exercicios munkres

8/12/2019 exercicios munkres

1/22

Exercise Problems

13 Basis for a Topology

1. LetXbe a topological space; let A be a subset ofX. Suppose that for each x A thereis an open set U containingxsuch that UA. Show that A is open.

sol) For eacha A let Ua be an open set with a Ua A. Then A=aA

Ua, soA is open.

3. Show that the collectionTc of all subsets U ofX, where X Ueither is countable or X,is a topology on the set X.

sol) (i)X X= is countable andX = X, soX, Tc. (ii) LetU Tc for J. Then

XJ

U=J

(X U) =

X if X U= X J

countable otherwise

ThusJ

U Tc. (iii) LetUi Tc for 1 i n. Then

Xi

Ui=

i(X Ui) =

X if X Ui= X for some i

countable otherwise

This shows

iUi Tc.

4. Let{T} be a family of topologies on X.

(a) Is

U a topology on X?

sol) No. T1 T2 is not a topology since {a, b} {b, c}= {b} / T1 T2 (See (c) below).

(b) Show that there is a unique smallest topology on X containing all T, and a uniquelargest topology contained in allT.

These are based on student solutions.

1


2/22

sol) The unique largest topology contained in allT is

T. Noting

T is a subbasis for atopology onX, letTs be the topology generated by

T. IfU Ts thenUcan be written

asU= U,1 U,n where U,1, . . . , U ,n T.

Thus U T for any topologyT that contains

T, i.e. Ts T. Suppose Ts is a smallesttopology containing all T. In general we cant say eitherTs Ts or T

s Ts. However,

since Ts contains

T, so Ts T

s. ThenTs Ts since T

s is a smallest topology that

contains all T. This complete the proof. The unique largest topology contained in all Tis Tl =

T. The proof of uniqueness is similar to the above.

(c) If X = {a,b,c}, let T1 = {, X, {a}, {a, b}} and T2 = {, X, {a}, {b, c}}. Find thesmallest topology containing T1 and T2, and the largest topology contained in T1 and T2.

sol) By (b), Ts = {, X, {a}, {b}, {a, b}, {b, c}}and Tl = {, X, {a}}.

5. Show that ifA is a basis for a topology on X, then the topology generated by A equalsthe intersection of all topologies on X that contain A.

sol) Let {T} be a family of all topologies on X that contains the basis A and TA be thetopology generated by the basis A. Note

T is a topology on X and

T TA since

A TA. On the other hand, any U TA is a union of elements in A, so UT for all and thus TA

T. This completes the proof.

16 The Subspace Topology

1. Show that ifY is a subspace ofX, and A is a subset ofY, then the topology A inheritsas a subspace ofY is the same as the topology it inherits as a subspace ofX.

sol) Let Tbe the topology ofX and TYdenote the subspace topology ofY. We also denotebyTA/Y (resp. TA/X) the subspace topology ofA as a subspace ofY (resp. X). Then

W TA/Y W =V A for some V TY W = (U Y) A for some U T

W =U A for some U T W TA/X.

4. A mapf :X Y is said to be an open map if for every open setU ofX, the setf(U) isopen in Y . Show that 1: X Y Xis an open map.

sol) LetWbe an open set in X Y. ThenW =

(U V) where U andV are open in Xand Y respectively. So,

1(W) =1(

(U V) ) =

1(U V)) =

U

This shows 1 is an open map.

2


3/22

17 Closed Sets and Limit Points

6. LetA, B and Cdenote subsets of a space X. Prove the followins:

(a) IfA B , then A B.

sol) Recall thatA is the intersection of all closed sets that contain A. Since B is closed and itcontainsA, we have A B.

(b) A B= A B.

sol) A B A B sinceA B is closed set containing A B. On the other hand, A B is aclosed set that contains both A and B , soA B A B.

(c)

A

A; give an example where equality fails.

sol)A is a closed set that contains all A, so it contains all A, i.e.,A A (Notethat

A is, in general, not closed !) Let An = {

1n } for n N. Then 0

An but

0 /

An.

8. LetA, B andA denote subsets of a space X. Determine whether the following equationshold ; if an equality fails, determine whether one of the inclusions or holds.

(a) A B A B.

sol) A Bis a closed set that containsA B, soA B A B. LetA = (0, 1) andB = (1, 2).Then A B= , while A B= {1}.

(b)AAsol) Similarly as above,

A is a closed set that contains

A. So,

A

A.

(c) A B (A B).

sol) Let x (A B) = A (XB) A (XB). Then for any neighborhood U of x,U A= (since x A) and U (X B)= (since x U (X B)). Thus

x A (X B) =A B.

LetA = [0, 1] and B = (0, 1). Then A B= {0, 1}, while A B=.

9. LetA Xand B Y. Show that A B=A B in X Y.

sol) A B A Bsince the latter is a closed set (see prob 3) that containsA B. Conversely,let (x, y) A B. Then for any basiselement U V containing (x, y) we have

(U V) (A B) = (U A) (V B)=.

Thus x A B.

13. Show that Xis Hausdorff iff the diagonal = {(x, x)|x X} is closed in X X.

3


4/22

sol) Suppose X is Hausdorff and let (x, y) XX . Since x = y, there are disjointneighborhoods U andV ofx and y respectively. Then (x, y) U V X X . Thisimplies X X is open, or is closed. Conversely, ifX X is open then for anyx=y there is a basis element U Vsuch that (x, y) U V X X . This implies

U andVare disjoint neighborhoods ofx and y respectively.

19. IfA X, we define the boundary ofAby the equation Bd A= A X A.

(a) Show that IntAand BdAare disjoint, and A = IntA BdA.

sol) Suppose there is x IntA BdA. Then IntA is a neighborhood ofx and x X A, soIntA (X A)= which is impossible since IntA A. On the other hand, note that Acontains both IntA and BdA. Letx A IntA. Suppose there is a neighborhood U ofxsuch that U A. Thenx U IntA, which contradicts to the assumption on x. Anyneighborhood ofx thus intersectsX Aand hencex BdA. This showsA = IntA BdA.

(b) Show that BdA= A is both open and closed.

sol) If BdA = then IntA A A = IntA where the last equality follows from (a). Con-versely, ifA is both open and closed, then IntA = A = A, so again by (a) BdA = .

(c) Show that U is open BdU=U U.

sol) U is open U= IntU BdU =U Uwhere the last equivalence follows by (a).

(d) IfUis open, is it true thatU= Int(U)? Justify your answer.

sol) Not true; let U= (1, 0) (0, 1). Then Int(U) = (1, 1).

20. Find the boundary and the interior of each of the following subsets ofR2 :

(d) D = {(x, y)|xis rational}.

sol) IntD= (since D contains no open sets in R2) and hence BdD= D = R2.

18 Continuous Functions

2. Suppose thatf : X Y is continuous. Ifx is a limit point of the subset A ofX, is itnecessarily true that f(x) is a limit point off(A)?

sol) Not true; constant functions.

3. LetXandX denote a single set in two topologiesT andT, respectively. leti : X Xbe the identity function.

4


5/22

(a) Show that i is continuous T is finer than T.

sol) i is continuous i1(U) =U T for allU T T T .

6. Find a functionf : R R is continuous ar precisely one point.

sol) Define a functionf : R R by

f(x) =

x ifxis rational,x ifxis irrational

Then,fis continuous at 0 since f(/2, /2) (, ) for any >0. On the other hand,ifx = 0 then the interval (f(x) |f(x)|/2, f(x) + |f(x)|/2) does not contain f(x , x + )for any >0. This shows fis not continuous at x= 0.

20 and 22

3. LetXbe a metric space with metric d. Show that d : X X R is continuous.

sol) Triangle inequality shows if (x, y) Bd(x0, /2) Bd(y0, /2) then

|d(x, y) d(x0, y0)|< .

This shows d is continuous at any (x0, y0) X X.

2. (a)Letp : XYbe a continuous map. Show that if there is a continuous map f :XY

such thatp fequals to the identity map ofY, then p is a quotient map.

sol) Let idY be the identity map ofY. ThenU = id1Y (U) = f

1(p1(U)). Thusp1(U) isopen in X = U is open Y. This impliesp is a quotient map since p is continuous andsurjective subjectivity ofp follows fromp f=idY.

4. (a) Define an equivalence relation on the plane X= R2 as follows,:

(x0, y0) (x1, y1) if x0+ y20 =x1+ y

21.

LetX be the corresponding quotient space. It is homeomorphic to a familiar space. Whatis it?

sol) Letp : X X be the quotient map induced from the equivalence relation and define afunctiong : X R byg(x, y) =x + y2. Theng is constant on the preimagep1([(x0, y0)])for each point [(x0, y0)] in X

. Thus there is a continuous function f : X R withg =f p. Since g is onto and g(x0, y0) =g(x1, y1) (x0, y0) (x1, y1), f is onto andone-to-one. In order to prove fis a homeomorphism, we will show thatfis an open map(i.e. its inverse function is continuous). LetG: XXdefined by G(x, y) = (g(x, y), y).Then G is a homeomorphism with the inverse function G1(x, y) = (x y2, y). Since

5


6/22

g= 1 G, where : X X Xis the projection map onto the first factor, and both 1and G are open map, so is g . On the other hand, subjectivity ofp andg= f p give

g(p1(U)) =f(p(p1(U))) =f(U).

Consequently, U is open = p1(U) is open = g (p1(U)) =f(U) is open.

23 Connected Spaces

2. Let{An} be a sequence of connected subspaces ofX, such that An An+1 = for all n.Show that An is connected.

sol) We will use induction together with the following claim :Claim : LetAand B be connected subsets ofXwithA B=. ThenA Bis connected.

proof : Suppose A B is a disjoint union of open sets C and D of A B. IfC = ,since C is both open and closed in A B, C A and C B are respectively nonemptyboth open and closed in A and in B (check using subspace topology!). Thus C A = Aand C B = B , or equivalently, A C and B C. This impliesD =, so there is noseparation ofA B, namelyA B is connected.

The claim showsA1 A2is connected. SupposeBk = ik

Aiis connected. ThenBk Ak+1

is also connected by the claim. The induction principle thus completes the proof.

6. Let A X. Show that if C is a connected subspace of X that intersects both A andX A, then C intersects Bd A.

sol) First, recall Bd A= A (X A) and A = Int A Bd A. Suppose C Bd A= . Then

C A= C A and C (C A) =C (X A) =C (X A).

C A is thus both open and closed in C. Since C is connected, either C A = orC A = C where the latter is equivalent to C (X A) = . This contradicts to theassumption C intersects both A andX A.

11. Let p : X Y be a quotient map. Show that if each p1

({y}) is connected, and ifY isconnected, then X is connected.

sol) LetA be both open and closed inX. Then for eachy Y,p1({y}) Ais both open andclosed in p1({y}). Since p1({y}) is connected, p1({y}) A or p1({y}) (X A).This implies p(A) p(X A) =, so we have

p1p(A)

p1

p(X A)

= .

6


7/22

This shows A= p1p(A)

and X A= p1

p(X A)

since

X = p1(Y) = p1p(A) p(X A)

= p1

p(A)

p1

p(X A)

A p

1p(A) and X A p1p(X A)Consequently, p(A) is both open and closed in Y since p is a quotient map. By Connect-edness ofY we have either A= or A= X.

12. LetY X; letXandYbe connected. Show that ifA and B form a separation ofX Y,then Y A and Y B are connected.

sol) Suppose Y A is not connected. Then there is a separation of Y A, i.e. Y A is adisjoint union of nonempty open sets C and D ofY A. Since Y is connected, applyingLemma 23.2, we can assume Y C. This, together with the factY A=C D, givesD A. On the other hand, Lemma 23.1 shows

C D = and C D = (0.1)

Similarly, together with the fact D A, again by Lemma 23.1 we have

B D B A = and B D B A = . (0.2)

Now, note that X is a disjoint union of (B C) and D :

X = Y (X\ Y) = Y A B = C D B = (B C) D. (0.3)

By (0.1) and (0.2), (B C) D = (B C) D = , so C B = C B by (0.3). ThusC B is closed inX. Repeating the same argument givesD = D is also closed inX. Thiscontradicts to the fact X is connected.

Similar solution without using Lemma 23.1 and Lemma 23.2 :

SinceA and B are nonempty open sets in X Y, there exist nonempty open sets UA andUB ofXsuch that A = UA (X Y) andB =UB (X Y), soX Ycan be written as

X Y = A B = [UA (X Y)] [UB (X Y)]

Suppose Y A is not connected. Then Y A has a separation, so (similarly as above)there exist nonempty open sets UC and UD ofX such that Y A can be written as :

Y A = [UC (Y A)] [UD (Y A)]

where both UC (Y A) and UD (Y A) are not empty. It then follows that

Y = Y (Y A) = (UC Y) (UD Y).

Since Y is connected, we can assume UC Y =Y, i.e. Y UC, and UD Y =. In thiscase,A UD = and

A = (UC A) (UD A).

7


8/22

This, together with the facts Y UC, A UA and B UB , implies

X = Y A B (UB UC) A (UB UC) (UD A)

(UB UC) (UA UD) X

Note that (i) =A UD UA UD and (ii) (UB UC) (UA UD) = since

(UA UD) UB UA UB = and (UA UD) UCUD UC=.

We have contradiction to the fact X is connected.

24 Connected Subspaces of the Real Line

1. (a) Show that no two of the spaces (0, 1), (0, 1], and [0, 1] are homeomorphic.

sol) Suppose there exists a homeomorphismf : (0, 1] (0, 1). Then the restriction map

g = f|(0,1) : (0, 1) (0, 1) {f(1)}

is also a homeomorphism. (0,1) is connected, but its image g(0, 1) = (0, 1) f(1) is notconnected. Contradiction!

(c) Show that Rn and R are not homeomorphic ifn >1.

sol) Suppose there exists a homeomorphismf : Rn

R. Then the restriction map

f|Rn\{f1(0)}

: Rn \ {f1(0)} R \ {0}

is also a homeomorphism. Rn \ {f1(0)}is path connected, while its image R \ {0}is not.Contradiction!

2. Let f : S1 R be a continuous map. Show there exists a point x of S1 such thatf(x) =f(x).

sol) Suppose not. Consider a continuous function

g: S1

R defined by g(x) = f(x) f(x).

The image g(S1) is connected since S1 is connected and g is continuous. Sinceg(x)= 0for all x S1, we have either g(x) > 0 for all x S1 or g(x) < 0 for all x S1. This isimpossible since g(x) =g(x).

3. Let f : X X be continuous. Show that ifX = [0, 1], then f(x) = x for some x X.What happens ifXequals to [0, 1) or (0, 1]?

8


9/22

sol) Define g : XX byg(x) =f(x) x. Then g is continuous and

g(0) =f(0) 0 0 and g(1) =f(1) 1 0.

Since g([0, 1]) is connected, g(x) = 0 for some x [0, 1], i.e., f has a fixed point. IfX= [0, 1) (resp. X= (0, 1]) then the function

f(x) = 12x + 12 (resp. f(x) =

12x)

has no fixed point.

8. (b) IfA X and A is path connected, is A necessarily path connected?

sol) No. (cf. p157 Example 6) Let S = {( x, sin( 1x ) ) | 1 < x 1}. Since S is the image of(0, 1] under the continuous map g : (0, 1] S defined by g(x) = ( x, sin( 1x ) ), S is pathconnected. In particular, Sis connected, so is its closure S in R2. Recall

S = V S where V = {0} [1, 1] R2

.

Suppose there is a path f : [a, c] Swithf(a) V and f(c) S. SinceV is closed in S,its preimagef1(V) is closed in [a, c] ; f1(V) is bounded and closed in R. The preimagef1(V) thus has a maximum b and b < c since f(c) S. Writeh(t) = (x(t), y(t)) for thecomposition map

h: [0, 1] [b, c] S

where the first map is t

(1 t)b + tc

/2 and the second map is the restriction off to[b, c]. WLOG, we may assumeh(0) = (0, 0). Sincex(t)>0 for t > 0, for each n > 0 wecan choose u satisfying

0< u < x( 1

n

) and sin( 1

u

) = (1)n.

Applying the Intermediate Value Theorem to the function t x(t), we can then find tnwith 0 < tn 0. Since any point in B(x, ) can be joinedto xby a line in B (x, ) U, by pasting two paths, any point in B(x, ) can be joined tox0 by a path in U. This implies P is open.

In order to prove P is closed in U, one can show U P is open as done in class. Letsdo differently. First recall that the closure ofP in U is P U, namely the closure ofPw.r.t the subspace topology ofU is P U. Let x P U. Since U is open in R2 andx U, we can choose an open ball B(x, ) U for small >0. Since x P, there existsy B(x, ) P. We can then joinx to x0 by pasting a path in U from x0 to y with aline segment in B (x, ) U fromy to x. Thus,x P. Since x is arbitrary, P =P U isclosed in U.

11. IfA is a connected subspace ofX, does it follows from that Int Aand Bd Aare connected?Does the converse hold?

sol) (i) X= R and A = (0, 1): Bd A= {0, 1}.

(ii) X= R2 and A = B

(0, 0), 1

B

(2, 0), 1

: Int A= B((0, 0), 1) B((2, 0), 1).

(iii) X= R2 and A= B((0, 0), 1) Xwhere X={((x, 0)|1< x


11/22

9. LetG be a topological group; let Cbe the component ofG containing the identity elemente. Show that Cis normal subgroup ofG.

sol) For eachx G,xCx1 =Rx1Lx(C) is connected whereLxandRxare homeomorphisms

defined by Lx(g) = xg and by Rx1(g) = gx1. On the other hand, since e = xex1 xCx1 and C is a component containing e, xCx1 C. A similar argument shows thatfor any a, b C, ab1C Cwhich implies ab1 C.

26 Compact Spaces

5. Let A and B be disjoint compact subspaces of the Hausdorff space X. Show that thereexists disjoint open sets U and V containing A and B , respectively.

sol) Since A is compact and A B = , for each b B there are disjoint neighborhoods UbandVb ofA and b respectively. By Compactness ofB there are finitely manyVb1, , Vbnwhose union contains B. We set

UA= Ub1 Ubn and VB =Vb1 Vbn .

ThenA UA, B VB and UA UB = sincex VB x Vbi for some bi x /Ubi x /UA.

6. Show that iff :XY is continuous, where X is compact and Y is Hausdorff, then f isa closed map.

sol) Let A be a closed set in X. ThenA is compact and hence f(A) is compact. Since Y isHausdorff,f(A) is closed.

7. Show that ifY is compact, then the projection 1: X Y X is a closed map.

sol) We will show thatX 1(A) is open ifA is closed in X Y. Letx X 1(A). Noting(x, y) / A for any y Y and X Y A is open, for each y Ychoose neighborhoodsUx,y ofx and Vy ofy satisfying

(x, y) Ux,y Vy X Y A.

Since {Vy | y Y} is an open covering ofY and Y is compact, there is a finite subcover{Vyi | i= 1, , n}. Now, set

U = Ux,y1 Ux,yn .

U is then a neighborhood ofx such that U Y X Y A. This shows U 1(A) =and hence X 1(A) is open.

11


12/22


13/22

sol) We will use the fact every topological group is regular (see p146 #7 (c)). Letc /A B.Then cB1 A = and hence by regularity of G for each b B there exist disjointneighborhoods Ub and Vb of cb

1 and A respectively. Note that cB1 = Lc (B) iscompact where Lc is the left translation by c and : G G defined by (g) =g1. One

can thus find finitely many Ub1, , Ubn whose union contains cB1. We set

U=Ub1 Ubn and V =Vb1 Vbn .

Then cB1 U, A V and U V = (cf. # 5 above). So, W =U B is a neighborhoodofc such that W AB U B V B = . This shows G AB is open.

(b) Let Hbe a subgroup ofG; let p: G G/Hbe the quotient map. IfH is compact,show that p is a closed map.

sol) Recall that p1(p(C)) = C H. IfC is closed, then p1(G/H p(C)) = G CH is openby (a). Thus G/H p(C) is open, i.e. p(C) is closed.

(c)Let Hbe a compact subgroup ofG. Show that ifG/His compact, thenG is compact.

sol) The proof directly follows from #12 and (b) since p1(gH) = gH is compact for anygHG/H.

27 Compact Subspaces of the Real Line

5. LetXbe a compact Hausdorff space; let {An} be a countable collection of closed sets of

X. Show that if each An has empty interior in X, then the union An has empty interiorin X.

sol) We will show that U An = for any open set U (this implies Int(An) = ). Wefirst show that given A1 and nonempty open set U there exists a nonempty open set V1satisfying

V1 U and V1 A1= .

Since IntA1= , there exists y U A1. NotingA1 is compact, choose a neighborhood ofW1 ofy and an open set W2 containingA1 withW1 W2= (by Lemma 26.4). Similarly,since X (W1 U) is closed and hence compact, choose a neighborhood W1 ofy and anopen set W2 containing X (W1 U) with W

1 W

2 = (again by Lemma 26.4). Let

V1= W

1

. The nonempty open set V1 then satisfies V1 W1 UU and V1 A1= .

Now, applying the same arguments inductively, given Anand Vn1one can find a nonemptyopen setVn satisfying Vn Vn1 and Vn An = . Consider the nested sequences

V1 V2

of nonempty closed sets ofX. Since X is compact, there exists x Vn (by Thm 26.9)such that x /An for each n since Vn An = for each n. Thus, x U An.

13


14/22

31 The Separation Axioms

2. Show that if X is normal, every pair of disjoint closed sets have neighborhoods whoseclosures are disjoint.

sol) Let A, B are disjoint closed sets. ThenX B is open and A X B and hence thereexists an open setUA withA UA andUA X B (see Lemma 31.1 (b)). In particular,B (XUA) sinceUA B=. Again by Lemma 31.1 (b) applied toB (XUA) givesUB open inXwithB UB andUB (X UA). The latter is equivalent to UB UA= .Thus UA and UB is the desired neighborhoods ofA and B respectively.

5. Letf, g : XYbe continuous; assume that Y is Hausdorff. Show that {x|f(x) =g(x)}is closed in X.

sol) Consider the graphGg ={(x, y) X Y | y= g(x)} ofg . Since g is continuous and Y isHausdorff, the graphGg is closed in X Y (cf. p171 ex 8). Define a function

F :X X Y by F(x) = (x, f(x)).

Then,F is continuous and hence F1(Gg) ={x|f(x) =g(x)} is closed in X.

6. Letp: XYbe a closed continuous surjective map. Show that ifX is normal, then sois Y.

sol) Lets first prove the given Hint :

Claim : IfUis an open set containing p1(y), then there is a neighborhood W ofy such

that p1

(W) U.

Proof : LetW=Yp(X U). ThenW is open inY sincep is a closed map and y W;suppose y P(X U). Then there existsx (X U) such that p(x) =y. This impliesx p1({y}) U. Contradiction! On the other hand, we have

p1(W) =p1(Y p(X U)) =X p1p(X U) X (X U) =U.

This completes the proof of the claim.

Let A and B be disjoint closed sets in Y. Thenp1(A) and p1(B) are disjoint closedsets inX. Using normality ofX, choose disjoint neighborhoodsUA andUB ofp

1(A) andp1(B) respectively. The claim then shows that for each a A (resp. b B) there is a

neighborhood Wa ofa (resp. Wb of b) with p1(Wa) UA (resp. p1(Wb) UB). Theopen sets

WA =

aAWa and WB =

bB

Wb

are desired neighborhoods ofAand B respectively;

p1(WA WB) =p1(WA) p

1(WB) UA UB =.

This together with subjectivity ofp shows WA WB =.

14


15/22

32 Normal Spaces

1. Show that a closed subspace of a normal space is normal.

sol) LetXbe normal and A be a closed subset ofX. LetA1 andA2 be disjoint closed subsetsofA. Since A is closed in X,A1 and A2 are also closed in X. Normality ofXthen showsthere are disjoint neighborhoods U1 and U2 of A1 and A2 respectively in X. Thus theopen sets U1 Aand U2 A in A are disjoint neighborhoods ofA1 and A2 in A.

33 The Urysohn Lemma

1. Examine the proof of the Urysohn lemma, and show that for givenr,

f1(r) =p>r

Upq r x /Ur,

(2) x /Ur f(x) r or equivalently f(x)< r x Ur.

Suppose x f1(r), i.e. f(x) = r. Thenf(x) < p for any rational p > r, so x

p>rUp

by (2). Similarly, for any rationalq < r, f(x) > q, so x / Uq Uq by (1). This impliesx /

qr

Upqr

Upq r, x Up Up, by (1) f(x) p. This showsf(x) r. Therefore, f(x) =r, i.e. x f1(r).

2. (a) Show that a connected normal space having more that one point is uncountable.

sol) LetXbe a connected normal space andxandy are distinct points ofX. Then by UryshonLemma, there is a continuous function f :X[0, 1] with f(x) = 0 and f(y) = 1. Sincethe image f(X) is connected, f(X) = [0, 1]; suppose not. Then f(X) is a disjoint unionof nonempty open sets f(X) [0, r) andf(X) (r, 1] for some r /f(X) where 0< r


16/22

sol) LetA and B are disjoint closed sets ofXand let

f(x) = d(x, A)

d(x, A) + d(x, B).

Recall that (i) d(x, A) = inf{ d(x, a) | a A }, (ii) it is a continuous function on X intoR and (iii) for closed A, d(x, A) = 0 iffx A (see p175-177). It thus follows thatf is acontinuous function on X (since A B = ) such that f(x) = 0 iffx A, f(x) = 1 iffx B and 0 f(x) 1 for all x X.

4. Recall A is a G set in X ifA is the intersection of a countable collection of open set ofX. Show that : Let Xbe normal. Then there exists a continuous function f :X[0, 1]such that f(x) = 0 for x A, and f(x)> 0 for x /A, if and only ifA is a closed G setin X.

sol) (by Jonathan) Suppose there exists a continuous functionf :X [0, 1] such thatf(x) = 0for x A, and f(x)> 0 for x /A. Then A is closed since A= f1(0) and A is a G

set

since

A =

nf1

[0,

1

n)

.

Conversely, suppose A is a closed G set. Then A = Un for some open sets Un. SinceX is normal, for closed A and open U1 A, there is an open set V1 with A V1 andV1 U1 (see Lemma 31.1). Inductively, for eachn N one can define an open set V 1

n+1

satisfyingA V 1

n+1and V 1

n+1

U1 U2 Un+1

V1n

.

Then we have A V1n

and V 1n+1

V1n

for all n N and

V1n

=A;

A V 1n+1 (U1 Un+1) V1n (U1 Un+1) Un = A.Now, similarly as in the proof of Urysohn lemma, writing rational numbers in the set(0, 1) { 1n |n N}as a sequence{p1, p2, }and using the open sets V1

nand induction on

finite subsets of{p1, p2, }, for any rational p (0, 1) one can define an open set Up Asatisfying Uq Up ifq < p and Up = V1

nifp = 1n . The functionf(x) = inf{p | x Up }

is then continuous and f(x) = 0 for x A. Moreover, f(x) > 0 for x / A; in this case,x /V1

nfor some n, sof(x) 1n (see p210 (2)).

(Another) Let A = Un. Then by Urysohn lemma, for each n there is a continuousfunction fn : X[0, 1] with fn 0 on A andfn 1 on X Un. Define a function

F :X[0, 1] by F(x) = fn(x)2n .Since 0 fn(x) 1 for all n, F is a well-defined function with F 0 on A. Let x / A.

Thenx X Un for some n (sinceA= Un) and hence F(x) fn(x)2n =

12n >0. Thus it

remains to show F is continuous. LetFn=n

i=1fi(x)2i

. Note that Fn is continuous for alln and

0 F(x) Fn(x)

i=n+1

1

2i =

1

2n+2 (0.4)

16


17/22


18/22

sol) LetC =C {(0, 0)}and (R2) = R2 {(0, 0)}and define continuous functions

f : R C by f(t) = (et cos t, et sin t),

G: (R2

)

R

by G(x, y) =

1

2ln (x

2

+ y

2

).

Then f is bijective and the restriction of G to C is the inverse function of f. Thecomposition f G : (R2) C is thus a retraction. Note that for any seq (xn, yn)in (R2) converging to (0,0) the sequence fG(xn, yn) also converges to (0,0). So, theextensionr: R2 C off Gdefined by r(0, 0) = (0, 0) is continuous (cf. Theorem 21.3)and thus r is a retraction, i.e. Cis a retract ofR2.

(b) Show that the knotted x-axis of Figure 35.2 is a retract ofR3.

sol) Noting K is homeomorphic to R, fix a homeomorphism f : K R. Then, since K isclosed in R3, by Tietze extension theorem, there is a continuous extension F : R3 R off. Then the composition f1 F : R3 Kis a retraction.

36 Imbeddings of Manifolds

2. LetXbe a compact Hausdorff space. Suppose that for each x X, there is a neighborhoodU ofxand a positive integer k such that Ucan be imbedded in Rk. Show that Xcan beimbedded in RN for some positive integer N.

sol) Since Xis compact, there is a finite open covering { Ui | i= 1, , m } ofXsuch that for

each i there is an imbedding i :Ui Rki for some ki >0. Note that X is normal sinceX is compact Hausdorff. So, there is a partition of unity{i} subordinated to {Ui} thatgives rise to an imbedding :

:X R R m times

Rk1 Rkm

defined by (x) =

1(x), , m(x), 1(x) 1(x), , m(x) m(x)

where for x inX supp(i) we define i(x) i(x) = 0 so that it is a continuous function on X.

5. LetXbe the union of the set R{0}and the two-point set {p, q}. TopologizeXby takingas basis the collection of all open intervals in R that do not contain 0, along with all sets

of the form (a, 0) {p} (0, a) and all sets of the form (a, 0) {q} (0, a), for a >0.

(b) Show that each of the space X {p} and X {q} is homeomorphic to R.

sol) Noting the collection of all open intervals in R that do not contain 0, along with all setsof the form (a, a) = (a, 0) {0} (0, a) for a >0 is a basis for the standard topologyon R, we define f :X {q} Rby f(x) =x for x R {0}and f(p) = 0. This functionf is bijective and both f and its inverse f1 send basis elements to basis elements. Thusfis a homeomorphism.

18


19/22

43 Complete Metric Spaces

2. Let (X, dX) and (Y, dY ) be metric spaces; let Y be complete. Let A X. Show thatiff : A Y is uniformly continuous, then f can be uniquely extended to a continuousfunction g : A Y, and g is uniformly continuous.

sol) Let a A. Then there is a sequence{an} in A converging to a (see Lemma 21.2). Sincef is uniformly continuous, {f(an)} is a Cauchy sequence in Y; given > 0 there exists >0 (depending only on ) such that ifdX(an, am)< thendY(f(an), f(am))< . Sincean a, in particular {an} is a Cauchy sequence in X, one can choose N >0 satisfying :ifn,m > N then dX(an, am)< . UsingY is complete, set g (a) = lim f(an). Then

(i) g :A Yis a well-defined function; suppose we have two sequences {an} and {bn}both converging toa A. Let {cn} be a sequence defined byc2n1= anand c2n = bn.Then a similar argument as above and the inequality

dX(an, bm) dX(an, a) + dX(a, bm) n, m N

imply that{f(cn)}is a Cauchy sequence in a complete metric space Y, so by unique-ness of the limit we have lim f(an) = lim f(c2n1) = lim f(c2n) = lim f(bn).

(ii) g is an extension off; for a A letan = a for all n. Then g(a) = lim f(an) =f(a).

(iii) g is uniformly continuous; given > 0 choose > 0 corresponding to the uniformcontinuity off and let a, b A with dX(a, b) < /2. One can then find sequences{an}and {bn}respectively converging to a and b such that there isN >0 satisfying :if n , m > N then dX(an, bm) < , dY(g(a), f(am)) < and dY(f(bm), g(b)) < .Consequently,

dY(g(a), g(b)) < dY(g(a), f(am)) + dY(f(an), f(bm)) + dY(f(bm), g(b)) < 3.

(iv) g is the unique continuous extension of f to A; suppose h is another continuousextension of f to A. Then for any a A and any sequence {an} converging to a,h(a) = lim h(an) = lim f(an) =g(a) (cf. Theorem 21.3).

4. Show that the metric space (X, d) is complete if and only if for every nested sequenceA1 A2 of nonempty closed sets ofX with diamAn 0, the intersection of thesets An is nonempty.

sol) The necessary condition for completeness of (X, d) is given by Lemma 48.3. Lets provesufficient condition. Let{xn} be a Cauchy sequence in (X, d). There exists N1 >0 such

that ifn,m > N1 then d(xn, xm)< 1. Choose n1> N1 and let A1= B(xn1, 1). Similarly,there exists N2 > N1 such that ifn,m > N2 then d(xn, xm) N2 and

let A2 =A1 B(xn2,12). Note that A2 A1 and ifn > N2 then xn A2 since N2 > N1.

Repeating the same argument with induction then shows for all k Nthere existsNk andclosed setAk such that (i)Ak Ak1 and diamAk 0 and (ii) xn Ak ifn > Nk. Now,the assumption and (i) imply that there is a point x in the intersection of the sets Ak.Then for any k >0, ifn > Nk thend(x, xn)0 there exists n > 0 such that

f(x n, x + n) (f(x) 13n , f(x) +

13n ).

Thus x Un. Conversely, let x Un. Then given > 0 and for 1

n < since

x Un there exists an open set U containing x with diam f(U) 0 choose N >0 such that

||An Am||2 =

i

|Ain Aim|

2 <

for all n, m > N . Taking limit n then gives ||A Am||2 . This implies (i)A Am Hand hence A= (A Am) + Am Hand (ii)Am A in H.

HW Find an example of a metric space that is not second countable.

sol) LetXbe an uncountable set and define a metricd by d(x, y) = 1 ifx =y andd(x, y) = 0

ifx= y. Then the metric topology is discrete, so Xis not second countable.

22

exercicios munkres

Documents