the collection
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W ′ = f (U ′ ) = {f (U ) ¯ U ∈ U ′ } is finite, covers f (A), and is a subcollection of sets from W.
Hence W ′ is a finite open subcover of f (A), and so f (A) is compact.
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Lemma 1 Let (X, τ ) be T2 , A a compact set in X, y ∈ X − A. Then there are disjoint open
sets U and V such that A ⊆ U and y ∈ V .
Proof : By T2 (and the Axiom of Choice), for every x ∈ A, we can choose disjoint nbhds Ux
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and Vx such that y ∈ Ux and x ∈ Vx . U = {Vx ¯ x ∈ A} is an open cover of A, so there is
a finite subcover U ′ = {Vx1 , Vx2 , . . . , Vxn }, where each Vxi is a nbhd of some xi ∈ A. Then
n
n
\
[
U=
Uxi is a nbhd of y, and V =
Vxi is an open set containing A, and by construction,
i=1
U ∩ V = φ.
i=1
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Prop A52 Let (X, τ ) be a compact topological space, A ⊆ X.
(a) If A is closed, then A is compact.
(b) If (X, τ ) is T2 , then A ⊆ X is compact iff A is closed.
Proof :
(a)
Let A ⊆ X be closed. Let U be an open cover of A. Then U ∪ {X − A} is an open
cover of X, which (by compactness of X) must have a finite subcover U ′ , and U ′ − {X − A}
is a finite open subcover of A.
(b) Let (X, τ ) be T2 . From (a), we already have (A closed) ⇒ (A compact). Suppose A is
compact. By the previous Lemma, for every y 6∈ A, there exist disjoint nbhds Uy , Vy with
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y ∈ Vy and A ⊆ Uy . Then X − A =
Vy , which is an open set, and hence A is closed. ¥
y∈X−A
In analysis, one becomes accustomed to the notion that compact sets are closed, and this is
indeed true in analysis, since all spaces considered are metrizable, and hence T2 . However, in
an indiscrete space, all subsets are compact, but only φ and X are closed. In any topological
space, all finite sets (including φ) are compact. In a discrete space, φ and X are the only
compact sets.
Prop A53 Let X be a set and µ ≤ τ in Π(X). If A ⊆ X is τ -compact, then A is µ-compact.
In particular, if (X, τ ) is compact, so is (X, µ).
Proof : Let U be a µ-open cover of A. Then U is also a τ -open cover of A, and hence has a
finite subcover U ′ , which is also a µ-open subcover, since all members of U are µ-open.
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Examples For our seven standard topologies on R, the only ones which are compact are τcof
and τi , and in both cases, every subset of R is compact. Let’s discuss this a bit.
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To see that (R, τcoc ) is not compact, for each n ∈ N, let An = {n + k ¯ k ∈ N}, and let
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Un = R − An . Then each Un is open, and U = {Un ¯ n ∈ N} is a τcoc -open cover of R with
no finite subcover. Note also that if D ⊆ R is denumerable, then D also fails to be τcoc compact, since the subspace topology on D inherited from τcoc is discrete, and no infinite
discrete space can be compact. Thus (R, τcoc ) has in common with any discrete space the
property that the only compact subsets are finite sets.
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(R, τo ), where τo = {φ, R} ∪ {(−∞, a) ¯ a ∈ R} is not compact, since U = {(−∞, a) ¯ a ∈ R}
is a τo -open cover of R with no finite subcover. Sets of the form [a, b] and (−∞, a] are
τo -compact, but the τo -closed sets of the form [a, ∞) are not.
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(R, τu ) is not compact (consider the open covering U = {(n−0.6, n+0.6) ¯ n ∈ Z}). We know
from analysis that the compact subsets of (R, τu ) are thaose which are closed and bounded
(relative to the usual metric d(x, y) = |x − y|).
Since τu < τs , (R, τs ) is also not compact. Somewhat surprisingly, the set [0, 1] is not τs ¯
compact, since the collection U = {[0, 21 − n1 ) ¯ n ≥ 3}∪{[ 21 , 32 )} is a τs -open cover of [0, 1] with
no finite subcover. The same reasoning applies to show that no interval in R is τs -compact.
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However (R, τs ) does have some infinite compact subsets, like A = { n1 ¯ n ∈ N} ∪ {0}.
Prop A54 Let f : (X, τ ) → (Y, µ) be continuous and bijective, where (X, τ ) is compact and
(Y, µ) is T2 . Then f is a homeomorphism.
Proof : It suffices to prove that f −1 : (Y, µ) → (X, τ ) is continuous, which can be accomplished by showing that (f −1 )−1 (A) = f (A) is µ-closed whenever A is τ -closed. If A is
τ -closed, then A is τ -compact by A52(a), f (A) is µ-compact by A51, and hence f (A) is
µ-closed by A52(b).
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Here’s an interesting consequence of A54. Suppose (X, τ ) is compact and T2 . We showed
in A53 that µ ≤ τ ⇒ (X, µ) is compact, and τ ≤ µ ⇒ (X, µ) is T2 . If µ < τ , then (X, µ)
is compact but not T2 , for if (X, µ) were T2 , then idX : (X, τ ) → (X, µ) would be a homeomorphism, and by problem 3 in §18, µ = τ . Likewise, if τ < µ, then (X, µ) is T2 but not
compact, for if (X, µ) were compact, idX : (X, µ) → (X, τ ) would be a homeomorphism, and
again µ = τ . In summary, if (X, τ ) is compact and T2 , then under any coarser topology, the
space will not be T2 , and under any finer topology, the space will not be compact.
Tychanov Theorem If (X, τ ) =
Y
(Xi , τi ), then (X, τ ) is compact iff (Xi , τi ) is compact for
i∈I
all i ∈ I.
Proof : The forward implication follows because the projection maps pi from the product
space to each (Xi , τi ) are continuous, and the continuous image of a compact space is compact. The proof of the reverse implication will be postponed. Meantime, we’ll assume the
validity of Tychanov Theorem, since it has many important consequences.
A familiar result which depends on compactness is
The Extreme Value Theorem
Let f : (X, τ ) → (Y, µ) be continuous, where Y is a totally ordered set and µ is the order
topology. If X is compact, then there exist points c and d in X such that f (c) ≤ f (x) ≤ f (d)
for all x ∈ X.
Proof : Since f is continuous and X is compact, A = f (X) is compact. We now show
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that A has a largest element M . If not, then the collection U = {(−∞, a) ¯ a ∈ A}
forms an open covering of A.
By compactness of A, there must be a finite subcover
U ′ = {(−∞, a1 ), (−∞, a2 ), . . . , (−∞, an )}. If ai is the largest of the elements a1 , a2 , . . . , an ,
then ai belongs to none of the sets in U ′ , which is a contradiction, since ai must be in A.
It can be similarly shown that A = f (X) has a smallest element m. The conclusion of the
theorem follows, since M and m are in f(X), which means there exist c, d ∈ X such that
f (c) = m, f (d) = M , and all f (x) in f (X) must be in [m, M ].
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