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FUNDAMENTOS DE TOPOLOGIA E ANALISE REAL

1o Semestre 2010/2011

1o Teste 5-11-2010 Durac ao 1h 30m

1. (2 val.) Let X= Z+ with the topology generated by the basis { 2n 1, 2n }, n Z+.Show that X is not Hausdorff. How many limits can a convergent sequence in Xhave?

Answer: Let z Z+. Any neighborhood ofz contains a basis element at z, that is,

ifz Uwith Uopen, then {z,z+

1} U, z odd, or {z 1,z} U, z even. Hence, zand z + 1 (or z and z 1, z even) do not have disjoint neighborhhods, and Xis notHausdorff.

Let zn z in X, then for n N, zn {z,z + 1} (or zn {z 1,z}, z even). It followsthat we have also zn z+ 1 (or z 1, z even) and, moreover, that ifw z,z+ 1 thenzn does not converge to w. We conclude that any convergent sequence has exactly2 limits.

2. (1.5 val.) Let R be the space of real sequences with the product topology and R

be the subspace of sequences such that for some N N, xn = 0 for n > N. Show

that:

a) R is dense in R.

Answer: Want to show that R = R. Let x = (xn) R and Ube a neigh-

borhood of x, U = nZ+Un, where Un = R, for n n1, ...,np and Uni is neigh-borhood ofxni . Ifyn = xni ,forn = ni andyn = 0 otherwise, theny = (yn) R

U.

Hence any neighborhood ofx intersects R, so x R and the result follows.

b) R is connected. Does it follow that R is also connected?

Answer: Write R =

NNRN, where RN is the space of sequences with xn = 0for n > N. Since RN is homeomorphic to R

N, under the map x (x1, ...,xN),and RN is connected, RN is also connected. Moreover, (0, 0, ...)

NN RN .

Therefore, R is connected, being the union of connected sets with one pointin common.

Since the closure of a connected set is connected, R is also connected.

3. (3 val.) Let Xbe a topological space. Show that:

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a) For any A, B X, A B = A B.

Answer: Since A B A B, have A B A B = A B (the union of closedsets is closed). On the other hand, A AB, hence A A B and the same for

B. Hence A B A B.

b) Any finite union of compact sets is compact.

Answer: Let A = K1 ...Kp, with Kp compact, and take an open cover {U} ofA. Then {U} covers each Ki so that we can take a finite subcover of Ki. The(finite) union of these finite subcovers yields a finite subcover for A. Hence Ais compact.

c) Let A Xbe closed and f : A [0, 1] be continuous such that f = 0 on A \ U,for some U A, open in X. Then the function g : X [0, 1] extending f andsuch that g = 0 on X\A is continuous. (Hint: pasting lemma.)

Answer: Let h(x) = 0 be defined on the closed set X\ U. Then on A (X\ U) =A \ U, we have f = h. Since A is also closed, A (X\ U) = X, and f, h arecontinuous, by the pasting lemma we have that g : X [0, 1] with g = f on Aand g = h = 0 on X\ U X\A is continuous.

d) Let now Xbe locally compact Hausdorff and K Xcompact. Then:

i) There exists an open set Vsuch that K Vand Vis compact.

Answer: Since X is locally compact Hausdorff, any x Xhas a neigh-

borhood Vx with Vx compact. Then {Vx}xK is an open cover for K andsince K is compact, we can take a finite subcover Vx1 , ...,Vxp . If we let

V= Vx1 ... Vxp , then Vis open, K Vand V= Vx1 ... Vxp , by a), whichis a compact set by b).

ii) There exist a continuous function g : X [0, 1] and a compact set K Ksuch that g(K) = 1 and g(x) = 0 for x K.

Answer: Let K = Vas in i). Since K is compact Hausdorff(it is a subspaceof a Hausdorff space), it is normal, so we can apply Urysohns lemma. 1

Have that Kis closed in K (compact and K Hausdorff), and Vis open in X,hence inK V, so K \Vis also closed. It follows that there is a continuousf : K [0, 1] such that f(K) = 1 and f(K \ V) = 0. Then take g extendingf as in c).

1Note that we cannot apply Urysohns lemma directly to X, since locally compact Hausdorff spacesare not necessarily normal.

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4. (2 val.) Prove or disprove:

a) Any regular second countable space is normal.

Answer: True. By Urysohns metrization theorem, any regular second counta-ble space is metrizable, hence normal.

b) Any first countable space is second countable.

Answer: False. Let Xbe any uncountable discrete space: it is first countable,since {x} is a basis for neighborhoods of x, but it does not have a countablebasis, since any basis must contain all the open sets {x}, for x X, and X isuncountable.

(Another example: Rl = R with the topology generated by [a, b[, a, b R: it isfirst countable, take [r, q[, with r x < q, r, q Q, but not second countable, sincefor any a R, [a, b[ contains a basis element which is also a neighborhood ofa,so any basis contains at least one set [a, ba[ for some ba. Since R is uncountable,any basis is uncountable. Note that in this case Q is dense in Rl, so it is alsoseparable.)

5. (1.5 val.) Let (X, d) be a metric space. Show that:

a) A Xis totally bounded if and only ifA is totally bounded.

Answer: Recall that a set is totally bounded if for any > 0, there is a finitecover by balls.

IfA is totally bounded, then A A is also totally bounded (any cover ofA isalso a cover ofA).

Now assume A totally bounded, and let > 0. First note that if < then

B(x, ) B(x, ): let y B(x, ), so can take x B(x, ) B(y, ). Thend(x,y) d(x, x) + d(x,y) < .

If we take then < , since A is totally bounded, can cover A by finite balls,A B(x1,

) ... B(xp, ) and we have

AB(x1

,

) ...

B(xp,

)B(x1

,

) ...

B(xp,

).

Hence A is totally bounded.

b) If (X, d) is complete, then A Xhas compact closure if and only if it is totallybounded.

Answer: In a complete metric space, a subspace is compact if and only if it

is closed and totally bounded. Hence, A is compact if and only ifA is totallybounded (it is closed) if and only ifA is totally bounded, by a).

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