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I've been working my way through Rudin's PoMA, and I was thinking about subsequential limits.

If you take a sequence with finitely many subsequential limits, then the set of subsequential limits is clearly closed and compact.

Theorem 3.17 in Rudin states that the upper limit of a sequence is the limit of some subsequence. This gives me the feeling that perhaps the set of subsequential limits for a divergent sequence with infinitely many subsequnetial limits is also closed or compact. Admittedly, however, I don't think this follows directly from theorem 3.17.

I've toyed with this idea for a bit, but I haven't been able to come up with a proof or a counter-example. Is this true or false, and what's the proof? Does this have any other interesting extensions/implications?

UYasher
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  • Rudin, Principles of Mathematical Analysis (3rd ed. 1976), Theorem 3.7: "The subsequential limits of a sequence ${p_n}$ in a metric space $X$ form a closed subset of $X.$" Also, I searched Maths.SE and found several duplicates or near-duplicates of this question. Most of the answers are complicated, but Hagen von Eitzen's answer to Proof that the set of subsequential limits is closed is very similar to my answer here. I shall flag the present question as a duplicate of that one, even though it may not be an exact duplicate. – Calum Gilhooley Jul 13 '20 at 11:56

2 Answers2

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If $x$ is not a subsequential limit, then (i) $x$ occurs in the sequence finitely often or not at all, and (ii) $x$ has an open neighbourhood, $N,$ whose intersection with the set of points in the sequence is $\varnothing$ or $\{x\}.$ Every point of $N \setminus \{x\},$ therefore, has a neighbourhood disjoint from the sequence, therefore it is not a subsequential limit. It follows that the complement of the set of subsequential limits is open, i.e. the set of subsequential limits is closed. As has already been pointed out, it need not be bounded, therefore it need not be compact.

Calum Gilhooley
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  • Actually, I think point (i) was irrelevant, but never mind. – Calum Gilhooley Jul 12 '20 at 15:59
  • The only relevance of point (i) is that if $y \in N\setminus{x},$ then not only is $y$ not a limit point of the image set, $S,$ of the sequence, but also $y \notin S,$ therefore a fortiori $y$ does not occur infinitely often in the sequence; and one needs both these observations (which correspond to (ii) and (i) respectively) in order to conclude that $y$ is not a subsequential limit. I've probably made it clear as mud now. I'll get me coat. $\ \ddot\frown.$ – Calum Gilhooley Jul 12 '20 at 16:27
  • Help me understand how you concluded the set $L$ of all subsequential limits is closed. Actually closed in what space ? You have shown $S\setminus L$ is an open set as it is the union of open sets, namely the collection of neighborhoods $N\setminus {x}$ in your notation. So the complement of the set $\cap_{x;\text{is not a subsequential limit }} N_x $ is closed. But this complement contains $L$. What am I missing ? – Medo Jul 12 '20 at 17:47
  • change $\cap$ to $\cup$ in my comment above. – Medo Jul 12 '20 at 18:02
  • I only had $\mathbb{R}$ in mind, because I had too hastily excluded the case of unbounded sequences. It would have been both clearer and more informative to have written an answer along the lines of Dave Renfro's roughly simultaneous comment on the previous answer. I hope someone will do so. (It would be unethical of me to rewrite this answer.) Still, if I'm not mistaken, my argument also applies to $\overline{\mathbb{R}},$ except for the final passing remark about compactness. Thinking aloud (I'll rewrite these comments if necessary): Let $X$ be either $\mathbb{R}$ or $\overline{\mathbb{R}}.$ – Calum Gilhooley Jul 12 '20 at 18:44
  • ... As in a previous comment of mine, let $S$ be the set of points in the sequence. (All in $\mathbb{R},$ of course!) As in your comment, let $L$ be the set of subsequential limits (in $X,$ whichever of the two possible spaces that is). I argued that if $x \in X \setminus L,$ then $x$ has an open neighbourhood $N$ in $X$ such that $N \cap S = \varnothing$ or $N \cap S = {x}.$ If $y \in N \setminus {x},$ then (a) $y \notin S,$ and (b) $y$ has an open neighbourhood (in $X$) disjoint from $S.$ Therefore $y \notin L.$ This shows directly that $X \setminus L$ is open, i.e. $L$ is closed in $X.$ – Calum Gilhooley Jul 12 '20 at 19:00
  • (I'm sorry for the lack of clarity. I'm struggling to carry on as normal, but it has been impossible for me to concentrate on mathematics since Tuesday 30 June, for reasons which have everything to do with Mathematics Stack Exchange and nothing to do with mathematics. Clearly I'm not back to normal yet, but at least I'm back in the saddle!) – Calum Gilhooley Jul 12 '20 at 19:12
  • An authoritative answer (cough) unlike mine (cough) is given by Kenneth A. Ross, Elementary Analysis: The Theory of Calculus (2nd ed. 2013), Theorem 11.9: "Let $S$ denote the set of subsequential limits of a sequence $(s_n).$ Suppose $(t_n)$ is a sequence in $S \cap \mathbb{R}$ and that $t = \lim t_n.$ Then $t$ belongs to $S.$" – Calum Gilhooley Jul 12 '20 at 23:13
  • This theorem does not say $S$ is closed (because it looks at the limits of convergent sequences in $\mathbb{R}\cap S$ not $S$ ). However, I finally understood and agree with your argument clarified in your comments. I upvoted your answer a while ago too. If this is actual "coughing" you have... well, I hope you are fine. Please, shut down your devices and get some rest. – Medo Jul 13 '20 at 01:17
  • @Medo Don't worry, it wasn't that kind of cough, thank goodness! When I can think more clearly I'll try to add some clarification to the answer and cut these comments down a bit. – Calum Gilhooley Jul 13 '20 at 10:59
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The upper limit is an infimum of a monotonically decreasing sequence, and that is what guarantees its existence (can be infinity). If $(a_{n})$ is a sequence then $$\lim\sup a_n =\lim_{n\rightarrow1}\sup_{k\geq n}a_k =\inf_{n\geq 1}\sup_{k\geq n}a_k.$$ Observe that the sequence $(\sup_{k\geq n}a_k)_{n\geq 1}$ is decreasing.

This is equal to the supremum of the set of the real numbers that consists of the limits of all convergent subsequences.

So, we should not worry about the set of subsequential limits being closed or bounded. The upper limit can in fact be $\infty$.

Medo
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  • The set of subsequential limits is compact in the extended reals, which is the "more appropriate space" to work with when dealing with subsequential limits. The set of non-infinite subsequential limits is also closed in the reals. These results can be useful since, for example, if you want the subsequential limits to include the set of all reals (equivalently, equal the set of all extended reals), it suffices to show they contain all rational numbers, a countable collection of requirements, which is useful in proving Baire category results such as this. – Dave L. Renfro Jul 12 '20 at 15:55
  • @DaveL.Renfro Sorry, my answer "crossed in the post" with your comment, which renders it redundant. – Calum Gilhooley Jul 12 '20 at 15:57