Solutions To Exercises In Munkres EX 26 S26
User Manual: EX-26
Open the PDF directly: View PDF 
.
Page Count: 3
1st December 2004
Munkres §26
Ex. 26.1 (Morten Poulsen).
(a).Let Tand T0be two topologies on the set X. Suppose T0⊃ T .
If (X, T0) is compact then (X, T) is compact: Clear, since every open covering if (X, T) is an
open covering in (X, T0).
If (X, T) is compact then (X, T) is in general not compact: Consider [0,1] in the standard
topology and the discrete topology.
(b).
Lemma 1. If (X, T)and (X, T0)are compact Hausdorff spaces then either Tand T0are equal
or not comparable.
Proof. If (X, T) compact and T0⊃ T then the identity map (X, T0)→(X, T) is a bijective
continuous map, hence a homeomorphism, by theorem 26.6. This proves the result.
Finally note that the set of topologies on the set Xis partially ordered, c.f. ex. 11.2, under
inclusion. From the lemma we conclude that the compact Hausdorff topologies on Xare minimal
elements in the set of all Hausdorff topologies on X.
Ex. 26.2 (Morten Poulsen).
(a).The result follows from the following lemma.
Lemma 2. If the set Xis equipped with the finite complement topology then every subspace of X
is compact.
Proof. Suppose A⊂Xand let Abe an open covering of A. Then any set A0∈ A will covering all
but a finite number of points. Now choose a finite number of sets from Acovering A−A0. These
sets and A0is a finite subcovering, hence Acompact.
(b).Lets prove a more general result: Let Xbe an uncountable set. Let
Tc={A⊂X|X−Acountable or equal X}.
It is straightforward to check that Tcis a topology on X. This topology is called the countable
complement topology.
Lemma 3. The compact subspaces of Xare exactly the finite subspaces.
Proof. Suppose Ais infinite. Let B={b1, b2, . . .}be a countable subset of A. Set
An= (X−B)∪ {b1, . . . , bn}.
Note that {An}is an open covering of Awith no finite subcovering.
The lemma shows that [0,1] ⊂Rin the countable complement topology is not compact.
Finally note that (X, Tc) is not Hausdorff, since no two nonempty open subsets Aand Bof X
are disjoint: If A∩B=∅then X−(A∩B) = (X−A)∪(X−B), hence Xcountable, contradicting
that Xuncountable.
Ex. 26.3 (Morten Poulsen).
Theorem 4. A finite union of compact subspaces of Xis compact.
Proof. Let A1, . . . , Anbe compact subspaces of X. Let Abe an open covering of Sn
i=1 Ai. Since
Aj⊂Sn
i1Aiis compact, 1 ≤j≤n, there is a finite subcovering Ajof Acovering Aj. Thus
Sn
j=1 Ajis a finite subcovering of A, hence Sn
i=1 Aiis compact.
1
2
Ex. 26.5. For each a∈A, choose [Lemma 26.4] disjoint open sets Ua∈aand Va⊃B. Since A
is compact, Ais contained in a finite union U=U1∪ · · · ∪ Unof the Uas. Let V=V1∩ · · · Vnbe
the intersection of the corresponding Vas. Then Uis an open set containing A,Vis an open set
containing B, and Uand Vare disjoint as U∩V=SUi∩V⊂SUi∩Vi=∅.
Ex. 26.6. Since any closed subset Aof the compact space Xis compact [Thm 26.2], the image
f(A) is a compact [Thm 26.5], hence closed [Thm 26.3], subspace of the Hausdorff space Y.
Ex. 26.7. This is just reformulation of The tube lemma [Lemma 26.8]: Let Cbe a closed subset
of X×Yand x∈Xa point such that the slice {x} × Yis disjoint from C. Then, since Yis
compact, there is a neighborhood Wof xsuch that the whole tube W×Yis disjoint from C.
In other words, if x6∈ π1(C) then there is a neighborhood Wof xwhich is disjoint from π1(C).
Thus The tube lemma says that π1:X×Y→Xis closed when Yis compact (so that π1is an
example of a perfect map [Ex 26.12]). On the other hand, projection maps are always open [Ex
16.4].
Ex. 26.8. Let G⊂X×Ybe the graph of a function f:X→Ywhere Yis compact Hausdorff.
Then
Gis closed in X×Y⇔fis continuous
⇐: (For this it suffices that Ybe Hausdorff.) Let (x, y)∈X×Ybe a point that is not in the
graph of f. Then y6=f(x) so by the Hausdorff axiom there will be disjoint neighborhoods V3y
and W3f(x). By continuity of f,f(U)⊂W⊂Y−V. This means that (U×V)∩G=∅.
⇒: Let Vbe a neighborhood of f(x) for some x∈X. Then G∩(X×(Y−V)) is closed in X×Y
so [Ex 26.7] the projection π1(G∩(X×(Y−V))) is closed in Xand does not contain x. Let U
be a neighborhood of Xsuch that U×Ydoes not intersect G∩(X×(Y−V)). Then f(U) does
not intersect Y−V, or f(U)⊂V. This shows that fis continuous at the arbitrary point x∈X.
Ex. 26.12. (Any perfect map is proper; see the January 2003 exam for more on proper maps.)
Let p:X→Ybe closed continuous surjective map such that p−1(y) is compact for each y∈Y.
Then p−1(C) is compact for any compact subspace C⊂Y.
For this exercise we shall use the following lemma.
Lemma 5. Let p:X→Ybe a closed map.
(1) If p−1(y)⊂Uwhere Uis an open subspace of X, then p−1(W)⊂Ufor some neighborhood
W⊂Yof y.
(2) If p−1(B)⊂Ufor some subspace Bof Yand some open subspace Uof X, then p−1(W)⊂
Ufor some neighborhood W⊂Yof B.
Proof. Note that
p−1(W)⊂U⇔p(x)∈W⇒x∈U⇔x6∈ U⇒p(x)6∈ W⇔p(X−U)⊂Y−W
⇔p(X−U)∩W=∅
(1) The point ydoes not belong to the closed set p(X−U). Therefore a whole neighborhood
W⊂Yof yis disjoint from p(X−U), i.e. p−1(W)⊂U.
(2) Each point y∈Bhas a neighborhood Wysuch that p−1(Wy)⊂U. The union W=SWyis
then a neighborhood of Bwith p−1(W)⊂U.
We shall not need point (2) here.
Let C⊂Ybe compact. Consider a collection {Uα}α∈Jof open sets covering of p−1(C).
For each y∈C, the compact space p−1(y) is contained in a the union of a finite subcollection
{Uα}α∈J(y). There is neighborhood Wyof ysuch that p−1(Wy) is contained in this finite union. By
compactness of C, finitely many Wy1, . . . , Wykcover Y. Then the finite collection Sk
i=1{Uα}α∈J(yi)
cover p−1(C). This shows that p−1(C) is compact.
Ex. 26.13. Let Gbe a topological group and Aand Bsubspaces of G.
3
(a).Aclosed and Bcompact ⇒AB closed
Assume c6∈ AB =Sb∈BAb. The regularity axiom for G[Suppl Ex 22.7] implies that there are
disjoint open sets Wb3cand Ub⊃Ab separating cand Ab for each point b∈B. Then A−1Ubis
an open neighborhood of b. Since Bis compact, it can be covered by finitely many of these open
sets A−1Ub, say
B⊂A−1U1∪ · · · ∪ A−1Uk=A−1U
where U=U1∪ · · · ∪ Uk. The corresponding open set W=W1∩ · · · ∩ Wkis an open neighborhood
of cthat is disjoint from AB since W∩AB ⊂SW∩Ui⊂SWi∩Ui=∅.
(b).Hcompact subgroup of G⇒p:G→G/H is a closed map
The saturation AH of any closed subset A⊂Gis closed by (a).
(c).Hcompact subgroup of Gand G/H compact ⇒Gcompact
The quotient map p:G→G/H is a perfect map because it is a closed map by (b) and has compact
fibres p−1(gH) = gH. Now apply [Ex 26.12].
References
