Real Variables II Homework Set #5
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1 Real Variables II Homework Set #5 Name: Due Friday /0 by 4pm (at GOS-4) Istructios: () Attach this page to the frot of your homework assigmet you tur i (or write each problem before your solutio). () Please write (or prit) o oe side of the paper oly. () Keep it sequetial. Problems should ot be out of order. (4) Leave eough room i the margis ad after each problem so that the grader ca provide helpful commets whe ecessary. Problem 5. Costruct a Cator-like set by removig the middle third from the uit iterval, removig the middle ith (NOT the middle third) from each of the remaiig itervals, removig the middle twety-seveth from each of the remaiig itervals after that, ad so o. What is the legth of the set that you costruct i this fashio? Problem 5. Let f : R R be defied by f (x) = x. Prove that f is cotiuous. Problem 5. Give a example of a cotiuous fuctio f ad a coected set E such that f (E) is ot coected. Is there a coditio you ca add that will force f (E) to be coected? Problem 5 4. Give a example of a cotiuous fuctio f ad a compact set K such that f (K) is ot compact. Is there a coditio you ca add that will force f (E) to be compact? Defiitio: For two oempty sets A, B R, defie the distace betwee A ad B, deoted dist(a,b), as dist(a,b) = if{ x y : xa, yb}. [Istructor s Note: The distace is always defied sice the set { x y : xa, yb} is bouded below by 0.] Problem 5 5. Fid two oempty disoit closed sets A, B R such that dist(a,b) = 0. Problem 5 6. Let A, B R be disoit with A compact ad B closed. Prove that dist(a,b) > 0.
2 Solutios to Homework #5 Problem 5. This is a special type of Cator set called a fat Cator set (so I guess the origial should be called a skiy Cator set). Solutio #: Follow the outlie give i Problem -5. Let k be the legth of the iterval remaiig at step. (So k 0 = ad k = /.) Let l be the legth of the iterval removed at step. (So l = /.) Sice the legth of the iterval removed from step is / times the legth of the iterval remaiig i the previous step, the l k, where. The legth of the iterval remaiig at step is the legth of the previous iterval remaiig mius the legth of the iterval removed at step ad that result divided by. That is, k ( k l) ( k k ) k ( ) where. Usig the iitial coditio ad the recurrece relatio, we ca look for a iterative solutio of k by writig k0 k k0( ) ( ) k k ( ) ( )( ) ( )( ) k k ( ) ( )( )( ) k ( )( )( ) ( ) ( ) Note, however, that this formula for k is ot valid whe = 0. Rewritig ow i terms of l, we have that l k ( ). This formula is ot valid for the case whe =, so we eed to specify what happes there. Overall,, if ; l ( ), if. Sice there are itervals removed at step, the the total legth of itervals removed is l l l ( ) ( ). The legths of itervals remaiig is ( ) l. This last expressio caot be evaluated with aythig we kow about summatios or products, but we ca use a computer algebra system to approximate the aswer to be Solutio #: Let C [0,] be this Cator-like set. Let l be the legth of C remaiig at step. (So l 0 =, l = / ad l = 6 / 7.) [Istructor s Note: l is defied here differetly tha i the first solutio.] The legth of C remaiig at step is the legth of the previous iterval remaiig mius the legth of the iterval removed at step, which is / the legth of the previous iterval. That is, l l l ( ) l where. Usig the iitial coditio ad the recurrece relatio, we ca look for a iterative solutio of l by writig
3 Thus the legth of C is l l l ( ) ( ) 0 0 l l( ) ( )( ) l l ( ) ( )( )( ) l l ( ) ( )( )( ) ( ) ( ). l lim lim ( ) ( ) This last expressio caot be evaluated with aythig we kow about summatios or products, but we ca use a computer algebra system to approximate the aswer to be It was oted i class that this expressio has bee studied i combiatorics literature ad is oe of a special type of umbers called the q-pochhammer symbol (also called a q-shifted factorial). The particular oe we have is deoted as (/) (/; /). Problem 5. Proof #: Show f is cotiuous at pr. Choose > 0. Let δ p p. This term is always positive. For ay xr with x p <, we have x p = x p x + p < x + p = x p + p ( x p + p ) < ( + p ) = = p p p p [triagle iequality] p p p p p [substitute i our choice for ] = (p + ) p = Thus f is cotiuous at pr. This implies f is cotiuous at all poits i its domai. Proof #: Show f is cotiuous at pr. Choose > 0. Let δ mi,. p This term is always positive ad it allows us to say that δ ad. p For ay xr with x p <, we have x p = x p x + p < x + p
4 = x p + p ( x p + p ) < ( + p ) ( + p ) p p = [triagle iequality] [usig the secod iequality for above] [usig the first iequality for above] Thus f is cotiuous at pr. This implies f is cotiuous at all poits i its domai. Proof #: Show f is cotiuous at pr. Choose > 0. Let δ mi,. 4 p This term is always positive ad it allows us to say that δ ad δ. 4 p For ay xr with x p <, we have x p = x p x + p < x + p = x p + p ( x p + p ) < ( + p ) = + p p δ = p δ p 4 p < = [sice [triagle iequality] [usig the secod iequality for above] [usig the first iequality for above] p 4 p ] Thus f is cotiuous at pr. This implies f is cotiuous at all poits i its domai. Problem 5. Let f : R R be defied by f (x) = x. The set E = [,4] is coected, but f (E) = [,] [,] is ot coected. Yes, the coditio that f (E) is coected is that f must also be iective. Problem 5 4. Let f : R + R be defied by f (x) = /x. The set E = [0,] is compact, but f (E) = [,] is ot compact. No, there is ot a coditio. Problem 5 5. Take A = Z + ad B = { + / : Z + }. Both sets are closed because the each cosist etirely of isolated poits, either have ay accumulatio poits. The sequeces {a } = {} is i A ad {b } = { + / } is i B. These sequeces are arbitrary close to each other for large values of.
5 Problem 5 6. Proof: Suppose ot. That is, assume dist(a,b) = 0. [We show that A ad B are ot disoit.] I order for the distace to be 0, there must exist sequeces {x } i A ad {y } i B such that the sequece { x y } coverges to 0. Sice A is compact, the it is sequetially compact (by the Heie-Borel theorem) ad so there exists a subsequece { x k } of {x } that coverges i A. Let s say it coverges to. This value must be i A sice A is closed. Claim: { y k } coverges to. Choose > 0. Set = /. Sice { x y } coverges to 0, the there exists N Z + such that > N implies x y 0 = x y <. Sice { x k } coverges to, the there exists N Z + such that k > N implies x k α. Set N = max{n, N }. By defiitio of a subsequece, we always have k. For > N (ad so this implies that k > N), we have y α y x x α k k k k y x x α [triagle iequality] k k k < + = / + / =. Thus the subsequece { y k } also coverges to. Thus alpha is a accumulatio poit of {y } ad, hece, a accumulatio poit of B. Sice B is closed, the, by defiitio, it cotais all of its accumulatio poits. So B. Thus A B. This is a cotradictio to the coditio that A ad B are disoit. Therefore, the assumptio that dist(a,b) = 0 is false, so dist(a,b) > 0.
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