Cauchy SequencesDate: 09/11/97 at 02:49:34 From: Suzanne Mooney Subject: Cauchy Sequences Hi, I'm trying to prove that every Cauchy sequence has a sequential limit point. I don't have a real proof, but an idea, but don't know if it's right, and if it is, how to formulate it. My idea is that, for some N a positive integer, then for every m,n>N, a Cauchy sequence has |psubn - psubm| < epsilon for some epsilon >0. Then - epsilon < psubn - psubm < epsilon, so psubm - epsilon < psubn < psubm + epsilon which means that psubm - epsilon and psubm + epsilon are, respectively, lower and upper bounds on {psubn|n>N}. So (psubm - epsilon, psubm + epsilon) would be a segment containing all psubn for every n>N. Then, no matter how small this segment becomes, i.e. for bigger and bigger values of N, it still will contain its midpoint, psubm and all psubn for n>N, so wouldn't that make psubm a sequential limit point? And then I'm thinking psubm is just an arbitrary member of the set {psubn|n>N}, so would that mean that every member of that set would then be a sequential limit point of the Cauchy sequence? Does that make sense? It seems to me so now, but it's very late and I've been messing with this thing too long, so forgive me if my reasoning (or lack thereof) is somewhat fuzzy. Ever appreciative of your help, Suzanne Date: 09/11/97 at 09:07:42 From: Doctor Jerry Subject: Re: Cauchy Sequences COMMENT: "Wouldn't that make p_m a sequential limit point?" For which m? I don't know the tools you have available. For example, do you know or can you assume that every bounded infinite set has a limit point? Or do you know or can you assume that each subset of R that is bounded above has a least upper bound? It's possible to show that the set {p_1,p_2,...}, where {p_m} is a Cauchy sequence, is bounded. For example, take epsilon = 1; then all p_m are within 1 of p_N. The entries p_1,...,p_{N-1} can't be too far away since they are only finite in number. So the entries of a Cauchy sequence form a bounded set. So, this set has a limit point, call it x. You can show that x is the sequential limit of the Cauchy sequence. Since it is a limit point, then, given any eps>0, there is at least one entry nearby. But, being a Cauchy sequence, the entries tend to hang around each other for large n. Etc. Rough, but the gist is here, I think. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/11/97 at 17:50:07 From: Suzanne Mooney Subject: Re: Cauchy Sequences Thanks again, as always. Yes, the gist is there so that even I see it (yay!). And I think may be I can even 'unrough' enough. Thanks, thanks, thanks. Suzanne :-) Date: 09/21/97 at 16:36:32 From: Suzanne Mooney Subject: Re: Cauchy Sequences Okay. Let's see if I came up with something that works. We show that x is the limit point of the sequence, along the lines you said. Then we pick an arbitrary segment (a,b) containing x, so it has some p_m in it. Then we choose a subsegment of (a,b) that contains x but not p_m, so this subsegment must also have some p_n in it. Now there exists some N_1 s.t. m,n > N_1 ==> |p_m - p_n| < epsilon_1. Now this epsilon_1 must be less than |b-a|, since both are contained in (a,b). Then (a,b) contains all p_m s.t. m>N_1, and since (a,b) is an arbitrarily selected segment containing x, x is the sequential limit point of the Cauchy sequence. Is this an okay proof or does it have some gaping hole in it that I haven't seen? Be kind, it took me a long time to come up with this! Thanks as always, Suzanne :-) P.S. If it does have holes, just point them out to me, but please don't tell me how to fix them... I'm trying to get to where I can do this on my own! There's just not enough thanks to give you for all the help you've given me! Maybe thanks raised to the millionth power? (I have to save some for the next time you help me!) Date: 09/22/97 at 09:34:48 From: Doctor Jerry Subject: Re: Cauchy Sequences Hi Suzanne, Since I'm retired, I can vacation anytime. I was in northern Minnesota last week listening to the loons and watching the leaves turn and so didn't see your reply until today. I'll respond on the chance that you are still interested in these questions. I made a mistake in my last reply. It's possible that the set of entries of the Cauchy sequence is finite (1,2,3,3,3,3,3,3,...). If this is so, let x be a p_j that is repeated for infinitely many values of j. If the set of entries is infinite, then let x be the limit point. Okay, now for comments on your argument. I'll start my comments in capital letters, not telling all. >We show that x is the limit point of the sequence, along the lines >you said. Then we pick an arbitrary segment (a,b) containing x so it >has some p_m in it. Then we choose a subsegment of (a,b) that >contains x but not p_m. IT COULD HAPPEN THAT X IS THE SAME AS THE p_m. I DON'T THINK THAT THIS IS A BIG PROBLEM. >So this subsegment must also have some p_n in it. Now there exists >some N_1 s.t. m,n > N_1 ==> |p_m - p_n| < epsilon_1. Now this >epsilon_1 must be less than |b-a|, since both are contained in (a,b). IT'S FUZZY HERE. GIVEN epsilon_1, there is N_1 and so on; however, you haven't said how big epsilon_1 is at this point and so can't say that epsilon_1 <|b-a|. Also, m was prescribed and we don't know how it relates to N_1. DON'T READ WHAT I'm going to say next if you want to work entirely on your own (this is often a good idea). I'll give a small hint. Choose epsilon so that the interval (x-epsilon,x+epsilon) is inside (a,b). Next, choose N so that m,n>N implies |p_m-p_m|<epsilon/2. Now, convince yourself that you can find a p_r within (x-epsilon/2,x+ epsilon/2) and r>N. >Be kind, it took me a long time to come up with this! MANY PEOPLE TAKE A LONG TIME TO GET THEIR MATHEMATICS STRAIGHT. WHAT COUNTS IN THE LONG RUN IS WHETHER THE PROOF IS CORRECT OR NOT. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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