Willard - General Topology (Solutions)

7/22/2019 Willard - General Topology (Solutions)

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General Topology

A Solution Manual forWillard(2004)

Jianfei Shen

School of Economics, The University of New South Wales

Sydney, Australia October 15, 2011


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Preface

Sydney, Jianfei Shen

October 15, 2011

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Acknowledgements

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Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

1 Set Theory and Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Neighborhoods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Bases and Subbases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 New Spaces from Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.1 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 Continuous Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3 Product Spaces, Weak Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.4 Quotient Spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.1 Inadequacy of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2 Nets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.3 Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Separation and Countability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.1 The Separation Axioms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.2 Regularity and Complete Regularity. . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5.3 Normal Spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.4 Countability Properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

6 Compactness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.1 Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

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Acronyms

R the set of real numbers

I 0; 1

P RX

Q

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1SET THEORY AND METRIC SPACES

1.1 Set Theory

1A. Russells Paradox

I Exercise 1. The phenomenon to be presented here was first exhibited by

Russell in 1901, and consequently is known asRussells Paradox.

Suppose we allow as sets things A for which A2 A. LetP be the set of allsets. Then P can be divided into two nonempty subsets, P1D

A 2 P W A A

and P2 DfA 2 P W A 2 Ag. Show that this results in the contradiction: P1 2P1 () P1 P1. Does our (naive) restriction on sets given in 1.1 eliminate thecontradiction?

Proof. IfP12 P1, then P12 P2, i.e., P1 P1. But ifP1 P1, then P12 P1. Acontradiction.

ut

1B. De Morgans laws and the distributive laws

I Exercise2. a. A X T2 B D S2 .A X B/.b. B[ T2 B D T2.B[ B/.c. IfAnm is a subset ofA fornD 1; 2 ; : : :andmD 1; 2 ; : : :, is it necessarily true

that1

[nD12

41

\mD1Anm

3

5D

1

\mD12

41

[nD1Anm

3

5

Proof. (a) If x2 AX T2 B, then x2 A and x T2 B; thus, x2 Aandx B for some , so x2 .A X B/ for some ; hence x2

S2 .A X B/.

On the other hand, if x2 S2 .A X B/, then x2 AXB for some 2 ,i.e.,x2 A and x B for some 2 . Thus, x2 A and x

T2 B; that is,

x2 A X T2 B.

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2 CHAPTER 1 SET THEORY AND METRIC SPACES

(b) Ifx2B[ T2 B, then x2B for all, then x2.B[ B/ for all, i.e.,x2 T2.B[ B/. On the other hand, ifx2 T2.B[ B/, thenx2.B[ B/for all, i.e.,x2 B orx2 B for all; that is, x2 B[

T2 B

.

(c) They are one and the same set. ut

1C. Ordered pairs

I Exercise 3. Show that, if .x1; x2/ is defined to befx1g; fx1; x2g, then

.x1; x2/ D .y1; y2/iffx1D y1 andx2D y2.

Proof. If x1 D y1 and x2 D y2, then, clearly, .x1; x2/ Dfx1g; fx1; x2g Dfy1g; fy1; y2gD .y1; y2/. Now assume that fx1g; fx1; x2gD fy1g; fy1; y2g.

Ifx1 x2, thenfx1g D fy1g andfx1; x2g D fy1; y2g. So, first, x1D y1 and thenfx1; x2g D fy1; y2g implies thatx2D y2. Ifx1D x2, then

fx1g; fx1; x1gDfx1g

.

Sofy1g D fy1; y2g D fx1g, and we get y1D y2D x1, so x1D y1 and x2D y2holds in this case, too. ut

1D. Cartesian products

I Exercise 4. Provide an inductive definition of the orderedn-tuple .x1; : : : ; xn/

of elementsx1; : : : ; xn of a set so that.x1; : : : ; xn/ and.y1; : : : ; yn/ are equal iff

their coordinates are equal in order, i.e., iffx1D y1; : : : ; xnD yn.

Proof. Define.x1; : : : ; xn/ D f.1;x1/ ; : : : ; . n ; xn/gas a finite sequence. ut

I Exercise5. Given setsX1; : : : ; X n define the Cartesian productX1 Xna. by using the definition of orderedn-tuple you gave inExercise 4,

b. inductively from the definition of the Cartesian product of two sets,

and show that the two approaches are the same.

Proof. (a)X1 XnD ff2 .Sn

iD1 Xi /n W f.i/ 2 Xig.

(b) From the definition of the Cartesian product of two sets, X1 XnDf.x1; : : : ; xn/ W xi2 Xi g, where.x1; : : : ; xn/ D ..x1; : : : ; xn1/; xn/.

These two definitions are equal essentially since there is a bijection between

them.

utI Exercise6. Given setsX1; : : : ; X nletXD X1 Xnand letX be the set ofall functionsf from f1 ; : : : ; ng intoSnkD1 Xk having the property thatf.k/ 2 Xkfor eachkD 1 ; : : : ; n. Show thatX is the same set asX.

Proof. Each function f can be written as f.1;x1/ ; : : : ; . n ; xn/g. So define FWX !X asF.f / D .x1; : : : ; xn/. ut


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SECTION 1.2 METRIC SPACES 3

I Exercise7. Use what you learned inExercise 6to define the Cartesian prod-

uctX1 X2 of denumerably many sets as a collection of certain functionswith domain N.

Proof. X1X2 consists of functionsfW N ! S1nD1 Xnsuch thatf.n/ 2 Xnfor alln 2 N. ut

1.2 Metric Spaces

2A. Metrics onRn

I Exercise8. Verify that each of the following is a metric on Rn:

a. .x;y/Dp nXiD1.xi yi /2.

b. 1.x;y/ DPn

iD1 jxi yi j.

c. 2.x;y/ D maxfjx1 y1j; : : : ; jxn ynjg.

Proof. Clearly, it suffices to verify the triangle inequalities for all of the three

functions. Pick arbitrary x ;y; z 2 Rn.(a) By Minkowskis Inequality, we have

.x; z/ Dp nXiD1.xi zi /2 Dp n

XiD1.xi yi / C .yi zi /2

6

p nXiD1

.xi yi /2 Cp

nXiD1

.yi zi /2

D .x;y/ C .y; z/:

(b) We have

1.x; z/ DnXiD1

jxi zi j DnXiD1

.jxi yi j C jyi zi j/ D 1.x;y/ C 1.y; z/:

(c) We have

2.x; z/ D maxfjx1 z1j; : : : ; jxn znjg6 maxfjx1 y1j C jy1 z1j; : : : ; jxn ynj C jyn znjg6 maxfjx1 y1j; : : : ; jxn ynjg C maxfjy1 z1j; : : : ; jyn znjgD 2.x;y/ C 2.y; z/: ut


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2B. Metrics onC.I/

I Exercise9. LetC.I/denote the set of all continuous real-valued functions on

the unit intervalI and letx0 be a fixed point ofI .

a. .f; g/ D supx2I jf.x/ g.x/jis a metric on C.I/.

b. .f;g/ D R10jf.x/ g.x/j dx is a metric on C.I/.

c. .f; g/ D jf .x0/ g.x0/jis a pseudometric on C.I/.

Proof. Let f; g; h2 C.I/. It is clear that , , and are positive, symmetric; itis also clear that and satisfy M-b.

(a) We have

.f; h/ D supx2I

jf.x/ h.x/j 6 supx2I

.jf.x/ g.x/j C jg.x/ h.x/j/

6 supx2I

jf.x/ g.x/j C supx2I

jg.x/ h.x/j

D .f;g/ C .g; h/:

(b) We have

.f;h/ DZ 10

jf.x/ h.x/j 6Z 10

jf.x/ g.x/j CZ 10

jg.x/ h.x/j

D .f;g/ C .g; h/:

(c) For arbitraryf; g2 C.I/withf .x0/ D g.x0/we have.f; g/ D 0, so .f; g/ D0does not imply that fD g. Further,.f; h/ D jf .x0/h.x0/j 6 jf .x0/g.x0/jCjg.x0/ h.x0/j D .f;g/ C .g;h/. ut

2C. Pseudometrics

I Exercise 10. Let .M;/ be a pseudometric space. Define a relation on Mbyx y iff.x; y/ D 0. Then is an equivalence relation.

Proof. (i) x x since .x;x/D 0 for all x2 M. (ii) x y iff.x;y/D 0 iff.y; x/D 0 iff y x. (iii) Suppose x y and y z. Then .x;z/ 6 .x;y/ C.y;z/ D 0; that is, .x; z/ D 0. Sox z. ut

I Exercise 11. IfM

is he set of equivalence classes in M under the equiva-lence relationand if is defined on M by.x; y/D.x; y/, then is awell-defined metric onM.

Proof. is well-defined since it does not dependent on the representative of

x: letx 0 2 x and y 0 2 y. Then

.x0; y0/ 6 .x0; x/ C .x;y/ C .y;y0/ D .x; y/:


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Symmetrically,.x; y/ 6 .x 0; y0/. To verify is a metric on M, it suffices to

show that satisfies the triangle inequality. Let x; y; z 2 M. Then

.x; z/ D .x;z/ 6 .x;y/ C .y;z/ D .x; y/ C .y; z/: ut

I Exercise 12. Ifh W M! M is the mappingh.x/Dx , then a setA in M isclosed (open) iffh.A/is closed (open) in M.

Proof. Let A be open in M andh.x/D x2 h.A/ for some x2 A. Since A isopen, there exist an "-disk U.x;"/ contained in A. For each y2 U.x;"/, wehaveh.y/Dy 2h.A/, and.x; y/D.x; y/ 6 ". Hence, for eachx2h.A/,there exists an "-diskU.x; "/ D h.U.x; "//contained in h.A/; that is, h.A/isopen in M. Since h is surjective, it is now easy to see that h.A/ is closed in

M wheneverA is closed in M. ut

I Exercise 13. Iff is any real-valued function on a setM, then the distance

functionf.x;y/ D jf.x/ f.y/jis a pseudometric on M.Proof. Easy. ut

I Exercise14. If.M;/is any pseudometric space, then a function fW M! Ris continuous iff each set open in .M;f/is open in.M; /.

Proof. Suppose thatfis continuous andG is open in.M;f/. For eachx2 G,there is an" > 0such that ifjf.y/f.x/j < "theny2 G. The continuity off atx implies that there exists > 0such that if.y;x/ < thenjf.y/ f.x/j< ",and soy2 U. We thus proved that for each x2 Uthere exists a-diskU.x;/contained inG ; that is, G is open in .M; /.

Conversely, suppose that each set is open in .M;/ whenever it is open in.M;f/. For each x2 .M;f/, there is an "-diskUf.x;"/ contained in M sinceMis open under f; then Uf.x;"/ is open in .M;/ since Uf.x;"/ is open in

.M;f/. Hence, there is an-diskU.x;/such that U.x;/ Uf.x;"/; that is,if.y; x/ < , thenjf.y/ f.x/j < ". Sof is continuous onM. ut

2D. Disks Are Open

I Exercise 15. For any subsetA of a metric space M and any" > 0, the set

U.A;"/ is open.

Proof. Let A M and " > 0. Take an arbitrary point x2 U.A; "/; take anarbitrary pointy2 A such that .x; y/ < ". Observe that every "-diskU.y;"/ iscontained inU.A;"/. Sincex2 U.y;"/andU.y;"/is open, there exists an-diskU.x;/contained in U.y;"/. Therefore,U.A;"/is open. ut


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2E. Bounded Metrics

I Exercise 16. If is any metric on M, the distance function .x;y/ Dmin

f.x;y;/;1

gis a metric also and is bounded.

Proof. To see is a metric, it suffices to show the triangle inequality. Let

x ; y; z2 M. Then

.x;z/ D minf.x;z/;1g 6 minf.x;y/ C .y;z/;1g6 minf.x;y/;1g C minf.y;z/;1gD .x;y/ C .y;z /:

It is clear that is bounded above by 1. ut

I Exercise 17. A function f is continuous on .M;/ iff it is continuous on

.M;/.

Proof. It suffices to show that and are equivalent. IfG is open in .M;/,

then for each x2 G there is an "-diskU.x;"/ G. Since U.x;"/ U.x;"/,we knowG is open in .M;/. Similarly, we can show that G is open in .M;/

whenever it is open in .M; /. ut

2F. The Hausdorff Metric

Let be a bounded metric on M; that is, for some constant A, .x; y/ 6 A for

allx and y inM.

I Exercise 18. Show that the elevation of to the power setP.M / as definedin 2.4 is not necessarily a pseudometric on P.M /.

Proof. Let M f.x1; x2/2 R2 W x21C x22 6 1g, and let be the usual metric.Then is a bounded metric on M. We show that the function W .E; F /7!infx2E;y2F.x;y/, for all E; F 2 P.M /, is not a pseudometric on P.M / byshowing that the triangle inequality fails. Let E ; F ; G 2 P.M /, where E DU..1=4; 0/;1=4/, G D U..1=4; 0/;1=4/, and F meets both E and G. Then.E; G/ > 0, but. E; F / D .F;G/ D 0. ut

I Exercise 19. LetF.M / be all nonempty closed subsets ofM and forA; B2F.M /define

dA.B/ D supf.A;x/ W x2 Bgd.A;B/ D maxfdA.B/;dB.A/g:

Then d is a metric on F.M / with the property that d.fxg; fyg/D .x;y/. It iscalled theHausdorff metricon F.M /.


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Proof. Clearly, d is nonnegative and symmetric. Ifd.A; B/D0, then dA.B/DdB.A/ D 0, i.e., supy2B.A;y / D supx2A .B;x/ D 0. But then.A;y/ D 0for ally2 B and .B;x/D 0 for all x2 A. Since A is closed, we have y2 A for ally

2B; that is, B

A. Similarly,A

B. Hence, A

DB .

We next show the triangle inequality ofd. Let A; B; C 2 F.M /. For an ar-bitrary pointa2 A, take a point b2 C such that .a; b/D .B; a/(since B isclosed, such a point exists). Then

.a; b/ 6 supx2A

.B; x/ D dB.A/ 6 d.A;B/:

For thisb2 B, we take a point c2 Csuch that.b; c/ 6 d.B;C/. Therefore,

.a;c/ 6 .a; b/ C .b;c/ 6 d.A;B/ C d.B;C/:

We thus proved that for every a2 A, there exists c2 C (depends on a), suchthat .a; c/ 6 d.A;B/

Cd.B;C/. In particular, we have

.a;C/ D infz2C

.a; z/ 6 d.A;B/ C d.B;C/:

Since the above inequality holds for all a2 A, we obtain

dC.A/ D supx2A

.a;C/ 6 d.A;B/ C d.B;C/: (1.1)

Similarly, for each c2 C there exists b2 B with .c;b/ 6 d.B;C /; for this b,there exists a2A with .a; b/ 6 d.A;B/. Hence .a; c/ 6 d.A; B/ C d.B;C/ forallc2 C. The same argument shows that

dA.C / 6 d.A;B/

Cd.B;C/: (1.2)

Combining(1.1) and(1.2) we get the desired result.

Finally, notice that dfxg.fyg/ D dfyg.fxg/ D .x;y/; hence, d.fxg; fyg/ D.x;y/. ut

I Exercise 20. Prove that closed sets A and B are close in the Hausdorff

metric iff they are uniformly close; that is, d.A;B/ < " iff A U.B;"/ andB U.A; "/.

Proof. Ifd.A; B/ < ", then supy2B.A;y/DA.B/ < "; that is, .A;y/ < " forally2 B, soB U.A; "/. Similarly, A U.B;"/.

Conversely, ifA

U.B;"/, then .B; x/ < " for allx

2A. SinceA is closed,

we have dB.A/ < "; similarly, B U.A; "/ implies that dA. B/ < ". Hence,d.A;B/ < ". ut


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2G. Isometry

Metric spaces .M; /and .N; / are isometriciff there is a one-one function f

from M onto N such that .x;y/

D .f.x/;f.y// for all x and y in M; f is

called anisometry.

I Exercise 21. If f is an isometry from M to N, then both f and f1 are

continuous functions.

Proof. By definition, f is (uniformly) continuous on M: for every " > 0, let

D "; then.x;y/ < implies that.f.x/;f.y// D .x;y/ < ".On the other hand, for every " > 0 and y2 N, pick the unique f1.y/2

M (since f is bijective). For each z 2 N with . y ; z / < ", we must have.f1.y/;f1.z//D.f.f1.y//;f.f1.z///D.y; z/ < "; that is, f1 is con-tinuous. ut

I Exercise22. Ris not isometric toR2 (each with its usual metric).

Proof. Consider S1 D f.x;y/2 R2 W x2 C y2 D 1g. Notice that there are onlytwo points aroundf1.0;0/with distance 1. ut

I Exercise 23. I is isometric to any other closed interval in R of the same

length.

Proof. Consider the functionfW I! a;a C 1defined byf .x/ D a C x for allx2 I. ut


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2TOPOLOGICAL SPACES

2.1 Fundamental Concepts

3A. Examples of Topologies

I Exercise 24. IfF is the collection of all closed, bounded subset of R (in its

usual topology), together with R itself, then F is the family of closed sets for a

topology on R strictly weaker than the usual topology.

Proof. It is easy to see that Fis a topology. Further, for instance, .1; 0is aclosed set ofR, but it is not in F. ut

I Exercise25. IfA X, show that the family of all subsets ofXwhich containA, together with the empty set, is a topology on X. Describe the closure and

interior operations. What topology results when A D ? whenA D X?

Proof. Let

ED fE XW A Eg [ fg :Now suppose that E 2 E for each 2 . Then A

S E X and soS

E2 E. The other postulates are easy to check.For any setB X, ifA B, thenB2 E and so B B D B; if not, thenB B D .IfA D , then E is the discrete topology; ifA D X, then ED f; Xg. ut

3D. Regularly Open and Regularly Closed Sets

An open subsetG in a topological space is regular open iffG is the interior of

its closure. A closed subset is regularly closediff it is the closure of its interior.

I Exercise26. The complement of a regularly open set is regularly closed and

vice versa.

Proof. SupposeG is regular open; that is, GD . xG/. Then

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10 CHAPTER 2 TOPOLOGICAL SPACES

XX GD XX . xG/ D XXxGD .XX G/:

Hence,XX G is regularly closed. IfF is regular closed, i.e., FD F, then

XX FD XX F

D .XX F

/

D .XX F /

Ithat is,XX F is regularly open. ut

I Exercise27. There are open sets in R which are not regularly open.

Proof. Consider Q. We have . xQ/ D R D R Q. So Q is not regularly open.ut

I Exercise 28. IfA is any subset of a topological space, then int.cl.A// is reg-

ularly open.

Proof. LetA be a subset of a topological spaceX. We then have

int.cl.A// cl.int.cl.A/// H) int.cl.A// D int.int.cl.A/// int.cl.int.cl.A////;

and

int.cl.A// cl.A/ H) cl.int.cl.A/// cl.cl.A// D cl.A/H) int.cl.int.cl.A//// int.cl.A//:

Therefore, int.cl.A//D int.cl.int.cl.A////; that is, int.cl.A// is regularly open.ut

I Exercise29. The intersection, but not necessarily the union, of two regularly

open sets is regularly open.

Proof. LetA and B be two regularly open sets in a topological space X. Then

.A \ B/ .xA \xB/ D .xA/ \ . xB/ D A \ B;

and

.xA \xB/ D .xA/ \ . xB/ D A \ B A \ BH) A \ BD .xA \xB/ D

h.xA \xB/

i .A \ B/:

Hence,A \ BD .A \ B/.To see that the union of two regularly open sets is not necessarily regularly

open, considerA D .0; 1/and BD .1; 2/in R with its usual topology. Then

.A [ B/ D 0; 2 D .0; 2/ A [ B: ut


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SECTION 2.1 FUNDAMENTAL CONCEPTS 11

3E. Metrizable Spaces

LetXbe a metrizable space whose topology is generated by a metric .

I Exercise 30. The metric 2 defined by 2.x;y /D 2.x; y/ generates thesame topology on X.

Proof. Let O be the collection of open sets in .X;/, and let O2 be the col-

lection of open sets in.X;2/. IfO2 O, then for every x2 O , there exists anopen ball B.x;"/O ; but then B2.x; "=2/O . Hence, O2 O2. Similarly, wecan show that O2 O. In fact, and 2 are equivalent metrics. ut

I Exercise 31. The closure of a setE Xis given byxED fy2 XW .E;y/ D 0g.

Proof. Denote zE fy2 X W .E;y/D 0g. We first show that zE is closed(see Definition 2.5, p. 17). Take an arbitrary x2 Xsuch that for every n2 N,there exists yn2 zE with .x;yn/ < 1=2n. For each yn2 zE, take zn2 E with.yn; zn/ < 1=2n. Then

.x;zn/ 6 .x;yn/ C .yn; zn/ < 1=n; for alln 2 N:

Thus, .x; E/ D 0, i.e.,x2zE. Therefore, zE is closed. It is clear that EzE, andsoxEzE.

We next show that zE xE. Take an arbitrary x2 zE and a closed set Kcontaining E . Ifx2 XXK, then.x;K/ > 0(seeExercise 35). But then.x; E/ >0sinceE Kand so

infy2E

.x;y/ > infz2K

.x; z/:

Hence,zExE. utI Exercise 32. The closed diskU.x; x"/D fyW .x;y/ 6 "g is closed in X, butmay not be the closure of the open diskU.x;"/.

Proof. Fix x2 X. We show that the function .x; / W X! R is (uniformly)continuous. For anyy; z2 X, the triangle inequality yields

j.x;y/ .x;z/j 6 .y; z/:

Hence, for every" > 0, take D ", and .x; / satisfies the "- criterion. There-fore,U.x; x"/is closed sinceU.x; x"/ D 1.x;0; "/and 0;"is closed in R.

To see it is not necessary that U.x;x"/

DU.x; "/, consider"

D1and the usual

metric on n.x;y/ 2 R2 W x2 C y2 D 1

o[n

.x;0/ 2 R2 W 0 6 x 6 1oI

seeFigure 2.1.Observe that .0; 0/U.x;1/, but .0; 0/2U.x; x1/. It follows fromExercise 31that.0;0/ U.x;1/. ut


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0x

Figure2.1. U.x;x1/ U.x;1/.

3H.G andF Sets

I Exercise33. The complement of aG is anF, and vice versa.

Proof. If A is a G set, then there exists a sequence of open sets fUng suchthat A D T1nD1 Un. ThenAc DS1nD1 Ucn isF. Vice versa. utI Exercise 34. An Fcan be written as the union of an increasing sequence

F1 F2 of closed sets.

Proof. Let B

D S1nD1 En, where En is closed for all n

2 N. Define F1

D E1

and Fn D SniD1 Ei for n > 2. Then each Fn is closed, F1 F2 , andS1nD1 FnD

S1nD1D B . ut

I Exercise35. A closed set in a metric space is aG .

Proof. For an arbitrary setA Xand a point x2 X, define

.x; A/ D infy2A

f.x;y/g:

We first show that .; A/ W X! R is (uniformly) continuous by showing

j.x; A/ .y; A/j 6 .x; y/; for allx; y2 X: (2.1)

For an arbitraryz2 A, we have

.x; A/ 6 .x;z/ 6 .x;y/ C .y; z/:

Take the infimum over z2 Aand we get

.x;A/ 6 .x;y/ C .y; A/: (2.2)


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SECTION 2.2 NEIGHBORHOODS 13

Symmetrically, we have

.y;A/ 6 .x;y/ C .x; A/: (2.3)

Hence, (2.1)follows from(2.2) and (2.3). We next show that ifA is closed, then.x; A/D 0 iff x2 A. The if part is trivial, so we do the only if part. If.x; A/D0, then for everyn2 N, there exists yn2 A such that .x; yn/ < 1=n;that is,yn! x. Sincefyng Aand A is closed, we must have x2 A.

Therefore,

A D1\nD1

fx2 XW .x; A/ < 1=ng :

The continuity of.; A/ implies thatfx2 XW .x; A/ < 1=ng is open for all n.Thus, A is aG set. ut

I Exercise36. The rationals are an F in R.

Proof. Q is countable, and every singleton set in R is closed; hence, Q is an

F. ut

3I. Borel Sets

2.2 Neighborhoods

4A. The Sorgenfrey Line

I Exercise37. Verify that the setx;z/, forz > x, do form a nhood base atx

for a topology on the real line.

Proof. We need only check that for each x2 R, the familyBx fx;z/ W z > xgsatisfies V-a, V-b, and V-c in Theorem 4.5. V-a is trivial. If x; z1/2 Bx andx;z2/2 Bx , then x;z1/\x; z2/D x;z1 ^ z2/2 Bx and is in x;z1/\x; z2/.For V-c, let x;z/2 Bx . Let z0 2 .x;z. Then

x; z0

2 Bx , and ify2 x; z0, theright-open interval

y; z0

2 By and y; z0 x; z/.Then, define open sets using V-d: G R is open if and only ifG contains a

setx;z/of each of its points x . ut

I Exercise38. Which intervals on the real line are open sets in the Sorgenfrey

topology?

Solution.

Sets of the form.1; x/,x;z/, orx; 1/are both open and closed.

Sets of the form.x; z/or .x; C1/are open in R, since

.x;z/ D[

fy;z/ W x < y < zg: ut


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I Exercise 39. Describe the closure of each of the following subset of the Sor-

genfrey line: the rationalsQ, the setf1=n W n D 1 ; 2 ; : : : ; g, the setf1=n W n D 1 ; 2 ; : : :g,the integersZ.

Solution. Recall that, by Theorem 4.7, for each E R, we havexED x2 R W each basic nhood ofx meetsE :

Then xQD R since for any x2 R, we have x; z/ \ Q for z > x. Similarly,f1=n W n D 1 ; 2 ; : : :g D f1=n W n D 1 ; 2 ; : : :g, andxZ D Z. ut

4B. The Moore Plane

I Exercise40. Verify that this gives a topology on .

Proof. Verify (V-a)(V-c). It is easy. ut

4E. Topologies from nhoods

I Exercise 41. Show that if each pointx in a setXhas assigned a collection

Ux of subsets ofXsatisfying N-a through N-d of 4.2, then the collection

D G XW for eachx in G ,x2 U G for some U2 Uxis a topology forX, in which the nhood system at each x is justUx .

Proof. We need to check G1G3 in Definition 3.1. Since G1 and G3 are evi-

dent, we focus on G2. Let E1; E22 . Take any x2 E1\ E2. Then there existsome U1; U22 Ux such that x2 U1 E1 and x2 U2 E2. By N-b, we knowthat U1 \ U22 Ux . Hence,

x2 U1 \ U2 E1 \ E2;

and soE1 \ E22 . The induction principle then means that is closed underfinite intersections. ut

4F. Spaces of Functions

I Exercise42. For eachf2 RI

, each finite subsetF ofI and each positive ,let

U.f;F;/ Dn

g2 RI W jg.x/ f.x/j < , for eachx2 Fo

:

Show that the setsU.f; F; /form a nhood base atf, making RI a topological

space.

Proof. Denote


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SECTION 2.2 NEIGHBORHOODS 15

BfD

U.f;F;/ W F I; jFj < 1; > 0 :(V-a) For each U.f; F; /2 Bf , we havejf.x/f .x/j D 0 < for all x2 F;hence, f2 U.f;F;/.(V-b) LetU.f;F1; 1/;U.f;F2; 2/ 2 Bf. DefineU.f; F3; 3/by letting

F3D F1 [ F2; and 3D minf1; 2g:

Clearly,U.f; F3; 3/ 2 Bf. Ifg2 U.f;F3; 3/, then

jg.x/ f.x/j


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Proof. Denote VfD fV.f;"/ W " > 0g. We verify the following properties.(V-a) IfV .f; "/ 2 Vf, thenjf.x/ f.x/j D 0 < "; that is, f2 V.f;"/.(V-b) LetV.f; "1/;V.f;"2/

2Vf. Let"3

Dmin

f"1; "2

g. Ifg

2V.f;"3/, then

jg.x/ f.x/j < "3D minf"1; "2g; for allx2 I:

Hence,V.f; "3/ V.f;"1/ \ V.f;"2/.(V-c) For an arbitrary V .f; "/2 Vf, pickV .f; "=2/2 Vf. For each g2 V .f; "=2/,pickV .g; "=2/ 2 Vg . Ifh 2 V .g; "=2/, then jh.x/ g.x/j < "=2for allx2 I. Hence

jh.x/ f.x/j 6 jh.x/ g.x/j C jg.x/ f.x/j < "I

that is,V .g; "=2/ V.f;"/. ut

I Exercise45. Compare the topologies defined in 1 and 3.

Proof. It is evident that for every U.f; F; /2 Bf, there exists V.f;/2 Vfsuch that V.f;/ U.f; F; /. Hence, the topology in 1 is weaker than in 3 byHausdorff criterion. ut

2.3 Bases and Subbases

5D. No Axioms for Subbase

I Exercise46. Any family of subsets of a setXis a subbase for some topology

on X and the topology which results is the smallest topology containing thegiven collection of sets.

Proof. Let be a family of subsets of X. Let ./ be the intersection of all

topologies containing . Such topologies exist, since 2X is one such. Also ./

is a topology. It evidently satisfies the requirements unique and smallest.

The topology ./can be described as follows: It consists of, X, all finite

intersections of the-sets, and all arbitrary unions of these finite intersections.

To verify this, note that since ./, then ./ must contain all the setslisted. Conversely, because

S distributes over

T, the sets listed actually do

from a topology containing , and which therefore contains ./. ut

5E. Bases for the Closed Sets

I Exercise47. F is a base for the closed sets inXiff the family of complements

of members ofFis a base for the open sets.


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SECTION 2.3 BASES AND SUBBASES 17

Proof. LetG be an open set in X. ThenGD XX E for some closed subsetE .Since EDTF2GF F, we obtain

GD XX0@\F2GF

F1A D [F2GF

Fc

:

Thus,fFc W F2 Fg forms a base for the open sets. The converse direction issimilar. ut

I Exercise 48. F is a base for the closed sets for some topology on X iff (a)

wheneverF1 andF2 belong toF, F1[ F2 is an intersection of elements ofF,and (b)

TF2F FD .

Proof. If F is a base for the closed sets for some topology on X, then (a)

and (b) are clear. Suppose, on the other hand, X is a set and F a collection

of subsets of Xwith (a) and (b). Let Tbe all intersections of subcollections

from F. Then any intersection of members of T certainly belongs to T, so T

satisfies (F-a). Moreover, ifF1 F and F2 F, so thatT

E2F1E and

TF2F2

F

are elements ofT, then0@\E2F1

E

1A[0@\F2F2

F

1A D \E2F1

\F2F2

.E[ F /:

But by property (a), the union of two elements of F is an intersection of el-

ements ofF, so .T

E2F1E/ [ .TF2F2F / is an intersection of elements ofF,

and hence belongs to T. Thus T satisfies (F-b). Finally, 2 Tby (b) andX2 Tsince Xis the intersection of the empty subcollection from F. Hence T sat-

isfies (F-c). This completes the proof that T is the collection of closed sets ofX. ut


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31/57

3NEW SPACES FROM OLD

3.1 Subspaces

3.2 Continuous Functions

7A. Characterization of Spaces Using Functions

I Exercise49. The characteristic function ofA is continuous iffA is both open

and closed in X.

Proof. Let 1A W X! Rbe the characteristic function ofA, which is defined by

1A .x/ D 1 ifx2 A0 ifx A:

First suppose that 1A is continuous. Then, say, 11A

1=2;2

D A is open,and11A

1;1=2 D XX Ais open. Hence, A is both open and closed in X.Conversely, suppose that A is both open and closed inX. For any open set

U R, we have

11A .U / D

A if1 2 U and0 UXX A if1 U and0 2 U if1 U and0 UX if1 2 U and0 2 U:

Then 1A is continuous. ut

I Exercise 50. X has the discrete topology iff wheneverY is a topological space

andfW X! Y, thenf is continuous.

Proof. Let Y be a topological space and fW X! Y. It is easy to see that fis continuous if X has the discrete topology, so we focus on the sufficiency

19


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20 CHAPTER 3 NEW SPACES FROM OLD

direction. For any A X, let YD R and f D 1A. Then by Exercise 49 A isopen. ut

7C. Functions Agreeing on A Dense Subset

I Exercise 51. If f and g are continuous functions from X to R, the set of

points x for which f.x/D g.x/ is a closed subset of X. Thus two continuousmaps onX toR which agree on a dense subset must agree on all ofX.

Proof. Denote AD fx2 X W f.x/ g.x/g. Take a point y 2 A such thatf.y/ > g.y/ (if it is not true then let g.y/ > f.y/). Take an " > 0 such that

f.y/" > g.y/C". Sincef andg are continuous, there exist nhoodsU1and U2ofy such that f U1." C f.y/;" C f.y//and gU2." C g.y/; " C g.y//.LetUD U1 \ U2. ThenUis a nhood ofx and for everyz2 U we have

f.z/ g.z/ > f.x/ " g.x/ C " > 0:

Hence, U A; that is, U is open, and sofx2 XW f.x/D g.x/g D XX U isclosed.

Now suppose that D fx2 XW f.x/D g.x/g is dense. Take an arbitraryx2 X. Sincef andg are continuous, for each n 2 N, there exist nhoodsVf andVg such thatjf.y/ f.x/j < 1=n for all y2 Vf andjg.y/ g.x/j < 1=n for ally2 Vg . Let VnD Vf\Vg . Then there exists xn2 Vn\Dwith jf .xn/f.x/j < 1=2nandjg.xn/ g.x/j < 1=2n. Sincef .xn/ D g.xn/, we have

jf.x/ g.x/j 6 jf.x/ f .xn/j C jf .xn/ g.x/j D jf.x/ f .xn/j C jg.xn/ g.x/j< 1=n:

Therefore,f .x/ D g.x/. ut

7E. Range Immaterial

I Exercise 52. IfY Z andfW X! Y, then f is continuous as a map fromX toY ifffis continuous as a map from X toZ .

Proof. LetfW X! Z be continuous. LetUbe open in Y. ThenUD Y\ V forsomeVwhich is open in Z . Therefore,

f1.U / D f1 .Y\ V / D f1 .Y / \ f1 .V / D X\ f1 .V / D f1 .V /is open in X, and sof is continuous as a map from X toY.

Conversely, let fW X ! Y be continuous and V be open in Z. Thenf1 .V /D f1 .Y\ V /. Since Y\ V is open in Y and f is continuous fromX toY, the setf1 .Y\ V /is open in Xand sofis continuous as a map fromX toZ . ut


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SECTION 3.2 CONTINUOUS FUNCTIONS 21

7G. Homeomorphisms within the Line

I Exercise53. Show that all open intervals in R are homeomorphic.

Proof. We have

.a;b/ .0; 1/by f1.x/ D .x a/=.b a/.

.a; 1/ .1; 1/by f2.x/ D x a C 1.

.1; 1/ .0; 1/by f3.x/ D 1=x.

.1; a/ .a; 1/by f4.x/ D x.

.1; 1/ .=2; =2/by f5.x/ D arctan x.

Therefore, by compositing, every open interval is homeomorphic to .0; 1/. ut

I Exercise54. All bounded closed intervals in R are homeomorphic.

Proof. a;b 0; 1by f .x/ D .x a/=.b a/. ut

I Exercise 55. The property that every real-valued continuous function on X

assumes its maximum is a topological property. Thus, I 0;1 is not homeo-morphic toR.

Proof. Every continuous function assumes its maximum on 0; 1; however,x2

has no maximum onR. Therefore, I6 R. ut

7K. Semicontinuous Functions

I Exercise 56. Iff is a lower semicontinuous real-valued function on X for

each 2 A, and ifsupf .x/ exists at each x2 X, then the function f.x/Dsupf .x/ is lower semicontinuous onX.

Proof. For an arbitrary a2 R, we have f.x/ 6 a iff f .x/ 6 a for all 2 A.Hence,

fx2 XW f.x/ 6 g D\2A

fx2 XW f .x/ 6 ag ;

and sof1.1; a is closed; that is,f is lower semicontinuous. ut

I Exercise57. Every continuous function fromX toR is lower semicontinuous.

Thus the supremum of a family of continuous functions, if it exists, is lower

semicontinuous. Show by an example that lower semicontinuous cannot be

replaced by continuous in the previous sentence.

Proof. Suppose that fW X! Ris continuous. Since .1; x is closed in R, thesetf1.1; x is closed in X; that is, f is lower semicontinuous.

To construct an example, letfW 0; 1/ ! Rbe defined as follows:


34/57


fn.x/ D

nx if0 6 x 6 1=n

1 ifx > 1=n:

Then

f.x/ D supn

fn.x/ D

0 ifxD 01 ifx > 0;

andf is not continuous. ut

I Exercise58. The characteristic function of a setA in X is lower semicontin-

uous iffA is open, upper semicontinuous iffA is closed.

Proof. Observe that

11

A

.1

; aD ifa < 0XX A if0 6 a < 1

X ifa > 1:

Therefore,1A is LSC iffA is open. Similarly for the USC case. ut

I Exercise 59. If X is metrizable and f is a lower semicontinuous function

from X to I, then f is the supremum of an increasing sequence of continuous

functions onX toI .

Proof. Letdbe the metric on X. First assumef is nonnegative. Define

fn.x/ D infz2X

ff.z/ C nd.x;z/g :

Ifx; y2 X, thenf .z/ C nd.x;z/ 6 f.z/ C nd.y;z/ C nd.x;y/. Take the inf over z(first on the left side, then on the right side) to obtain fn.x/ 6 fn.y/ C nd.x;y/.By symmetry,

jfn.x/ fn.y/j 6 nd.x;y/Ihence, fn is uniformly continuous on X. Furthermore, since f > 0, we have

0 6 fn.x/ 6 f.x/ C nd.x;x/ D f.x/. By definition, fn increases withn; we mustshow that limn fn is actuallyf.

Given" > 0, by definition offn.x/ there is a point zn2 Xsuch that

fn.x/ C " > f.zn/ C nd.x;zn/ > nd.x;zn/ (3.1)

sincef > 0. But fn.x/ C " 6 f.x/ C "; hence d.x; zn/! 0. Since f is LSC, wehave lim infn f .zn/ > f.x/(Ash,2009,Theorem 8.4.2); hence

f .zn/ > f .x/ " ev: (3.2)

By(3.1) and(3.2),

fn.x/ > f.zn/ " C nd.x;zn/ > f .zn/ " > f.x/ 2"


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SECTION 3.2 CONTINUOUS FUNCTIONS 23

for all sufficiently largen. Thus,fn.x/ ! f.x/.Ifjfj 6 M M. ut

7M.C.X/andC.X/

I Exercise60. Iff andg belong toC.X/, then so dofC g,f g anda f, fora2 R. If, in addition, f andg are bounded, then so arefC g,f g anda f.

Proof. We first do fC g. Since f; g2 C.X/, for each x2 X and each " > 0,there exist nhoodsU1 andU2 ofx such thatf U1."=2 C f.x/;"=2 C f.x//andgU2 ."=2 C g.x/;"=2 C g.x//. LetUD U1 \ U2. ThenUis a nhood ofx ,and for everyy2 U, we have

jf.y/

Cg.y/

f.x/

Cg.x/

j6

jf.y/

f.x/

j C jg.y/

g.x/

j< "

Ithat is,fC g is continuous.

We then do af . We suppose that a > 0 (all other cases are similar). Foreachx2 X and" > 0, there exists a nhood U ofx such that f U ."=a Cf .x/; "=a C f.x//. Then.a f /U 2 ." C a f.x/;" C a f.x//. Soa f2 C.X/.

Finally, to do f g, we first show that f2 2 C.X/ whenever f2 C.X/. Foreach x2 X and " > 0, there is a nhood U of x such that f U .p"Cf.x/;

p" C f.x//. Thenf2 U ." C f2.x/; " C f2.x//, i.e.,f2 2 C.X/. Since

f.x/ g.x/ D 14

hf.x/ C g.x/2 f.x/ g.x/2i ;

we know thatf

g

2C.X/ from the previous arguments.

utI Exercise61. C.X/andC.X/are algebras over the real numbers.

Proof. It follows from the previous exercise that C.X/is a vector space on R.

So everything is easy now. ut

I Exercise62. C.X/is a normed linear space with the operations of addition

and scalar multiplication given above and the normkfk D supx2Xjf.x/j.

Proof. It is easy to see that C.X/is a linear space. So it suffices to show that

k kis a norm on C.X/. We focus on the triangle inequality. Let f; g2C.X/.Then for every x2 X, we havejf.x/Cg.x/j 6jf.x/j C jg.x/j 6kfk C kgk;hence,kfC gk 6 kfk C kgk. ut


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3.3 Product Spaces, Weak Topologies

8A. Projection Maps

I Exercise63. Theth projection map is continuous and open. The projec-

tion1 W R2 ! Ris not closed.

Proof. Let U be open in X . Then 1

.U / is a subbasis open set of the

Tychonoff topology on X , and so is open. Hence, is continuous.Take an arbitrary basis open set U in the Tychonoff topology. Denote I

f1 ; : : : ; ng. ThenUD U ;

whereU is open inX for every2 A, andUjD X

jfor allj I. Hence,

.U / D U ifD i for somei2 IX otherwise.

That is, .U / is open in X in both case. Since any open set is a union of

basis open sets, and since functions preserve unions, the image of any open

set under is open.

0

F

Figure3.1. 1.F /D .0;1/

Finally, let FD epi.1=x/. ThenFis closed in R2, but1.F /D.0; 1/ is openinR; that is,1 is not closed. SeeFigure 3.1. ut

I Exercise64. Show that the projection ofI RontoR is a closed map.

Proof. Let W I R! R be the projection. Suppose A I R is closed, andsuppose y0 2 RXA. For every x2 I, since .x;y0/ A and A is closed,we find a basis open subset U.x/V .x/ of I R that contains .x;y0/, andU.x/ V.x/ \ AD . The collectionfU.x/W x2 Ig covers I, so finitely manyof them cover I by compactness, say U.x1/ ; : : : ; U . xn/ do. Now define V D


37/57

SECTION 3.3 PRODUCT SPACES, WEAK TOPOLOGIES 25TniD1 V .xi /, and note thatVis an open nhood ofy0, andV\ A D . SoA

is closed; that is, is closed. SeeLee(2011, Lemma 4.35, p. 95) for the Tube

Lemma.

Generally, if

WX

Y

!X is a projection may where Y is compact, then

is a closed map. ut

8B. Separating Points from Closed Sets

I Exercise65. Iff is a map (continuous function) ofX toX for each 2A,thenff W 2 Ag separates points from closed sets in X iffff1 V W 2A; V open inX g is a base for the topology on X.

Proof. Suppose thatff1 V W2A; V open in X g consists of a base for thetopology onX. Let B be closed in X andx B . Then x2 XX B andXX B isopen in X. Hence there exists f1 V such that x

2 f1 V

X

XB; that is,

f .x/2V . Since V\ f BD, i.e., f BX X V, andX X V is closed, wegetf B X X V. Thus,f .x/ f B.

Next assume thatff W 2Ag separates points from closed sets in X. Takean arbitrary open subset U X andx2 U. Then B XX U is closed in X,and hence there exists2A such that f .x/f B. Thenf .x/2X X f Band, sinceX X f B is open in X , there exists an open set V ofX such thatf .x/ 2 V X X f B. Therefore,

x2 f1 V f1h

X X f Bi

D XX f1h

f Bi

XX f1 f B

X

XB

D U:

Hence,ff1 V W 2 A; V open in X gis a base for the topology onX. ut

8D. Closure and Interior in Products

LetXandYbe topological spaces containing subsetsA andB , respectively. In

the product space X Y:

I Exercise66. .A B/B D AB BB.

Proof. Since AB A is open in A and B B B is open in B , the set AB BB A B is open inA B ; hence,AB BB .A B/B.

For the converse inclusion, let xD .a;b/2 .A B/B. Then there is an basisopen setU1 U2 such thatx2U1 U2A B , whereU1 is open inA and U2is open in B . Hence, a2 U1 A and b2 U2 B; that is, a2 AB and b 2 BB.Then x2 AB BB. ut


38/57


I Exercise67. A BDxA xB .

Proof. SeeExercise 68. ut

I Exercise68. Part 2 can be extended to infinite products, while part 1 can beextended only to finite products.

Proof. Assume that yD y2 A; we show that y2 SA for each ; thatis,y2 SA. Lety2 U , whereU is open inY ; sincey2 1 .U /, we musthave

1 .U / \AD .U \ A/

A

!;

and soU \A . This provesy2SA . The converse inclusion is establishedby reversing these steps: Ify2 SA , then for any open nhood

B

U1

U n Y W 1; : : : ; n ;eachU

i\ Ai so thatB\A . ut

I Exercise69. Fr.A B/ D xA Fr.B/ [ Fr.A/ xB.

Proof. We have

Fr.A B/ D A B\ .X Y / X .A B/D .xA xB/ \ .X Y / X .A B/D .xA xB/ \

hX .YX B/ [ .XX A/ Yi

D

xA

Fr.B/

[Fr.A/

xB:

utI Exercise 70. IfX is a nonempty topological space andA X , for each2 A, then A is dense in X iffA is dense in X , for each .Proof. It follows fromExercise 68that

AD xAIthat is, A is dense in X iffA is dense in X , for each . ut8E. Miscellaneous Facts about Product Spaces

LetX be a nonempty topological space for each 2 A, and let XD X .I Exercise 71. IfV is a nonempty open set in X, then .V /D X for all butfinitely many2 A.

Proof. Let T be the topology on X for each2 A. LetVbe an arbitrary openset in X. Then VD Sk2KBk, where for each k2 Kwe have BkD 2A Ek ,


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SECTION 3.3 PRODUCT SPACES, WEAK TOPOLOGIES 27

and for each2 Awe have Ek2 T while

Ak f2 A W Ek X g

is finite. Then Tk2KAk is finite. If0Tk2KAk , then there existsk02 K suchthat E0k0D X 0 . Then

10 .Bk0/ D 10

2A

Ek0

!D X

0;

and soX0D 10.Bk0/ 10.V /implies that 10.V / D X 0 . ut

I Exercise72. Ifb is a fixed point in X , for each2A, thenX00D fx2XWxD b whenever 0g is homeomorphic toX 0 .

Proof. Write an element in X00 as .x0 ;b0/. Then consider the mapping

.x0 ;b0/ 7! x0.

ut

8G. The Box Topology

LetX be a topological space for each 2 A.

I Exercise 73. In X , the sets of the form U , where U is open in Xfor each2 A, form a base for a topology.

Proof. Let B U W 2 A; U open in X . Then it is clear that X 2B sinceX is open for each 2A. Now take anyB1; B22 B , withB1D

U1

andB2D

U2 . Let

pD p1; p2; : : : 2 B1 \ B2D U1\ U2 :Then p2 U1\ U2 , and so there exists an open set B X such that p2B U1\ U2 . Hence, B2 B andp2 B B1 \ B2. ut8H. Weak Topologies on Subspaces

Let Xhave the weak topology induced by a collection of maps f W X! X ,for2 A.

I Exercise74. If eachX has the weak topology given by a collection of maps

g W X ! Y , for2 , then X has the weak topology given by the mapsg B f W X! Y for2 Aand 2 .

Proof. A subbase for the weak topology onX induced byfgW 2 gisng1 .U / W 2 ; U open in Y

o:


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Then a subbasic open set in X for the weak topology on X induced by

ff W 2 Agis

nf1 g

1 .U /

W

2A;

2; U open inY o :

Since f1 .g1

.U // D .g B f /1.U /, we get the result. ut

I Exercise75. AnyB Xhas the weak topology induced by the mapsf B .

Proof. As a subspace ofX, the subbase onB isnB\ f1 .U / W 2 A; U open in X

o:

On the other hand, .f B/1.U /DB\ f1 .U / for every 2 A and U openinX . Hence, the above set is also the subbase for the weak topology induced

byff BW 2 Ag. ut

3.4 Quotient Spaces

9B. Quotients versus Decompositions

I Exercise76. The process given in 9.5 for forming the topology on a decom-

position space does define a topology.

Proof. Let .X; T/ be a topological space; let Dbe a decomposition ofX. Define

F

D is open in D

() [ fFW F2 Fg is open in X: (3.3)Let Tbe the collection of open sets defined by ( 3.3). We show that.D ;T/is a

topological space.

Take an arbitrary collectionfFigi2I T; thenSfF W F2 Fig is open in X

for eachi2 I. Hence, Si2IFi2 Tsince[

F2Si2I Fi

FD[i2I

0@[F2Fi

F

1Ais open in X.

Let F1;F22 T; then SE2F1E and SF2F2Fare open in X. Therefore, F1\F22 Tsince [

F2F1\F2

FD0@[E2F1

E

1A \0@[F2F2

F

1Ais open in X.

2 Tsince S D is open in X; finally, D2 Tsince SDD X. ut


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SECTION 3.4 QUOTIENT SPACES 29

I Exercise 77. The topology on a decomposition space D ofXis the quotient

topology induced by the natural mapP W X! D. (See 9.6.)

Proof. Let Tbe the decomposition topology ofD, and let TPbe the quotient

topology induced by P. Take an open set F2 T; then SF2F F is open in X.Hence,

P1.F/ D P10@[F2F

F

1A D [F2F

P1.F / D[F2F

F

is open in X, and so F2 TP. We thus proved that T TP.Next take an arbitrary F2 TP. By definition, we haveP1.F/D

SF2F F is

open inX. But then F2 T.We finally prove Theorem 9.7 (McCleary, 2006, Theorem 4.18): Suppose

fW X! Y is a quotient map. Suppose is the equivalence relation definedonX byxx 0 iff .x/Df .x0/. Then the quotient space X= is homeomorphic

toY.By the definition of the equivalence relation, we have the diagram:

X

X=

Y

Y

hBP

Df

f

P

h

Define h W X= ! Y by letting h.x/D f.x/. It is well-defined. Notice thath B PD f since for eachx2 Xwe obtain

.h B P/.x/ D h.P.x// D h.x/ D f.x/:

Both f andPare quotient maps so h is continuous by Theorem 9.4. We show

that h is injective, subjective and h1 is continuous, which implies that h is a

homeomorphism. If h.x/D h.x0/, then f .x/D f .x0/ and so x x0; that is,xD x0, and h is injective. Ify 2 Y, then yD f .x/ since f is surjective andh.x/ D f.x/ D y so h is surjective. To see that h1 is continuous, observe thatsincefis a quotient map andPis a quotient map, this showsPD h1 B f andTheorem 9.4 implies thath1 is continuous. ut


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43/57

4CONVERGENCE

4.1 Inadequacy of Sequences

10B. Sequential Convergence and Continuity

I Exercise 78. Find spacesX andY and a function FWX! Y which is notcontinuous, but which has the property thatF .xn/ ! F.x/in Y wheneverxn!x inX.

Proof. Let X D RR and Y D R. Define FW RR ! R by letting F.f / Dsupx2R jf.x/j. ThenFis not continuous: Let

EDn

f2 RR W f.x/ D 0or 1 and f .x/ D 0only finitely ofteno

;

and let g2 RR be the function which is 0 everywhere. Then g2 xE. However,0 2 F xE sinceF.g/ D 0, andFE D f1g. ut

10C. Topology of First-Countable Spaces

LetX andYbe first-countable spaces.

I Exercise79. U Xis open iff wheneverxn! x2 U, then.xn/ is eventuallyinU.

Proof. IfUis open and xn!x2U, thenx has a nhood V such thatx2VU. By definition of convergence, there is some positive integer n0 such that

n > n0 impliesxn2 V U; hence,.xn/is eventually in U.Conversely, suppose that whenever xn! x2 U, then .xn/ is eventually in

U. IfUis not open, then there existsx2Usuch that for every nhood ofV ofx we haveV\ .XXU / . SinceXis first-countable, we can pick a countablenhood basefVnW n2 Ng at x. Replacing VnD

TniD1 Vi where necessary, we

may assume that V1 V2 . Now Vn\ .XX U / for each n, so wecan pick xn2 Vn\ .XX U /. The result is a sequence .xn/ contained in XX U

31


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32 CHAPTER 4 CONVERGENCE

which converges to x2 U; that is, xn! x but .xn/ is not eventually in U. Acontradiction. ut

I Exercise80. F Xis closed iff whenever.xn/is contained inF andxn! x,thenx2 F.Proof. Let Fbe closed; let .xn/ be contained in F andxn! x . Then x2 xFDF.

Conversely, assume that whenever .xn/ is contained in F andxn! x , thenx2 F. It follows from Theorem 10.4 that x2xFwith the hypothesis; therefore,xF F, i.e., xFD Fand soF is closed. ut

I Exercise 81. fW X! Y is continuous iff whenever xn ! x in X, thenf .xn/ ! f.x/ in Y.

Proof. Suppose f is continuous and xn ! x. Since f is continuous at x,for every nhood V of f.x/ in Y, there exists a nhood U of x in X such thatf .U/V. Sincexn!x , there exists n0 such that n > n0 implies that xn2 U.Hence, for every nhoodV off .x/, there existsn0 such thatn > n0 implies that

f .xn/ 2 V; that is, f .xn/ ! f.x/.Conversely, let the criterion hold. Suppose that fis not continuous. Then

there exists x2 X and a nhood V of f.x/, such that for every nhood baseUn, n2 N, of x, there is xn2 Un with f .xn/ V. By letting U1 U2 ,we have xn ! x and so f .xn/! f.x/; that is, eventually, f .xn/ is in V. Acontradiction. ut

4.2 Nets

11A. Examples of Net Converence

I Exercise82. In RR, let

EDn

f2 RR W f.x/ D 0or1; andf .x/ D 0only finitely ofteno

;

andg be the function in RR which is identically0. Then, in the product topology

on RR,g2xE. Find a net.f/in E which converges tog.

Proof. Let Ug

D fU.g;F;"/

W" > 0;F

R a finite set

gbe the nhood base ofg.

OrderUg as follows:

U.g;F1; "1/ 6 U.g;F2; "2/ () U.g;F2; "2/ U.g;F;"2/() F1 F2 and "2 6 "1:

Then Ug is a directed set. So we have a net .fF;"/converging tog. ut


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SECTION 4.2 NETS 33

11B. Subnets and Cluster Points

I Exercise83. Every subnet of an ultranet is an ultranet.

Proof. Take an arbitrary subset E X. Let .x/ be an ultranet in X, andsuppose that .x/ is residually in E, i.e., there exists some 02 such that> 0 implies that x2 E . If.x/ is a subnet of.x/, then there exists some0 such that0 > 0. Then for every > 0, we have > 0, and so > 0

implies thatx2 E; that is,.x/is residually in E . ut

I Exercise84. Every net has a subnet which is an ultranet.

Proof. SeeAdamson(1996,Exercise 127, p. 40). ut

I Exercise85. If an ultranet hasx as a cluster point, then it converges tox .

Proof. Let .x/ be an ultranet, and x be a cluster point of .x/. Let U be a

nhood ofx . Then .x/lies in U eventually since for any0 there exists > 0

such thatx2 U. ut

11D. Nets Describe Topologies

I Exercise86. Nets have the following four properties:

a. ifxD x for each 2 , thenx! x,b. ifx! x, then every subnet of.x/converges tox ,c. if every subnet of.x/ has a subnet converging tox , then .x/ converges to

x,

d. (Diagonal principal) ifx! x and, for each 2 , a net.x/u2M convergestox, then there is a diagonal net converging tox ; i.e., the net.x

/2;2M ,

ordered lexicographically by, then byM, has a subnet which converges to

x.

Proof. (a)If the net .x/is trivial, then for each nhood U ofx , we havex2 Ufor all 2 . Hence,x! x.(b) Let .x'.//2M be a subnet of .x/. Take any nhood U of x. Then there

exists 02 such that > 0 implies that x2 U since x! x. Since ' iscofinal in , there exists 02 M such that ' .0/ > 0; since ' is increasing, > 0 implies that './ > '.0/ > 0. Hence, there exists 02 M such that > 0 implies thatx'./2 U; that is, x'./! x.(c) Suppose by way of contradiction that .x/ does not converge to x. Then

there exists a nhoodU ofx such that for any 2 , there exists some'./ > with x'./ U. Then .x'.// is a subnet of.x/, but which has no convergingsubnets.


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(d) Orderf.;/ W 2 ; 2 Mg as follows:

.1; 1/ 6 .2; 2/ () 1 6 2, or1D 2 and1 6 2:

Let U be the nhood system ofx which is ordered by U1 6 U2 iffU2 U1 for allU1; U22 U. Define

Dn

.;U/ W 2 ; U2 Usuch that x 2 Uo

:

Order as follows: .1; U1/ 6 .2; U2/ iff 1 6 2 and U2 U1. For each.;U/2 pick 2 M so that x2 U for all > (such a exists sincex!x andx 2 U). Define ' W .;U /7!x for all.;U /2 . It now easy tosee that this subnet converges to x . ut

4.3 Filters

12A. Examples of Filter Convergence

I Exercise 87. Show that if a filter in a metric space converges, it must con-

verge to a unique point.

Proof. Suppose a filter F in a metric space .X;d/ converges to x; y2 X. Ifx y , then there exists r > 0such that B.x;r/ \ B.y;r/D . But since F!xand F! y, we must have B.x;r/2 F and B.y;r/2 F. This contradicts thefact that the intersection of every two elements in a filter is nonempty. Thus,

xD y. ut

12C. Ultrafilters: Uniqueness

I Exercise88. If a filterF is contained in a unique ultrafilterF0, then FD F0.

Proof. We first show: Every filterFon a non-empty setX is the intersection of

the family of ultrafilters which include F.

Let E be a set which does not belong to F. Then for each set F2 F wecannot haveF E and hence we must have F\Ec . So F[fEcg generatesa filter on X, which is included in some ultrafilter FE . SinceE

c 2 FE we musthave E FE . Thus E does not belong to the intersection of the set of allultrafilters which include F. Hence this intersection is just the filter F itself.

Now, ifF is contained in a unique ultrafilter F0, we must have FD F0. ut


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SECTION 4.3 FILTERS 35

12D. Nets and Filters: The Translation Process

I Exercise 89. A net .x/ has x as a cluster point iff the filter generated by

.x/hasx as a cluster point.

Proof. Supposex is a cluster point of the net .x/. Then for every nhoodU of

x, we have x2 U i: o: But then U meets every B0 fxW > 0g, the filterbase of the filter F generated by .x/; that is, x is a cluster point ofF. The

converse implication is obvious. ut

I Exercise90. A filterF hasx as a cluster point iff the net based on F hasx

as a cluster point.

Proof. Suppose x is a cluster point ofF. IfU is a nhood ofx , then U meets

every F 2 F. Then for an arbitrary .p;F/2 F, pick q 2 F\ U so that.q;F/

2F, .q;F / > .p;F /, and P.p;F /

Dp

2U; that is, x is a cluster point

of the net based on F.Conversely, suppose the net based on F hasx as a cluster point. Let U be

a nhood ofx . Then for every .p0; F0/2F, there exists .p;F / > .p0; F0/ suchthat p2 U. ThenF0 \ U , and sox is a cluster point ofF. ut

I Exercise91. If.x/is a subnet of.x/, then the filter generated by.x/is

finer than the filter generated by.x/.

Proof. Suppose .x/ is a subnet of .x/. Let F is the filter generated by

.x/, and F be the filter generated by .x/. Then the base generating Fis the sets B0 D fx W > 0g, and the base generating F is the setsB0

D fx

W > 0

g. For each such a B0 , there exists 0 such that 0 > 0;

that is,B0 B0 . Therefore, F F . ut

I Exercise 92. The net based on an ultrafilter is an ultranet and the filter

generated by an ultranet is an ultrafilter.

Proof. Suppose Fis an ultrafilter. LetE Xand we assume thatE2 F. Pickp2 E. If.q;F / > .p;E/, then q2 E; that is, P.p;F/2 E ev: Hence, the net

based on F is an ultranet.

Conversely, suppose.x/is an ultranet. LetE Xand we assume that thereexists0 such that x2E for all > 0. Then B0D fxW > 0g E and soE2 F, where F is the filter generated by.x/. Hence, F is an ultrafilter. ut

I Exercise 93. The net based on a free ultrafilter is a nontrivial ultranet.Hence, assuming the axiom of choice, there are nontrivial ultranets.

Proof. Let F be a free ultrafilter, and .x/ be the net based on F. It follows

from the previous exercise that.x/ is an ultranet. If.x/ is trivial, i.e.,xDxfor somex2Xand all2F, then for allF2 F, we must have FD fxg. Butthen

TFD fxg ; that is, F is fixed. A contradiction.


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Now, for instance, the Frechet filter F on R is contained in some free ultra-

filter Gby Example (b) when the Axiom of Choice is assumed. Hence, the net

based on Gis a nontrivial ultranet. ut


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5SEPARATION AND COUNTABILITY

5.1 The Separation Axioms

13B. T0- andT1-Spaces

I Exercise94. Any subspace of aT0- orT1-space is, respectively, T0 orT1.

Proof. Let X be a T0-space, and A X. Let x and y be distinct points in A.Then, say, there exists an open nhood U of x such that y U. Then U\ Ais relatively open in A, contains x , and y A \ U. The T1 case can be provedsimilarly. ut

I Exercise 95. Any nonempty product space isT0 orT1 iff each factor space

is, respectively,T0 orT1.

Proof. IfX is aT0-space, for each2 A, andx y in X , then for somecoordinate we have x y , so there exists an open set U containing, say,x but not y . Now

1 .U / is an open set in X containing x but not y.

Thus, X isT0.Conversely, ifX is a nonempty T0-space, pick a fixed pointb2 X , for

each2A. Then the subspace B fx2 X WxD b unlessDg is T0,byExercise 94, and is homeomorphic to X under the restriction to B of the

projection map. ThusX isT0, for each2 A. TheT1 case is similar. ut

13C. TheT0-Identification

For any topological spaceX, defineby x y ifffxg D fyg.I Exercise96. is an equivalence relation onX.

Proof. Straightforward. ut

I Exercise97. The resulting quotient space X= DX isT0.37


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38 CHAPTER 5 SEPARATION AND COUNTABILITY

Proof. We first show that X is T0 iff whenever x y thenfxgfyg. IfX isT0 andx y, then there exists an open nhood U ofx such that y U; thenyfxg. Since y2fyg, we havefxgfyg. Conversely, suppose that x yimplies that

fx

gfy

g. Take any x

y in Xand we show that there exists an

open nhood of one of the two points such that the other point is not in U. If

not, theny2fxg; sincefxg is closed, we havefygfxg; similarly,fxgfyg. Acontradiction.

Now take anyfxg fyg inX= . Then fxg D fxg fyg D fyg. Hence,X= isT0. ut

13D. The Zariski Topology

For a polynomial P in n real variables, let Z.P/ D f.x1; : : : ; xn/ 2 Rn WP .x1; : : : ; xn/ D 0g. Let Pbe the collection of all such polynomials.

I Exercise 98.fZ.P/W P2 Pg is a base for the closed sets of a topology (theZariski topology) on Rn.

Proof. Denote Z fZ.P/W P2 Pg. If Z.P1/ and Z.P2/ belong to Z, thenZ.P1/ [ Z.P2/D Z.P1 P2/2 Z since P1 P22 P. Further,

TP2PZ.P/D

since there are P2 P with Z.P/D (for instance, PD 1 C X21C C X2n ).It follows from Exercise 48 that Z is a base for the closed sets of the Zariski

topology on Rn. ut

I Exercise99. The Zariski topology on Rn isT1 but notT2.

Proof. To verify that the Zariski topology is T1, we show that every single-

ton set in Rn is closed (by Theorem 13.4). For each .x1; : : : ; xn/2 Rn, define apolynomialP2 P as follows:

PD .X1 x1/2 C .Xn xn/2:

Then Z.P/ D f.x1; : : : ; xn/g; that is,f.x1; : : : ; xn/gis closed.To see the Zariski topology is not T2, consider the R case. In R, the Zariski

topology coincides with the cofinite topology (seeExercise 100). It is well know

that the cofinite topology is not Hausdorff (Example 13.5(a)). ut

I Exercise100. OnR, the Zariski topology coincides with the cofinite topology;

in Rn,n > 1, they are different.

Proof. OnR, everyZ.P/is finite. So on R every closed set in the Zariski topol-

ogy is finite since every closed set is an intersection of some subfamily ofZ.

However, ifn > 1, thenZ.P/can be infinite: for example, consider the polyno-

mialX1X2 (letX1D 0, then allX22 Ris a solution). ut


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SECTION 5.2 REGULARITY AND COMPLETE REGULARITY 39

13H. Open Images of Hausdorff Spaces

I Exercise 101. Given any set X, there is a Hausdorff space Y which is the

union of a collection

fYx

Wx

2X

gof disjoint subsets, each dense inY.

Proof. ut

5.2 Regularity and Complete Regularity

Theorem5.1 (Dugundji 1966). a. Let PW X! Y be a closed map. Given anysubsetS Yand any openU containingP1.S/, there exists an open V Ssuch thatP1.V / U.

b. LetP W X! Y be an open map. Given any subsetS Y, and any closedAcontainingP

1

S, there exists a closedB Ssuch thatP1

.B/ A.Proof. It is enough to prove (a). Let VD YX P .XX U /. Then

P1.S/ UH) XX U XX P1.S/ D P1.YX S/H) P .XX U / P P1.YX S/H) YX P P1.YX S/ V:

Since P P1.YX S / YX S, we obtain

SD YX .YX S/ YX P P1.YX S/ VI

that is,S

V. BecauseP is closed,Vis open in Y. Observing that

P1.V / D XX P1P.XX U / XX .XX U / D U

completes the proof. ut

Theorem5.2 (Theorem 14.6). IfX isT3 andf is a continuous, open and closed

map ofX ontoY, thenY isT2.

Proof. By Theorem 13.11, it is sufficient to show that the set

A f.x1; x2/ 2 X XW f .x1/ D f .x2/g

is closed in X X. If .x1; x2/ A, then x1 f1

f.x2/. Since a T3-space isT1, the singleton setfx2g is closed in X; sincef is closed,ff .x2/g is closed inY; sincef is continuous, f1f.x2/is closed in X. BecauseX isT3, there are

disjoint open setsU andV with

x12 U; and f1f.x2/ V:


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Sincefis closed, it follows fromTheorem 5.1that there exists open set W Ysuch thatff .x2/g W, andf1.W / V; that is,

f1f.x2/ f1.W / V:

Then U f1.W / is a nhood of.x1; x2/. We finally show that U f1.W / \AD . If there exists .y1; y2/2 A such that .y1; y2/2 U f1.W /, then y12f1f.y2/ f1.W /; that is, y12 U f1.W /. However, U\ V D andf1.W / V imply that U\ f1.W / D . A contradiction. ut

Definition 5.3. If X is a space and A X, then X=A denotes the quotientspace obtained via the equivalence relation whose equivalence classes are A

and the single point setsfxg,x2 XX A.

Theorem5.4. IfX isT3 andY is obtained fromXby identifying a single closed

setA in Xwith a point, thenY isT2.

Proof. LetA be a closed subset of a T3-spaceX. ThenXX Ais an open subsetin both X and X=A and its two subspace topologies agree. Thus, points in

XXA X=Aare different fromAand have disjoint nhoods as Xis Hausdorff.Finally, for x2 XX A, there exist disjoint open nhoods V .x/ and W.A/. Theirimages,f .V /andf .W /, are disjoint open nhoods ofx and AinX=A, because

VD f1f.V /and WD f1f.W /are disjoint open sets in X. ut

5.3 Normal Spaces

15B. Completely Normal Spaces

I Exercise 102. X is completely normal iff wheneverA andB are subsets of

X withA \xBDxA \ BD , then there are disjoint open setsU AandV B.

Proof. Suppose that wheneverAandB are subsets ofXwithA\xBDxA\BD, then there are disjoint open setsU AandV B . LetY X, andC; D Y

be disjoint closed subsets ofY. Hence,

D clY.C / \ clY.D/ D xC\ Y \ xD \ Y DxC\ xD \ Y :

Since D clY.D/, we havexC\ DD . Similarly, C\xDD . Hence there are

disjoint open setsU0

andV0

inXsuch thatC U0

andD V0

. LetUD U0

\YandVDV0 \ Y. ThenU andVare open in Y,C U, andDV; that is, Y isnormal, and soXis completely normal.

Now suppose that X is completely normal and consider the subspace YXX .xA \xB/. We first show thatA; B Y. IfA Y, then there exists x2 Awithx Y; that is, x2 xA \xB. But then x2 A \xB . A contradiction. Similarly for B .In the normal spaceY, we have


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SECTION 5.4 COUNTABILITY PROPERTIES 41

clY.A/ \ clY.B/ D xA \ Y \ xB\ Y D .xA \xB/ \ XX .xA \xB/ D :

Therefore, there exist disjoint open sets U clY.A/ and V clY.B/. SinceA clY.A/and B clY.B/, we get the desired result. ut

I Exercise 103. Why cant the method used to show every subspace of a reg-

ular space is regular be carried over to give a proof that every subspace of a

normal space is normal?

Proof. In the first proof, ifA Y Xis closed inY andx2 YXA, then theremust exists closed set B in X such that x B . This property is not applied iffxgis replaced a general closed set B inY. ut

I Exercise104. Every metric space is completely normal.

Proof. Every subspace of a metric space is a metric space; every metric space

is normalRoyden and Fitzpatrick(2010,Proposition 11.7). ut

5.4 Countability Properties

16A. First Countable Spaces

I Exercise105. Every subspace of a first-countable space is first countable.

Proof. Let AX. Ifx2 A, then V is a nhood ofx in A iffVD U\ A, whereUis a nhood ofx2 X(Theorem 6.3(d)). ut

I Exercise 106. A productX of first-countable spaces is first countableiff each X is first countable, and all but countably many of the X are trivial

spaces.

Proof. IfX is first-countable, then each X is first countable since it ishomeomorphic to a subspace ofX . If the number of the family of untrivialsetsfX g is uncountable, then for x2 X the number of nhood bases isuncountable. ut

I Exercise 107. The continuous image of a first-countable space need not be

first countable; but the continuous open image of a first-countable space is first

countable.

Proof. LetXbe a discrete topological space. Then any function defined on X

is continuous.

Now suppose that Xis first countable, and fis a continuous open map of

X onto Y. Pick an arbitrary y2 Y. Let x2 f1.y/, and Ux be a countablenhood base ofx . IfW is a nhood ofy , then there is a nhood V ofx such that


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f . V /W sincef is continuous. So there exists U2 Ux withf. U /W. Thisproves thatff. U / W U2 Uxgis a nhood base ofy . Sinceff. U / W U2 Uxgis ut


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6COMPACTNESS

6.1 Compact Spaces

17B. Compact Subsets

I Exercise108. A subsetE ofXis compact iff every cover ofE by open subsets

ofXhas a finite subcover.

Remark(Lee 2011,p. 94). To say that a subsetof a topological space is com-

pact is to say that it is a compact space when endowed with the subspace

topology. In this situation, it is often useful to extend our terminology in the

following way. IfXis a topological space andA X, a collection of subsets ofXwhose union contains A is also called a cover of A; if the subsets are open

in X we sometimes call it an open cover of A. We try to make clear in each

specific situation which kind of open cover ofA is meant: a collection of open

subsets of A whose union is A, or a collection of open subsets of X whose

union containsA.

Proof. The only if part is trivial. So we focus on the if part. Let U be an

open cover ofE , i.e.,UD SfUW U2 Ug. For everyU2 U, there exists an openset VU inX such that UD VU\ E. ThenfVUW U2 Ug is an open cover ofE ,i.e.,U SfVUW U2 Ug. Then there exists a finite subcover, say VU1 ; : : : ; V UnoffVUW U2 Ug, such that E

SniD1 VUi . Hence, ED

SniD1.VUi\ E/; that is,E

is compact. ut

I Exercise 109. The union of a finite collection of compact subsets of X is

compact.

Proof. Let A and B be compact, and U be a family of open subsets of X

which covers A[ B . Then U covers A and there is a finite subcover, say,UA1 ; : : : ; U

Am of A; similarly, there is a finite subcover, say, U

B1 ; : : : ; U

Bn of B.

But thenfUA1 ; : : : ; U Am ; UB1 ; : : : ; U Bng is an open subcover ofA [ B , so A [ B iscompact. ut

43


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References

[1] Adamson, Iain T. (1996) A General Topology Workbook, Boston:

Birkhuser. [33]

[2] Ash, Robert B. (2009) Real Variables with Basic Metric Space Topology,

New York: Dover Publications, Inc. [22]

[3] Dugundji, James(1966)Topology, Boston: Allyn and Bacon, Inc. [39]

[4] Lee, John M. (2011) Introduction to Topological Manifolds, 202 of Grad-

uate Texts in Mathematics, New York: Springer-Verlag, 2nd edition. [25,

43]

[5] McCleary, John (2006) A First Course in Topology: Continuity and Di-

mension, 31 of Student Mathematical Library, Providence, Rhode Island:

American Mathematical Society. [29]

[6] Royden, Halsey and Patrick Fitzpatrick (2010) Real Analysis, New

Jersey: Prentice Hall, 4th edition. [41]

[7] Willard, Stephen (2004) General Topology, New York: Dover Publica-

tions, Inc. [i]

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