Complete Metric Spaces

1 Complete Metric SpacesDefinition (X, d) be a Metric space. A sequence (xn) inXis called aCauchy sequenceif for any >0, there is ann Nsuch thatd(xm, xn)< for anym n ,n n .Theorem convergent sequence in a Metric space is a Cauchy that (xn) is a sequence which converges tox. Let >0 begiven. Then there is anN Nsuch thatd(xn, x)< 2for alln N. Letm,n Nbe such thatm N,n N. Thend(xm, xn) d(xm, x) +d(xn, x)< 2+ 2= .Hence (xn) is a Cauchy converse of this theorem is not true. For example, letX= (0,1].Then (1n) is a Cauchy sequence which is not convergent Metric space (X, d) is said to becompleteif every Cauchysequence inXconverges (to a point inX).Theorem closed subset of a Complete Metric space is a Complete a closed subspace of a Complete Metric spaceX. Let (xn) be aCauchy sequence inS. Then (xn) is a Cauchy sequence inXand hence it mustconverge to a pointxinX. But thenx S=S.)

2 ThusSis Complete subspace of a Metric space is a closed a Complete subspace of a Metric spaceX. Letx S. Thenthere is a sequence (xn) inSwhich converges tox(inX). Hence (xn) is aCauchy sequence inS. SinceSis Complete , (xn) must converge to some point,say,yinS. By the uniqueness of limit, we must havex=y S. HenceS=S, a nonempty subset of a Metric space (X, d). ThediameterofAis defined to bediam(A) = sup{d(x, y)|x, y A}.We say thatAisboundedif diam(A) is (X, d) be a Complete Metric space. If (Fn) is a sequenceof nonempty closed subsets ofXsuch thatFn+1 Fnfor alln Nand(diam(Fn)) converges to 0, then n=1 Fnis a that (X, d) is a Complete Metric space. Let (Fn) be a sequenceof nonempty closed subsets ofXsuch thatFn+1 Fnfor everyn Nand(diam(Fn)) converges to 0. First, we show that n=1 Fnis nonempty. For eachn N, choosexn Fn. To show that (xn) is a Cauchy sequence, let > (diam(Fn)) converges to 0, there is anN Nsuch that diam(FN)<.

3 Letm,n Nwithn N,m N. Thenxm Fm FN,xn Fn (xm, xn) diam(FN)< .Thus (xn) is a Cauchy sequence inX. SinceXis Complete , it follows that(xn) is a convergent sequence inX. Letx= limn now show thatx n=1Fn. Letn N. Note thatxm Fm Fnforanym n. Thus the sequence (xn, xn+1, ..) is a sequence inFnand is asubsequence of (xn), so it converges tox. This implies thatx Fn= n=1Fn. Next, we show that n=1 Fnis a singleton. To see this, letx,y n=1 Fnand >0. Then there is anN Nsuch that diam(FN)< .Sincex,y FN, it follows thatd(x, y) diam(FN)< . This is true for any >0. Henced(x, y) = 0, which meansx= a function from a Metric space (X, d) into a metricspace (Y, ). We say thatfisuniformly continuousif given any >0, thereexists a >0 such that for anyx,y X,d(x, y)< implies (f(x), f(y))< .Theorem uniformly continuous function maps Cauchy sequences intoCauchy : (X, d) (Y, ) be a uniformly continuous function.

4 Let (xn) bea Cauchy sequence inX. To see that (f(xn)) is a Cauchy sequence, let > there is a >0 such that x, y X, d(x, y)< (f(x), f(y))< .Thus there exists anN Nsuch thatd(xm, xn)< for anym,n N. Itfollows that (f(xm), f(xn))< for anym,n N. Hence (f(xn)) is a Cauchysequence not uniformly continuous, then the theorem may not be example,f(x) =1xis continuous on (0, ) and (xn) = (1n) is a Cauchysequence in (0, ) but (f(xn)) = (n) is not a Cauchy a function from a Metric space (X, d) into a metricspace (Y, ). We say thatfis anisometryifd(a, b) = (f(a), f(b)) for anya,b : (X, d) (Y, ) be an isometry. Then it is injective anduniformly continuous. Moreover, its inversef 1: (f[X], ) (X, d) is also : (X, d) (Y, ) be an isometry. Let >0. Choose = > ,b Xbe such thatd(a, b)< . Then (f(a), f(b)) =d(a, b)< = .Hencefis uniformly continuous. Next, leta,b Xbe such thatf(a) =f(b).

5 Thusd(a, b) = (f(a), f(b)) = 0. This shows thata=b. Hencefis see thatf 1is an isometry, lety,z f[X] and leta,b Xbe such thatf(a) =yandf(b) =z. Thus (y, z) = (f(a), f(b)) =d(a, b) =d(f 1(y), f 1(z)).Hencef 1is an a dense subset of a Metric space (X, d). Letfbe auniformly continuous function (isometry) fromAinto a Complete Metric space(Y, ). Then there is a unique uniformly continuous function (isometry)gfromXintoYwhich will give a proof only for a uniformly continuous function. Theproof for an isometry is similar and somewhat (X, d) be a Metric space and (Y, ) a Complete Metric space. LetAbea dense subset ofXand letfbe a uniformly continuous 1: define a functiong:X eachx X=A, there is a sequence (xn) inAwhich converges (xn) is a Cauchy sequence inX. Thus (f(xn)) is a Cauchy sequence inY. SinceYis Complete , (f(xn)) is a convergent sequence. Defineg(x) = limn f(xn)for anyx X, where (xn) is a sequence inAwhich converges 2:gis well-defined, , independent of the choice of (xn).

6 Let (xn) and (yn) be any sequence inAwhich converges tox A=X. Then the sequence (x1, y1, x2, y2, .. , xn, yn, ..) must converge , the sequence (f(x1), f(y1), f(x2), f(y2), .. , f(xn), f(yn), ..) converges3to some pointz Y. Since (f(x1), f(x2), ..) and (f(y1), f(y2), ..) are itssubsequences, they must also converge toz. Hencez=g(x) does not dependon the choice of the 3:gis an extension Aand letan=afor eachn N. Then (an) is a sequence inAwhich converges toa. Henceg(a) = limn f(an) =f(a). This shows thatgis an extension 4:gis uniformly continuous >0. Then there is a >0 such that a, b A, d(a, b)< (f(a), f(b))< ,y Xbe such thatd(x, y)< . Then there are sequence (xn) and (yn)inAsuch thatxn xandyn y. Hencef(xn) g(x) andf(yn) g(y).ChooseN Nsuch thatd(xN, x)< d(x, y)2andd(yN, y)< d(x, y)2( )and (f(xN), g(x))< 3and (f(yN), g(y))< 3.( )Thend(xN, yN) d(xN, x) +d(x, y) +d(y, yN)< ,which implies that (f(xN), f(yN))< , by the uniform continuity , (g(x), g(y)) (g(x), f(xN)) + (f(xN), f(yN)) + (f(yN), g(y))< 3+ 3+ 3=.

7 This shows thatgis uniformly continuous 5:gis (uniformly) continuous functions onXwhich extendsfona dense subsetA. To see thatg=h, letx X. Then there is a sequence (xn)inAwhich converges tox. By continuity ofgandh,g(x) = limn g(xn) = limn h(xn) =h(x).Henceg= a Metric space (X, d) is a pair consisting ofa Complete Metric space (X , d ) and an isometry ofXintoX such that [X] is dense inX .4 Theorem Metric space has a (X, d) be a Metric space. Denote byC[X] the collection of allCauchy sequences inX. Define a relation onC[X] by(xn) (yn) limn d(xn, yn) = is easy to see that this is an equivalence relation onC[X]. LetX be theset of all equivalence classes for :X ={[(xn)] : (xn) C[X]}.Defined :X X [0, ) byd ([(xn)],[(yn)]) = limn d(xn, yn),where [(xn)], [(yn)] X . To show thatd is well-defined, let (x n) and (y n) betwo Cauchy sequences inXsuch that (xn) (x n) and (yn) (y n).]

8 Thenlimn d(xn, x n) = limn d(yn, y n) = the triangle inequality,d(xn, yn) d(xn, x n) +d(x n, y n) +d(y n, yn) andd(x n, y n) d(x n, xn) +d(xn, yn) +d(yn, y n).Hence,|d(xn, yn) d(x n, y n)| d(xn, x n) +d(yn, y n) both (d(xn, yn)) and (d(x n, y n)) are convergent, this shows thatlimn d(xn, yn) = limn d(x n, y n).Thusd is , we show thatd is a Metric onX . Let [(xn)], [(yn)], [(zn)] X .Thend ([(xn)],[(yn)]) = 0 limn d(xn, yn) = 0 (xn) (yn) [(xn)] = [(yn)].Also,d ([(xn)],[(yn)]) = limn d(xn, yn) = limn d(yn, xn) =d ([(yn)],[(xn)]).Sinced(xn, zn) d(xn, yn) +d(yn, zn),limn d(xn, zn) limn d(xn, yn) + limn d(yn, zn).5 Thusd ([(xn)],[(zn)]) d ([(xn)],[(yn)]) +d ([(yn)],[(zn)]).Henced is a Metric onX .For eachx X, let x= [(x, x, ..)] X , the equivalence classes of theconstant sequence (x, x, ..). Define :X X by (x) = x. Then for anyx,y X,d ( (x), (y)) =d ( x, y) = limn d(x, y) =d(x, y).

9 Hence is an isometry fromXintoX . To show that [X] is dense inX ,letx = [(xn)] X and let >0. Since (xn) is a Cauchy sequence, thereexists anN Nsuch that for anym,n N,d(xm, xn)< 2. Letz= z [X] andd (x , z) = limn d(xn, z) = limn d(xn, xN) 2< .Thus z Bd (x , ) [X]. Hence, [X] is dense inX .Finally we show that (X , d ) is Complete . To establish this, we apply thefollowing lemma of which proof is left as an exercise:Lemma (X, d) be a Metric space andAa dense subset such thatevery Cauchy sequence inAconverges inX. Prove thatXis , it suffices to show that every Cauchy sequence in the dense subspace [X] converges inX . Let ( zk) be a Cauchy sequence in [X], where each zkis represented by the Cauchy sequence (zk, zk, ..). Since is an isometry,d(zn, zm) =d ( zn, zm) for eachm, , (z1, z2, z3, ..) is a Cauchy sequence inX. Letz = [(z1, z2, z3, ..)] X . To show that ( zk) converges toz , let >0.

10 Then there is anN Nsuch thatd(zk, zn)< 2for anyk,n N. Hence, for eachk N,d ( zk, z ) = limn d(zk, zn) 2< .This shows that ( zk) converges to a pointz inX and thatX is proving that a completion of a Metric space is unique up to isometry,we will give an alternative definition of a completion in terms of a universalmapping property. We state this fact in the following definition:Theorem 16 (Universal Mapping Property).Let (X, d) be a metricspace, (X , d ) a Complete Metric space and : (X, d) (X , d ) an isom-etry. Then [X] is dense inX if and only if it satisfies the following universalmapping property:6 Given any Complete Metric space (Y, ) and an isomtryf:X Y, there existsa unique isometryF:X Ysuch thatF = //f ?????????X F that [X] is dense inX . We will show that it satisfiesthe universal mapping property. Let (Y, ) be a Complete Metric space andf:X Yan isometry fromXintoY.