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What is the limit superior of the following sequence of sets?

$\{X_n\}=\{\{1/2\},\{1/3\},\{1/4\},\{2/3\},\{1/3\},\{1/5\},\{3/4\},\{1/3\},\{1/6\}......\}(n\to∞)$

I.e., $X_1=\{1/2\}, X_2=\{1/3\}, X_3=\{1/4\}, \dots$ (I use the same notation as in the Wikipedia article linked below.)

I have taken the example from Wikipedia article on limit superior where two sequences are combined. Here I have combined three sequences:

  • the sequence $\{\frac{n}{n+1}\}$
  • the constant sequence $\{\frac13\}$
  • the sequence $\{\frac1n\}$

What's the value of $\limsup{X_n},\liminf{X_n}$ and why? If it's possible to visualize?

From Wikipedia, I learn that they find subsequence first while if I don't know the number of subsequence, it seems hard to get answer. So I am confused and hope to visualize to understand it.

Martin Sleziak
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Robert Jiang
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    I can't see a clear pattern in the behavior of $X_n$. Could you perhaps give the expression of $X_n$ explicitly? –  Jul 13 '15 at 05:08
  • @BolzWeir it's {(2*n-1)/(2^n)} {1/3} and {1/n} (n->∞) – Robert Jiang Jul 13 '15 at 05:23
  • So you mean $X_1={1/2}$, $X_2={1/3}$, ...? Since this is not what is written in the post. – Martin Sleziak Jul 13 '15 at 07:11
  • @MartinSleziak I followed wikipedia limit superior and limit inferior Sequences of sets part and I'm confused – Robert Jiang Jul 13 '15 at 07:22
  • @MartinSleziak So I make a little change and hope to complement my confusion – Robert Jiang Jul 13 '15 at 07:22
  • @MartinSleziak I am thinking if there are several subsequences, I don't know it's expression but follow a certain monotonicity then how to get it's limsup and liminf . From Wikipedia ,I learn that they find subsequence first while if I don't know the number of subsequence,it seems hard to get answer.So I am confused and hope to visualize to understand it – Robert Jiang Jul 13 '15 at 09:06
  • There are several reasons why your post is unclear. (Which might be a reason why your post was put on hold.) It is not clear whether you mean, $X_1={1/2}. X_2={1/3},\dots$ or $X_1={1/2}, X_2={1/2,1/3},\dots$. When I read your post you write $X_n$ as a set consisting of singleton sets, which seems like something completely different. – Martin Sleziak Jul 13 '15 at 13:01
  • It is also not clear which type of $\limsup$ you mean. Since you have mentioned you have looked in Wikipedia article, I will point out that there are two definitions there. One of them takes topology into account. The other one can be considered as a special case for discrete topology. – Martin Sleziak Jul 13 '15 at 13:02
  • @MartinSleziak it's in the special case:discrete metric. Using either the discrete metric or the Euclidean metric – Robert Jiang Jul 13 '15 at 13:12
  • Perhaps it might be useful for you to browse other questions tagged elementary-set-theory+limsup. Maybe seeing those questions and answers might clear some of the things you have problems with – Martin Sleziak Jul 13 '15 at 13:13
  • @MartinSleziak thank you very much but I'll still wait for the right answer – Robert Jiang Jul 13 '15 at 15:47
  • Re: I'll still wait for the right answer. It is difficult to answer a question which is not clearly stated. But it might be just me and maybe another user can understand your question. However, at the moment your question is closed/on-hold, so nobody can post an answer (only comments can be posted). Some guidance what to do if you want the question reopened is in the message below your post. I will also copy here a standardized message with some basic info. – Martin Sleziak Jul 13 '15 at 16:46
  • Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Jul 13 '15 at 16:47
  • @MartinSleziak I think I typed the {Xn} followed the examples in Wikipedia article .The difference is that I add a new subsequence {{1/3},{1/3},{1/3}......}and delete{0},{1}from {Xn} in Wikipedia article – Robert Jiang Jul 13 '15 at 17:06
  • @MartinSleziak What confused me is the way to get limsup. In the article ,Using the Euclidean metric , he find the "odd" and "even" elements of the sequence form two subsequences,which have limit points 1 and 0 , so limsup={0,1},But if I can't see subsequces clearly such as the limit points ,the number of subsequences and so on ,It's possible to get limsup{Xn} ? – Robert Jiang Jul 13 '15 at 17:16
  • @MartinSleziak thank you very much – Robert Jiang Jul 14 '15 at 07:37
  • It seems that you are interesting in Kuratowski limit superior. You have subsequence convergent to 0, subsequence convergence to 1/3 and subsequence convergent to 1. I did not notice other convergent subsequences (although I did not try to prove this rigorously.) So my guess is that limsup (in this sense) will be ${0,1/3,1}$. If you are interested in the other definition, i.e. $\limsup X_n = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty X_m$, then I suupose you will get $\limsup X_n={1/3}$. – Martin Sleziak Jul 14 '15 at 08:03
  • @MartinSleziak In my opinion the two definitions you said are the same thing in https://en.m.wikipedia.org/wiki/Limsup because they are all written in Sequences of sets – Robert Jiang Jul 14 '15 at 08:47
  • @MartinSleziak and the answer should be the same .I think the key is to find the subsequences .So picking "even" and "odd" elements of the sequence maybe a good way to find subsequences.If it's still hard to find convergent subsequence I can do the same work as mentioned above again until I find the obviously convergent subsequence =D – Robert Jiang Jul 14 '15 at 08:57
  • If I may add some advice on handling the situation when your post is put on hold: Such posts only get once into reopen queue through editing. Since you only added one sentence, it is not surprising that several users voted to leave this post closed. It is advisable to make more substantial edit. (I have tried to edit your post and then I cast reopen vote, which is how your post entered reopen quite the second time.) – Martin Sleziak Jul 14 '15 at 11:20

1 Answers1

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From the post (and the comments) it seems that you are interested in Kuratowski limit superior.

There are obvious subsequences $x_{n_k}\in X_{n_k}$ which are convergent to $0$, $\frac13$ and $1$. If you check that for other real numbers there is no such subsequence, then you get that $$\limsup X_n=\{0,\frac13,1\}.$$

There is also another definition of limit superior of sets, namely $$\limsup X_n = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty X_m.$$ It can be considered the special case of Kuratowski limit superior in the case you use discrete topology (or discrete metric).

In this case you will get $$\limsup X_n=\{\frac13\},$$ since $\frac13$ is the only number which appears in infinitely many $X_n$'s.

Since I've mentioned two different notions of limit superior of sets, I'll add some links to other posts on this site, where these notions are discussed:

(Maybe also some other posts linked here.)

Community
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Martin Sleziak
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  • There is also another definition of limit superior of sets, namely $$\limsup X_n = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty X_m.$$ It can be considered the special case of Kuratowski limit superior in the case you use dicrete topology (or discrete metric).

    In this case X_n={0} ,X_n+1={\frac13},X_n+2={1},X_n+3={0}....... $$\limsup X_n={0,\frac13,1},$$

     – Robert Jiang Jul 14 '15 at 11:36
  • You are aware that you wrote a different sequence in your comment, right? – Martin Sleziak Jul 14 '15 at 12:43
  • I don't think so $$\limsup X_n = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty X_m.$$ Firstly I wanna get \bigcup_{m=n}^\infty X_m ,namely X_m={1/2} ,X_m+1={1/3},X_m+2={1/4},......X_m+k={1},X_m+k+1={1/3},X_m+k+2={0},X_m+k+3={1}......(here m=n and n=1) I note A_1=\bigcup_{m=1}^\infty X_m. Then when n=2 X_m={1/3},X_m+1={1/4},X_m+2={2/3},.....X_m+k={1},X_m+k+1={1/3},X_m+k+2={0},X_m+k+3={1}..... I note A_2=\bigcup_{m=2}^\infty X_m....when n=p X_m={1},X_m+1={1/3},X_m+2={0},.....X_m+k={1},X_m+k+1={1/3},X_m+k+2={0},X_m+k+3={1}..... I note A_2=\bigcup_{m=p}^\infty X_m....... – Robert Jiang Jul 14 '15 at 13:56
  • After that \bigcap_{m=1}^\infty A_m . The value of that expression is {1 1/3 0}. Namely $$\limsup X_n={1 1/3 0} that's my opinion – Robert Jiang Jul 14 '15 at 14:01
  • When n=p It should be A_p. I am very sorry – Robert Jiang Jul 14 '15 at 14:03
  • Do you agree with my answer? – Robert Jiang Jul 15 '15 at 09:22
  • Which answer? To what question? I will try to repost your above comment with MathJax, so that it is readable. – Martin Sleziak Jul 15 '15 at 09:24
  • Robert's comment with math rendering: – Martin Sleziak Jul 15 '15 at 09:29
  • Firstly I wanna get $\bigcup_{m=n}^\infty X_m$, namely $X_m={1/2}, X_{m+1}={1/3},X_{m+2}={1/4},......X_{m+k}={1},X_{m+k+1}={1/3},X_{m+k+2}={0},X_{m+k+3}={1}$..‌​....(here $m=n$ and $n=1$) I note $A_1=\bigcup_{m=1}^\infty X_m$. Then when $n=2$ $X_m={1/3},$ $X_{m+1}={1/4},X_{m+2}={2/3},.....X_{m+k}={1},X_{m+k+1}={1/3},X_{m+k+2}={0},X_{m+k‌​+3}={1}.....$ I note $A_2=\bigcup_{m=2}^\infty X_m$....when $n=p$ $X_m={1},X_{m+1}={1/3},X_{m+2}={0},...,X_{m+k}={1},X_{m+k+1}={1/3},X_{m+k+2}={0},X_{m+k+3}={‌​1}.....$ I note $A_2=\bigcup_{m=p}^\infty X_m$ – Martin Sleziak Jul 15 '15 at 09:29
  • yes Then \bigcap_{m=1}^\infty A_m . The value of that expression is {1 1/3 0}. Namely $$\limsup X_n={1 1/3 0} that's my opinion so The two answers may be the same – Robert Jiang Jul 15 '15 at 09:41
  • Sorry Robert but I do not understand what you are talking about, so I am probably not able to help you. None of the sets given in your question contains $0$, so it cannot be contained in $\bigcup_{m=1}^\infty X_m$. It seems that most of your comments are about a different sequence than your original questions. (And also your comments are difficult to read, since you do not use MathJax for some reason.) – Martin Sleziak Jul 15 '15 at 10:27
  • Sorry for bothering you for a long time you are a warm heart and kind person thank you very much I will follow your advice to read more articles and learn to use Math Jax .But my account has been banned ,I have to open a new account I am still happy to meet you =D – Robert Jiang Jul 15 '15 at 10:55
  • @RobertJiang Clearly you were able to post a comment, so your account is not banned. Perhaps you mean post ban? In that case you might have a look in help or here. – Martin Sleziak Jul 16 '15 at 05:28