Is The Set Of Subsequential Limits Always Closed/Compact? 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?
 A: 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.
A: 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 $\infty$.
