Question

Have to get an idea of how i am doing on this problem. Whould be nice to get a good explaination for each part of the problem. d1 and d2 is the two different metrics, p ,Y.

Problem 2. Consider first the following definition: Definition. Let X be a set and let pand be two metrics on X. We say that p and are equivalent if the open balls in (X, p) and (x,y) are "nested". More precisely, p and are equivalent if for every reX and r>0 there exists,t> 0 such that both B. (X;8) CB, (2;r) and B, (;t) Belir) where Be(r;r) = {y EX | p(x,y) <r} and B, (r;r) = {y < X | 7r,y) <r}. (a) Prove that d, and d, are equivalent metrics on X = R2 (Draw a picture first!) (b) Prove that d2 (or dı, for that matter) is not equivalent to the discrete metric dc.) Jo ifr=y 11 if rhy for r, y ER2 For the rest of this problem we let X be some arbitrary set (not necessarily R?). (c) Prove that if p and 7 are equivalent metrics on X then (X,p) and (x,y) have the same open and closed sets. (d) Prove that if p and are equivalent metrics, then the following holds: If {{n}n is a sequence in X, then {{n}n converges in (X,p) if and only if it converges in (X,7). (e) Prove that if p and y are equivalent metrics then (X,p) and (x,y) have the same compact subsets. (f) Let p be any metric on X, and define p(x,y) 7(x, y) = (1) 1+p(x,y) (i) Prove that (X, 7) is bounded. (ii) Prove that yis a metric on X. (iii) Prove that p and 7 are equivalent. Remark. Equivalent metrics share many properties, such as convergence of sequences (as shown in (d)) and continuity of functions (as can be shown in a similar way). It is sometimes more convenient to work with bounded metrics, in which case the trick in problem (f) can be used - given an unbounded metric p, define the bounded metric in (1), prove whatever properties you wish using 7, and then use equivalence to deduce the same properties for p.

Answer 1

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IV En a b I lit de R² in (R? & B da (2) : hи с хео", uu we get can a bazie Ba (ar) open set of madini a B. (a, ro,). Sutt. x.e Bdi (n,r). Now from the figure we can get that a rectangular B disk containing a f which contained in Be (m, n) let us asscome that disk Barz (n.) & RE Bd₂ (2, 4) C B a (ne). Bdz (w, c.) e Bai(n,t) . Again, if yė R² in (12) d2), Then ja@, 's.t. ye Bofy, ). # By the figure, we come also get an open disko Baily, R₂) contained in Bolz (9, cz) ...34 (9, Iz) ¿ Bd (z q) Bely, a since a, y Scing arbitrary nence the open sets are in ( d) are in (R2, 2) and comensely . af d2 are equivalent Bez Y, C2) J Ô it dof eh are the metrices on X de R equivalent zuen.

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U becomes an Similarly we can open set which are opin set in (x, ) show that any in (x,M) also in Since closed sets are complements are sets, open the result is also there of closed sets. . Two metrices. P friare equivalent om X. It tue metrices, P & n are equivalent, then the collection of opensets in (x, e) is sanie as that in. (x, n) Tc. U is an open nod in (x,p) iff it is so of a in (x,n) This eim an ad in (xp) (a) the sequence nya is eventually in every open nobod of a in (XP) (2) the sequences eventually 'in every open mba of a in Xines linnax in (x, a) Conversely, dit A be an open in (x.pl suppose pan} is a sequence inxe in 1x N to a point a of a no eim anza in Gyr). Then by hypothesis limanaa in ( xp).. hane Now we the sequence eventually it ree in x of a converging neo liman we were که « sea

Since A is open. By the same logic life . follows that A is open in (X H) (1 . Similarly every open set of (x, n) is open in (xp) rence efn are equivalent - © .#(x, e) and (x,x) are equivalent metric spaces the oftea identity fanation i. (x,p) (x,x) is a homeomorphism ... If A being a compact set in . cap is (A) z A is also a compart set in (x,x) (since la continuous : map and continuous image of a compact - set is also compact) "B be pie compact set in Can then i1(B) zB is also compact sef in (x,p) (since alla continuous Lemap]. Hence the piroef Aguin its b

Ô. let p be any meterie onx, . arner N(n, y a plony no lep(m,g) O for boundness of a metric an, we have to show that any 3k for all nin EX. here lap (mig) > p (mig) rn sex. 2) P(4,4) 1:1 4P(1,9) : 2) (m, g) ckel, wa, yEx. in) is a bad metacie space. ( . To show s (m, g) is a metrics rex of) , o , since ? (ny) 70 KM, SEX die (tinee pis a meetuies 6 M6, ) 20 iff PCM 9' 20 off sey kn, yax, © nlimin. Plni; ) , plyaji leelor, y) lep (3,2). . z ⓇM(934) y ex . hчи роли :.0 is symmetric meathe symmetric) @ Now, P (1, 2) <P (0,4) eP(4, 2) implies Pln, 2) p (ny) 1.P. (2) (0,7) Leplande e(Y,Z) Plain P(y,t) HP (ny lep(y,z) 72(90,2)=

. = M(33,4) e 1 (19, 2) xz) <0 (ny) eh(3,2), wnyz.EX. .: Jo is transitive... irri- is a nutric. Cus) P f n are equivalent let sant ta in 6x, p). Then, corresponding to any bio, men s.fi anym, an E Be (n,r) 21 or (an; 2)? 25% (an, 9) (mu, n) 1-Р (*), ) . Spania) cro. ..gan} = 4 in (xp) . conversely, let yang xin (x,n). thin corresponeing to any poo, I meno Jef: onlanja) ch , tnom thom, plana) Teplan, a) er ie. PEXn, a) ch fren] ix in (x,p). op fat are equivalent matrices 7 ..lero