About Convergence of complex sequence
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From: Sharon on 17 Apr 2010 06:48 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
From: David Bernier on 17 Apr 2010 11:24 Eric wrote: > On 4?17?, ??4?54?, Tonico <Tonic...(a)yahoo.com> wrote: >> On Apr 17, 11:41 am, 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...) >>>>>>> By limit point, do you mean a cluster point? >>>>>>>> What did the problem actually ask? >>>>>>> What are the actual definitions being used? >>>>>> Tank you for your interest.My teacher talk about the "Conergence of >>>>>> complex sequences" >>>>>> Def.{Z_n} is convergent to the limit a iff >>>>>> for all E>0, there exist N>0:for all n>N =3D>|Z_n-a|<E >>>>>> this is almost mean that Z_n -> a as n -> infinite >>>>>> and he introduce (Cauchy criterion) and use it to prove that >>>>>> definition (Def) >>>>>> and then he gave us this problem (How to prove that has one limit >>>>>> point (Proof by contradiction ). >>>>> 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. >>>>> OK, let's get you started with the proof by contradiction. >>>>> Hint: how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction? >>>>> -- >>>>> Robert Israel isr...(a)math.MyUniversitysInitials.ca >>>>> Department of Mathematics http://www.math.ubc.ca/~israel >>>>> University of British Columbia Vancouver, BC, Canada- ���óQ�ޥΤ�r - >>>>> - ��ܳQ�ޥΤ�r - >>>> 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".- >> Perhaps so and perhaps, as somebody else alread pointed out, it is a >> language barrier. I think your first response was likely to cause >> confusion, as it seemingly did, in the OP due to his/her obvious >> beginner level. It was likely to think that he/she would confuse or >> wouldn't distinguish between "sequence having a limit" and "limit >> point(s) of a given sequence". >> >> So perhaps a more or less direct answer could clear doubts out: >> suppose {a_n} is a seq. that has two different limits A,B, say |A - B| >> = K, for some some constant K > 0 ==> taking >> e = K/2 > 0 we know there exist natural numbers N1, N2 s.t: >> >> 1) |a_n - A| < e for all n > N1 >> >> 2) |a_n - B| < e for all n > N2 >> >> Let now M:= Max(N1,N2) (the greatest between these two numbers N1, >> N2) ==> for all n > M both equalities (1)-(2) are true , but then we >> get for all n > M: >> >> |A - B| = |A - a_n + a_n - B| <= |A - a_n| + |a_n - B| < 2e ...and now >> can you ( the OP) see the contradiction we get? >> >> Tonio- ??????? - >> >> - ??????? - > > Thank you > You help me to solve a very big problem. > Thank you very much Do you use the Greek letters epsilon and delta when talking about convergence where you are? David Bernier
From: Link on 17 Apr 2010 18:22 On Apr 16, 10:37 am, 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...) > > > > By limit point, do you mean a cluster point? > > > > > What did the problem actually ask? > > > > What are the actual definitions being used? > > > Tank you for your interest.My teacher talk about the "Conergence of > > complex sequences" > > Def.{Z_n} is convergent to the limit a iff > > for all E>0, there exist N>0:for all n>N =3D>|Z_n-a|<E > > this is almost mean that Z_n -> a as n -> infinite > > and he introduce (Cauchy criterion) and use it to prove that > > definition (Def) > > > and then he gave us this problem (How to prove that has one limit > > point (Proof by contradiction ). > > 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. > > OK, let's get you started with the proof by contradiction. > > Hint: how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction? > -- > Robert Israel isr...(a)math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Results 1 - 10 for hint; how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction?. (0.38 seconds) Suppose that a, b and c are integers such that a | b and b | c. ... Prove (by contradiction) that there is no largest integer. ... number -- might as well call it s (for smallest) -- what can we do to produce an even smaller number? ... OK, let's get you started with the proof by contradiction. > >> Hint: how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction? ... Can you produce further solutions? Hint: Use the fact that 12 = 1. ... (e) Let A = x, B = 2v2 and C = u2. This is another Pythagorean triple of the ... uXY = 0 and that 0 <u<z, leading to a contradiction ... About Convergence of complex sequence - how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction? > Eric wrote: > > Thank you very much. ... Model-based diagnostics using hints observed, then these observations are clearly in contradiction with the expected ... hint can be seen as an uncertain relation, i.e. with a known probability pa the H {M(cj): cj e D}. (10). So the new hyper edge D has to be included can be either 0 or 1, whereas in case (B) we assume a faulty Hint: About Convergence of complex sequence ... how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction? ... contradiction.Sorry I have no idea. |Z_n - a| < E is the interior ... Science News Report hints at the existence of a new and massive elementary particle. ... positrons collide to produce millions of pairs of B mesons and anti-B mesons. ... seeming contradiction is permitted by the weird laws of quantum theory ... At D=0, E was = m and both E and m were, together, all the energy and matter ... (a) To show (bm : B(r). Suppose br > . Choose ... Follow the hint: For each positive integer N, there exist only finitely many .... Thus, changing the n + 1'th place of from 7 to 4 or from 4 to a 7, one can produce a 2 B ... imply \F n 6= ;, a contradiction. Thus E is infinite, ... [However, membership can be tightly-controlled, Davis and Moore tell us -- over expansion can produce unemployment even in the learned professions. ... (e) The degree of 'stratum solidarity' able to preserve the interests of ... of complexity or contradiction as necessary as in rival theories]. .
From: BURT on 17 Apr 2010 19:24 On Apr 17, 3:22 pm, Link <marty.musa...(a)gmail.com> wrote: > On Apr 16, 10:37 am, 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...) > > > > > By limit point, do you mean a cluster point? > > > > > > What did the problem actually ask? > > > > > What are the actual definitions being used? > > > > Tank you for your interest.My teacher talk about the "Conergence of > > > complex sequences" > > > Def.{Z_n} is convergent to the limit a iff > > > for all E>0, there exist N>0:for all n>N =3D>|Z_n-a|<E > > > this is almost mean that Z_n -> a as n -> infinite > > > and he introduce (Cauchy criterion) and use it to prove that > > > definition (Def) > > > > and then he gave us this problem (How to prove that has one limit > > > point (Proof by contradiction ). > > > 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. > > > OK, let's get you started with the proof by contradiction. > > > Hint: how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction? > > -- > > Robert Israel isr...(a)math.MyUniversitysInitials.ca > > Department of Mathematics http://www.math.ubc.ca/~israel > > University of British Columbia Vancouver, BC, Canada > > Results 1 - 10 for hint; how can |Z_n - a| < E and |Z_n - b| < E > produce a contradiction?. (0.38 seconds) > > Suppose that a, b and c are integers such that a | b and b | c. ... > Prove (by contradiction) that there is no largest integer. ... > number -- might as well call it s (for smallest) -- what can we do to > produce an even smaller number? ... > > OK, let's get you started with the proof by contradiction. > >> Hint: > how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction? ... > > Can you produce further solutions? Hint: Use the fact that 12 = > 1. ... > (e) Let A = x, B = 2v2 and C = u2. This is another Pythagorean triple > of the ... > uXY = 0 and that 0 <u<z, leading to a contradiction ... > > About Convergence of complex sequence - how can |Z_n - a| < E and |Z_n > - b| < E produce a contradiction? > Eric wrote: > > Thank you very > much. ... > > Model-based diagnostics using hints observed, then these observations > are clearly in contradiction with the expected ... > > hint can be seen as an uncertain relation, i.e. with a known > probability pa the H {M(cj): cj e D}. (10). > > So the new hyper edge D has to be included can be either 0 or 1, > whereas in case (B) we assume a faulty > Hint: > > About Convergence of complex sequence ... > how can |Z_n - a| < E and |Z_n - b| < E produce a contradiction? ... > contradiction.Sorry I have no idea. |Z_n - a| < E is the interior ... > > Science News Report hints at the existence of a new and massive > elementary particle. ... > positrons collide to produce millions of pairs of B mesons and anti-B > mesons. ... > seeming contradiction is permitted by the weird laws of quantum > theory ... > > At D=0, E was = m and both E and m were, together, all the energy and > matter ... > > (a) To show (bm : B(r). Suppose br > . Choose ... > > Follow the hint: For each positive integer N, there exist only > finitely many .... > > Thus, changing the n + 1'th place of from 7 to 4 or from 4 to a 7, one > can produce a 2 B ... > > imply \F n 6= ;, a contradiction. Thus E is infinite, ... > > [However, membership can be tightly-controlled, Davis and Moore tell > us -- over expansion can produce unemployment even in the learned > professions. ... > > (e) The degree of 'stratum solidarity' able to preserve the interests > of ... > > of complexity or contradiction as necessary as in rival theories]. .- Hide quoted text - > > - Show quoted text - Would you please move your I into the complex plane. Mitch Raemsch
From: Chip Eastham on 17 Apr 2010 20:26
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. regards, chip |