[PDF] The University of Sydney Pure Mathematics 3901 Metric Spaces

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The University of Sydney

Pure Mathematics 3901Metric Spaces 2000Tutorial 5

1.LetX= (X,d) be a metric space. Let (xn) and (yn) be two sequences inX

such that (yn) is a Cauchy sequence andd(xn,yn)→0 asn→ ∞.Prove that (i) (xn) is a Cauchy sequence inX, and (ii) (xn) converges to a limitxif and only if (yn) also converges tox.

Solution.

(i) Letε >0. Sinced(xn,yn)→0 asn→ ∞, there isN1such that d(xk,yk)< ε/3 for allk > N1. Since (yn) is a Cauchy sequence, there is N

2such thatd(ym,yn)< ε/3 for allm, n > N2. PutN= max{N1,N2}.

Then, by the triangle inequality, for allm, n > Nwe have

Hence (xn) is a Cauchy sequence.

(ii) Suppose that (yn) converges tox. Thend(yn,x)→0 asn→ ∞. Now by the triangle inequality, asn→ ∞; sod(xn, x)→0 asn→ ∞. So (xn) converges tox. Similarly, n→ ∞, whence (yn) converges toxalso.

2.Prove that every Cauchy sequence in a metric space (X,d) is bounded.

Solution.

(This was proved in lectures). Let (xn) be a Cauchy sequence of (X,d). By the definition of Cauchy sequence, applied withε= 1, there existsNsuch thatd(xm,xn)<1 for allm, n≥N; soxn?B(xN,1) for alln≥N. Now definer= 1 + max{1, d(x1,xN), d(x2,xN), ... , d(xN-1,xN)}. We see that x n?B(xN;r) for alln; so (xn) is bounded.

3.Show that the setXof all integers, with metricddefined byd(m,n) =|m-n|,

is a complete metric space.

Solution.

Note thatdis the metric induced by the Euclidean metric (the usual metric) onR. Since closed subspaces of complete spaces are complete, it suffices to2 show thatZis closed inR. The complement ofZinRis the union of all the open intervals (n,n+1), wherenruns through all ofZ, and this is open since every union of open sets is open. SoZis closed. Alternatively, let (an) be a Cauchy sequence inZ. Choose an integerNsuch thatd(xn,xm)<1 for alln≥N. Putx=xN. Then for alln≥Nwe have |xn-x|=d(xn,xN)<1. Butxn, x?Z, and since two distinct integers always differ by at least 1 it follows thatxn=x. This holds for alln > N. Soxn→xasn→ ∞(since for allε >0 we have 0 =d(xn,x)< εfor all n > N).

4.(i) Show that ifDis a metric on the setXandf:Y→Xis an injective

function then the formulad(a,b) =D(f(a),f(b)) defines a metricd onY, and use this to show thatd(m,n) =|m-1-n-1|defines a metric on the setZ+of all positive integers. (ii) Show that (Z+,d), wheredis as defined in Part (i), is not a complete metric space.

Solution.

(i) This is obvious, since we can regardfas identifyingYwithX. Neverthe- less, let us write out the details. Ifa, b, c?Y, thenf(a), f(b), f(c)?X.

SinceDis a metric onX, we have

and D(f(a),f(b) =D(f(b),f(a))≥0 with equality only iff(a) =f(b).

Thus for alla, b, c?Y,

which shows thatdsatisfies the triangle inequality. Similarly, for all a, b?Y d(a,b) =D(f(a),f(b) =D(f(b),f(a)) =d(a,b), and d(a,b) =d(f(a),f(b))≥0 with equality only iff(a) =f(b). Sincefis injective,f(a) =f(b) if and only ifa=b; so we deduce that d(a,b) =d(b,a)≥0 with equality only ifa=b, as required. The astute reader will have noticed that it was necessary only to assume thatfis injective, rather than bijective. The functionf:Z+→Rdefined byf(n) =n-1for alln?Z+is certainly injective, and if we takeDto be the usual metric onRand apply the principle we have been discussing, we obtain that d(m,n) =D(f(m),f(n)) =D(m-1,n-1) =|m-1-n-1| 3 defines a metric onZ, as claimed. (Or, observe thatn→n-1gives a bijection fromZ+to{n-1|n?Z+}, which has a metric induced from the usual metric onR.) (ii) The sequence (an)∞n=1defined byan=nis a Cauchy sequence with respect to the metric described in Part (i). To see this, letbn?Rbe defined bybn=f(an) =n-1for alln?Z+. Since (bn) is a convergent sequence inR(with limit 0), it is a Cauchy sequence. Furthermore, sinced(an,am) =D(f(an),f(am)) =D(bn,bm) for alln, m?Z+, the fact that (bn) is Cauchy implies that (an) is Cauchy also. Of course, a direct proof is trivial: givenε >0, if we defineN= 1/ε then it follows thatn-1, m-1?(0,ε), and so|n-1-m-1|< ε, for all n, m > N.

5.Letcbe the set of all sequencesx= (xk) of complex numbers that are

convergent in the usual sense, and letdbe the metric oncinduced from the space?∞. (That is,d(x,y) = supk?N|xk-yk|). Show that the metric space (c,d) is complete. [Hint: Show thatcis closed in?∞.]

Solution.

SinceCis complete, a sequence inCis convergent if and only if it is a Cauchy sequence. Soccan be described as the set of all Cauchy sequences inC. Recall that?∞is the set of all bounded sequences inC, with the sup metric. Every Cauchy sequence is bounded; so (c,d) is indeed a subspace of?∞. The space ∞is complete, by Example 2.6 on p. 41 of Choo"s notes. Since a closed subspace of a complete space is complete, it suffices to show thatcis a closed subset of?∞. So it suffices to show thatc?c. Letx?c. Then there exists a sequence (x(k))∞k=1of points ofcconverging in?∞to the pointx. Our task is to prove thatx?c. Since points of?∞are themselves sequences, let us writex(k) ifor thei-th term ofx(k)andxifor the i-th term ofx. That is, x (1)= (x(1)

1,x(1)

2,x(1)

3,...),

x (2)= (x(2)

1,x(2)

2,x(2)

3,...),

x (3)= (x(3)

1,x(3)

2,x(3)

3,...),

x= (x1, x2, x3, ...). We are given that eachx(k)is a Cauchy sequence, and the aim is to prove thatxis a Cauchy sequence. We are also given that (x(k)) converges in the ∞metric-that is, uniformly-tox. So our task can be restated as follows: prove that the uniform limit of a sequence of Cauchy sequences is Cauchy. This is somewhat analogous to the fact that the uniform limit of a sequence of continuous functions is continuous (cf. Q.4 of Tutorial 4.)4 Letε >0. ChooseK?Z+such thatd(x(k),x)< ε/3 for allk≥K. Choose N?Z+such that|x(K)m-x(K)n|< ε/3 for alln, m > N. Then for all n, m > Nwe have ε3+ε3+ε3=ε. This shows that (xi) is a Cauchy sequence, as required.

6.LetX= (0,1) with the Euclidean metricd. Give an example of a nested

sequence (An) of non-empty closed sets inXwith diam(An)→0 asn→ ∞, but n=1A n=∅. (Thediameter, diam(A), of a subsetAof a metric space, is the supremum of the set{d(x,y)|x, y?A}, if this set is bounded.)

Solution.

Note thatX= (0,1) is not complete, because it is not closed inR. For example, a sequence in (0,1) converging inRto the point 0 will be a Cauchy sequence in (0,1) with no limit in (0,1). PutAn= (0,1n]. This gives a nested sequence of subsets ofX. Each A n= [0,1]∩Xis closed inXas [0,1] is closedR. (Recall that ifYis a subspace of a topological spaceXthen the closed sets ofYare all sets of the formY∩C, whereCis a closed subset ofX). Alsod(An) =1n→0 as n→ ∞. However∞? n=1A n=∅.

7.LetX= (X,d) be a metric space and CS(X) the collection of all Cauchy

sequences inX. For (xn) and (yn) in CS(X), define (xn)≂(yn) if and only if limn→∞d(xn,yn) = 0. Show that≂is an equivalence relation on CS(X).

Solution.

If (xn) is any Cauchy sequence thend(xn,xn) = 0→0 asn→ ∞. So the relation is reflexive. It is symmetric, since if (xn) and (yn) are Cauchy sequences with (xn)≂(yn) thend(yn,xn) =d(xn,yn)→0 asn→ ∞. Finally, it is symmetric, since if (xn), (yn) and (zn) are Cauchy sequences with (xn)≂(yn) and (yn)≂(zn) then limn→∞d(xn,yn) = limn→∞d(yn,zn) = 0, so that by the triangle inequality asn→ ∞, giving limn→∞d(xn,zn) = 0 by the squeeze law.quotesdbs_dbs19.pdfusesText_25

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