Metric Space Properties

Transcription

1 Metric Space Properties Math 40 Fial Project Preseted by: Michael Brow, Alex Cordova, ad Alyssa Sachez We have already poited out ad will recogize throughout this book the importace of compact sets. All those cocered with geeral aalysis have see that it is impossible to do without them. Frechet, quoted i Aalysis by Its History by E. Hairer ad G. Waer.

2 Defiitio.: Let (X, d) be a metric space ad let Y be a subset of X. We say that Y is bouded iff there exists a ball B(x, r) i X such that Y B(x, r). Defiitio.: A metric space (X, d) is totally bouded if for every ε > 0, there exists a positive iteger ad a fiite umber of balls B(x (), ε),, B(x (), ε) which covers X, i.e. X = B(x (i), ε) i=. Propositio.: Every totally bouded metric space is bouded. Proof. Let (X, d) be a totally bouded metric space. Fixig ε > 0, there exists a fiite umber of balls B(x (), ε),, B(x (), ε) such that X = i= B(x (i), ε). We wat to show X is bouded, i.e. there M > 0 such that X B(x 0, M). It is sufficiet to show (x, y) M x, y X. Let x, y X = i= B(x (i), ε) B(x (m), ε), B(x (k), ε) such that x B(x (m), ε), y B(x (k), ε) for k, m. The, by use of triagle iequality, d(x, y) d(x, x (m) ) + d(x (m), x (k) ) + d(x (k), y) ε + d(x (m), x (k) ). We eed to cotrol d(x (m), x (k) ). Sice X is covered by a fiite amout of balls, we may take the maximum of the distace betwee the ceters of the balls to be the boud. We have the iequality, The, d(x (m), x (k) ) max i,j {d(x(i), x (j) )}. d(x, y) ε + d(x (m), x (k) ) ε + max i,j {d(x(i), x (j) )}. Defie M = ε + max i,j {d(x(i), x (j) )}. As defied, M > 0 satisfies d(x, y) M x, y X. Therefore, (X, d) is a bouded metric space. We ow kow that every totally bouded metric space is bouded. We caot say that every bouded space is totally bouded. Cosider the metric space (R, d disc ). Choose ay x, x R, ad choose r > ad d disc (x, x ), ad so the space is bouded by B(0,). Now give ε > 0, let ε =. The B (x 0, ) = {x R d disc(x 0, x) < } meaig that every ball would oly cotai its ceter. Thus it is ot possible to choose N fiite umber of balls B(x, ε) such that R = N B(x, ε), with N N, ad so the space is ot totally bouded. bouded totally bouded. Defiitio.3: Let (x () ) =m be a sequece of poits i a metric space (X, d). We say that this sequece is a Cauchy sequece iff for every ε > 0, there exists a N m such that d(x (i), x (k) ) < ε for all j, k N.

3 Lemma.: Let (x () ) = also a Cauchy sequece. be a sequece i (X, d) which coverges to some limit x0. The (x () ) = is Proof. Let (x () ) = be a sequece i (X, d) which coverges to a limit x0. Let ε > 0, so ε defiitio of covergece, N > 0 such that N, d(x (), x 0 ) < ε. > 0. By Let N = N > 0. The, m N, ad by use triagle iequality, d(x (), x (m) ) d(x (), x 0 ) + d(x 0, x (m) ) ε + ε = ε. Therefore, we foud a N = N > 0 satisfyig the defiitio of Cauchy. Lemma.: Give a sequece (x (m) ) m= i X, if (X, d) is a totally bouded metric space, the there exists at least oe ball that has a ifiite subsequece of the sequece. Proof. (X, d) is totally bouded ϵ > 0 a fiite umber of balls such that X = i= B(x (i), ϵ). Assume that each ball cotais a fiite subsequece of the sequece (x (m) ). The there exists a sequece such that (x (m k ) ) k= B(x (i), ϵ), i. The (x (m k ) ) i= B(x (i), ϵ) = X. But (x (m k ) ) (x (m) ) X. Therefore, there must be at least oe ball which cotais a ifiite amout of the sequece (x (m) ). Defiitio.4: A metric space (X, d) is said to be complete iff every Cauchy sequece i (X, d) is i fact coverget i (X, d). Propositio.: Let (X, d) be a metric space ad, ad (Y, d Y Y ) be a subspace of (X, d). If (Y, d Y Y ) is complete, the Y must also be closed i X. Proof. Give a complete metric space, every Cauchy sequece is coverget withi that metric space. This implies that the metric space must cotai all its limit poits ad is thus closed. We ow have that every complete metric space is a closed metric space. But ot every closed metric space is complete. Let us cosider the ratioal umbers Q. Q is closed but ot complete, sice we may fid a Cauchy sequece which coverges to a irratioal umber. Cosider the sequece,.4,.4,.44,.44, which coverges to Q. Closed Complete. 3

4 Defiitio.5: A metric space is said to be compact iff every sequece i (X, d) has at least oe coverget subsequece. Theorem.: If (X, d) is a compact metric space, the (X, d) is complete ad bouded. Proof. Compact bouded. Suppose that (X, d) is ot a bouded metric space. The r > 0 such that X B(x 0, r) with x 0 X. For each, with N, x X such that x B(x 0, ) or that d(x, x 0 ). Sice (X, d) is a compact metric space, the (x () ) = must have a coverget subsequece, say (x ( k ) ). Suppose that (x ( k ) ) coverges to some x X. By defiitio of covergece, give ε = > 0, there N 0 such that k N, d(x ( k ), x ) ε =, but also d(x, x 0 ) N 0, by Archimedea Priciple, for some N 0 N. The, k k d(x ( k ), x 0 ) d(x ( k ), x ) + d(x, x 0 ) + N 0. By choosig k max{n, N 0 + } N 0 +, the we coclude that, N 0 + N 0 +. This caot be true, therefore (X, d) must be bouded. Compact Complete Assume (X, d) is a compact metric space which is ot complete. The there exists a Cauchy sequece (x () ) = i (X, d) which does ot coverge i (X, d). Sice (X, d) is compact, (x () ) must have a coverget subsequece (x ( k ) ) k= which coverges to some x0 X. Let ε > 0, the ε > 0. By defiitio of covergece, N > 0 such that for k N, d(x ( k ), x 0 ) ϵ. By defiitio of Cauchy sequece, N > 0 such that, k N, with k k N, d(x (), x ( k ) ) ε. Let N = max{n, N }. The N, uder the coditio k N ad by usig triagle iequality, d(x (), x 0 ) d(x (), x ( k ) ) + d(x ( k ), x 0 ) ε + ε = ε. (x () ) = coverges to x0 with respect to the metric d. This cotradicts the assumptio that the sequece (x () ) = does ot coverge. Therefore, if (X, d) is a compact metric space, the (X, d) is a complete metric space. 4

5 We leared that every compact metric space is a complete ad bouded metric space. Also, every complete metric space is a closed metric space. Therefore, every compact metric space is bouded ad closed. It is ot true that every bouded ad closed space is compact. By replacig closed with complete, ad more importatly, bouded with totally bouded, the we are able to have that every complete ad totally bouded metric space is compact. This is the motivatio to our fial proof, to uderstad that if we have a totally bouded ad complete metric space, the we have a compact space. These stroger otios allow us to have a if ad oly if statemet which we state formally below. Theorem.: Let (X, d) be a metric space. The (X, d) is compact iff (X, d) it totally bouded ad complete. Proof. Compact Complete See Theorem.. Compact Totally Bouded Assume (X, d) is compact but ot totally bouded. The ε > 0 such that N ad x (i) X, i, X B(x (i), ε). Let x () X. The X B(x (), ε) x () X\B(x (), ε) so i= d(x (), x () ) > ε. Still, X B(x (), ε) B(x (), ε) x (3) X\B(x (), ε) B(x (), ε) so d(x (), x (3) ) > ε ad d(x (), x (3) ) > ε. Proceedig by iductio, we have: P(): {x (),, x () } X such that d(x (i), x (j) ) > ε for i j, i, j. Base case = give above. i= P( + ): X B(x (i), ε) x (+) X\ B(x (i), ε) x (+) B(x (i), ε) d(x (+), x (i) ) > ε, i. i= We have created a sequece (x () ) = i X such that d(x (i), x (j) ) > ε for all i, j N, i j. Sice X is compact, there exists a coverget subsequece (x ( k ) ) k= of (x () ) =. By Lemma., every coverget sequece is Cauchy. Give ε > 0, N > 0 such that m, k N, ε < d(x ( k ), x ( m ) ) ε. But ε ε. Therefore, if (X, d) is a compact metric space, the (X, d) is totally bouded. Totally bouded ad complete compact Suppose (X, d) is a totally bouded metric space ε > 0, a fiite umber of balls such that X = i= B(x (i), ε). Let (x (j) ) j= be a sequece i X ad let ε =. The X = i= B(x (i), ) for some N. By lemma., there exists at least oe ball which cotais a ifiite subsequece (x (j;) ) of (x (j) ). Say (x (j;) ) j= is cotaied i B(x (m ), ) with m. Now take ε =. The X = 5

6 B (x (i), ) for some N. Take the balls such that B(x (m ), i= ) B (x (i), ) for i. The there exists at least oe ball which cotais a ifiite subsequece (x (j;) ) of (x (j;) ). Say (x (j;) ) is cotaied i B (x (m ), ) with m. By takig (x (;) ) ad (x (;) ), the d(x (;), x (;) ) < = diam (B(x (m ), )). Proceedig by iductio, we have: P(k): Let ε =. The X = k k B (x(i), ). Take the balls such that B (x (m k ), i= ) B k k (x(i), ) k for i. The there exists at least oe ball cotaiig a ifiite subsequece (x (j;k) ) of k (x (j;k ) ) with the coditio that d(x (k ;k ), x (k;k) ) < = diam (B k (x(m k ), )). k Base case k = give above. P(k + ): Let ε =. The X = k+ k+ i= B (x(i), ) k+. If we take the balls where B (x (m k ), ) B k (x(i), ) for i k+ k+. The there exists at least oe ball cotaiig a ifiite subsequece (x (;k+) ) of (x (;k) ). Let (x (;k+) ) be cotaied i the ball B (x (m k+ ), ). By takig (x (k+;k+) ) ad (x (k;k) ), the d(x (k;k), x (k+;k+) ) < k = diam (B (x(m k ), k )). k+ Now we wat look at the diagoal sequece {x (;), x (;), x (3;3), }. WLOG, assume j > m. Sice x (j;j) belogs to the m th subsequece, x (m;m) ad x (j;j) are cotaied i the ball of radius m. Therefore, we may choose a M large eough such that < ε by Archimedea Priciples. Therefore, m j, k M, d(x (k,k), x (j;j) ) < ε. We ow have a Cauchy subsequece (x (j;j) ) j= of (x (j) ) j= i (X, d). Sice (X, d) is a complete metric space, every Cauchy sequece i (X, d) is coverget i (X, d) (x (j;j) ) coverges i (X, d). By defiitio.4, (X, d) is a compact metric space. (X, d) is a compact metric space (X, d) is complete ad totally bouded.. Remark: We used the diameter of the circle sice it possible for two poits to be iside a circle ad the distace betwee them be greater tha the radius.. Actually, it is ot ecessary to look at the diagoal sequece. Sice every sequece is a subsequece of the previous sequece, we may choose ay elemet i the give subsequece to satisfy the distace beig less tha ay ε > 0. 6