About Convergence of complex sequence
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From: David C. Ullrich on 18 Apr 2010 09:48 On Sat, 17 Apr 2010 02:28:56 -0700 (PDT), Eric <eric955308(a)yahoo.com.tw> wrote: >On 4?17?, ??4?41?, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: >> On Fri, 16 Apr 2010 10:53:04 -0700 (PDT), Eric >> >> >> >> >> >> <eric955...(a)yahoo.com.tw> wrote: >> >On 4��17��, �W��1��37��, Robert Israel >> ><isr...(a)math.MyUniversitysInitials.ca> wrote: >> >> Eric <eric955...(a)yahoo.com.tw> writes: >> >> > On 4=E6=9C=8816=E6=97=A5, =E4=B8=8B=E5=8D=886=E6=99=8215=E5=88=86, William >> >> > = >> >> > Elliot <ma...(a)rdrop.remove.com> wrote: >> >> > > On Fri, 16 Apr 2010, David C. Ullrich wrote: >> >> > > > <eric955...(a)yahoo.com.tw> wrote: >> >> >> > > >> How to prove the sequence just has one limit point (Proof by >> >> > > >> contradiction ). >> >> >> > > > _What_ sequence? Some sequences have more than one >> >> > > > limit point. >> >> >> > > > A given sequence has at most one limit (some sequences do not >> >> > > > have a limit, because they have more than one limit point...) >> >> >[...] >> >> >> I think there is a language barrier here. �"Limit point" is not the same as >> >> "limit". �The actual problem must have been to prove that a sequence has at >> >> most one limit. � >> >> >>[...] >> >> >Thank you very much. >> >Maybe we should suppose it has two limit point.And find the >> >contradiction.Sorry I have no idea. >> >> When you post questions about homework you should at least >> _read_ the replies! The problem was (almost certainly) to show that >> a sequence has at most one _limit_, not "limit point".- ??????? - >> >> - ??????? - > >Sorry,I'm not sure about it,so I can't replies this problem. You're not sure about what the problem asked you to prove? Look at the problem. Does it say "limit" or "limit point"? It seems incredibly likely that the original problem said "limit". Now, your saying "limit point" instead is reasonable, until more than one person points out that this is not the same. At that point, when you continue to say "limit point" instead of "limit" it's clear you're simply not reading the replies to your question. Or at least not reading them nearly as carefully as you should. >Sorry but thank you for your help >Thank you very much. > >Eric.
From: David C. Ullrich on 18 Apr 2010 09:50 On Sat, 17 Apr 2010 17:26:48 -0700 (PDT), Chip Eastham <hardmath(a)gmail.com> wrote: >On Apr 17, 10:48�am, Sharon <nicegirl_...(a)yahoo.com> wrote: >> Well, this is kinda confusing. I'm not sure if I >> have the standard definition, but, according to >> the one I have seen, we say z is a limt point of >> a sequence z_n if, for every eps > 0 and every >> positive integer k, there is n > k with |z_n - z| < eps. >> This is equivalent to (a) every neighborhood of z >> contains terms of z_n for infinitely many n's and >> (b) z is the limit of some subsequence of z_n. >> >> So, according to this definition, z_n can have no >> limit point, can have only one (if it's convergent), >> and can have more than one. It can have m limit >> points, or �infinitely many limit points, even >> uncountably many. � >> >> Sharon > >See Eric's clarification in response to my >earlier post (prompted by much the same >thought you've expressed above). He's only >interested in Cauchy sequences of complex >numbers. I seriously doubt that that's true. I know he said "yes" when you asked whether he was interested only in Cauchy sequences, but he's been saying a lot of things. I'm willing to bet that the problem he was told to solve is to show that any sequence has at most one limit. >regards, chip
From: Chip Eastham on 18 Apr 2010 11:47 On Apr 18, 9:50 am, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > On Sat, 17 Apr 2010 17:26:48 -0700 (PDT), Chip Eastham > > > > <hardm...(a)gmail.com> wrote: > >On Apr 17, 10:48 am, Sharon <nicegirl_...(a)yahoo.com> wrote: > >> Well, this is kinda confusing. I'm not sure if I > >> have the standard definition, but, according to > >> the one I have seen, we say z is a limt point of > >> a sequence z_n if, for every eps > 0 and every > >> positive integer k, there is n > k with |z_n - z| < eps. > >> This is equivalent to (a) every neighborhood of z > >> contains terms of z_n for infinitely many n's and > >> (b) z is the limit of some subsequence of z_n. > > >> So, according to this definition, z_n can have no > >> limit point, can have only one (if it's convergent), > >> and can have more than one. It can have m limit > >> points, or infinitely many limit points, even > >> uncountably many. > > >> Sharon > > >See Eric's clarification in response to my > >earlier post (prompted by much the same > >thought you've expressed above). He's only > >interested in Cauchy sequences of complex > >numbers. > > I seriously doubt that that's true. I know he said "yes" > when you asked whether he was interested only > in Cauchy sequences, but he's been saying a lot of > things. I'm willing to bet that the problem he was > told to solve is to show that any sequence has at > most one limit. > > >regards, chip > > Or (noting "Convergence" appears in the subject line but not the text of his original post) perhaps that any convergent sequence has exactly one limit point. --c
From: David Bernier on 18 Apr 2010 15:32 Chip Eastham wrote: > On Apr 18, 9:50 am, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: >> On Sat, 17 Apr 2010 17:26:48 -0700 (PDT), Chip Eastham >> >> >> >> <hardm...(a)gmail.com> wrote: >>> On Apr 17, 10:48 am, Sharon <nicegirl_...(a)yahoo.com> wrote: >>>> Well, this is kinda confusing. I'm not sure if I >>>> have the standard definition, but, according to >>>> the one I have seen, we say z is a limt point of >>>> a sequence z_n if, for every eps > 0 and every >>>> positive integer k, there is n > k with |z_n - z| < eps. >>>> This is equivalent to (a) every neighborhood of z >>>> contains terms of z_n for infinitely many n's and >>>> (b) z is the limit of some subsequence of z_n. >>>> So, according to this definition, z_n can have no >>>> limit point, can have only one (if it's convergent), >>>> and can have more than one. It can have m limit >>>> points, or infinitely many limit points, even >>>> uncountably many. >>>> Sharon >>> See Eric's clarification in response to my >>> earlier post (prompted by much the same >>> thought you've expressed above). He's only >>> interested in Cauchy sequences of complex >>> numbers. >> I seriously doubt that that's true. I know he said "yes" >> when you asked whether he was interested only >> in Cauchy sequences, but he's been saying a lot of >> things. I'm willing to bet that the problem he was >> told to solve is to show that any sequence has at >> most one limit. >> >>> regards, chip >> > > Or (noting "Convergence" appears in the subject line > but not the text of his original post) perhaps that > any convergent sequence has exactly one limit point. Not quite an "exegesis" exercise, but seems pretty close ... < http://en.wikipedia.org/wiki/Exegesis > . David Bernier
From: Chip Eastham on 18 Apr 2010 16:58
On Apr 18, 3:32 pm, David Bernier <david...(a)videotron.ca> wrote: > Chip Eastham wrote: > > On Apr 18, 9:50 am, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > >> On Sat, 17 Apr 2010 17:26:48 -0700 (PDT), Chip Eastham > > >> <hardm...(a)gmail.com> wrote: > >>> On Apr 17, 10:48 am, Sharon <nicegirl_...(a)yahoo.com> wrote: > >>>> Well, this is kinda confusing. I'm not sure if I > >>>> have the standard definition, but, according to > >>>> the one I have seen, we say z is a limt point of > >>>> a sequence z_n if, for every eps > 0 and every > >>>> positive integer k, there is n > k with |z_n - z| < eps. > >>>> This is equivalent to (a) every neighborhood of z > >>>> contains terms of z_n for infinitely many n's and > >>>> (b) z is the limit of some subsequence of z_n. > >>>> So, according to this definition, z_n can have no > >>>> limit point, can have only one (if it's convergent), > >>>> and can have more than one. It can have m limit > >>>> points, or infinitely many limit points, even > >>>> uncountably many. > >>>> Sharon > >>> See Eric's clarification in response to my > >>> earlier post (prompted by much the same > >>> thought you've expressed above). He's only > >>> interested in Cauchy sequences of complex > >>> numbers. > >> I seriously doubt that that's true. I know he said "yes" > >> when you asked whether he was interested only > >> in Cauchy sequences, but he's been saying a lot of > >> things. I'm willing to bet that the problem he was > >> told to solve is to show that any sequence has at > >> most one limit. > > >>> regards, chip > > > Or (noting "Convergence" appears in the subject line > > but not the text of his original post) perhaps that > > any convergent sequence has exactly one limit point. > > Not quite an "exegesis" exercise, but seems pretty close ... > <http://en.wikipedia.org/wiki/Exegesis> . > > David Bernier Hey, I'm doing "God's Work" don'tcha know... --c |