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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.

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    $\begingroup$ I can't see a clear pattern in the behavior of $X_n$. Could you perhaps give the expression of $X_n$ explicitly? $\endgroup$ – user230734 Jul 13 '15 at 5:08
  • $\begingroup$ @BolzWeir it's {(2*n-1)/(2^n)} {1/3} and {1/n} (n->∞) $\endgroup$ – Robert Jiang Jul 13 '15 at 5:23
  • $\begingroup$ So you mean $X_1=\{1/2\}$, $X_2=\{1/3\}$, ...? Since this is not what is written in the post. $\endgroup$ – Martin Sleziak Jul 13 '15 at 7:11
  • $\begingroup$ @MartinSleziak I followed wikipedia limit superior and limit inferior Sequences of sets part and I'm confused $\endgroup$ – Robert Jiang Jul 13 '15 at 7:22
  • $\begingroup$ @MartinSleziak So I make a little change and hope to complement my confusion $\endgroup$ – Robert Jiang Jul 13 '15 at 7:22
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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.)

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  • $\begingroup$ 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}....... $$\limsup X_n=\{0,\frac13\,1},$$ $\endgroup$ – Robert Jiang Jul 14 '15 at 11:36
  • $\begingroup$ You are aware that you wrote a different sequence in your comment, right? $\endgroup$ – Martin Sleziak Jul 14 '15 at 12:43
  • $\begingroup$ 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....... $\endgroup$ – Robert Jiang Jul 14 '15 at 13:56
  • $\begingroup$ 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 $\endgroup$ – Robert Jiang Jul 14 '15 at 14:01
  • $\begingroup$ When n=p It should be A_p. I am very sorry $\endgroup$ – Robert Jiang Jul 14 '15 at 14:03

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