Characterisation of sequences such that every bounded subsequence converges

Question

Call a sequence of real numbers semiconvergent if every bounded subsequence converges. Clearly, a bounded sequence is semiconvergent if and only if it is convergent. Also, all monotone sequences are semiconvergent (and indeed so are any sequences with no bounded subsequences). One can also do things such as alternate between terms of a monotone sequence and terms of a convergent sequence to get a semiconvergent sequence. I'm not sure if the sum of two semiconvergent sequences is necessarily semiconvergent, though I haven't given it much thought.

I hope this question isn't too open-ended, but seeing as I can't find anything about this notion online, I would like to know whether there is some alternative characterisation for (unbounded) semiconvergent sequences, or even interesting properties or sufficient conditions for semiconvergence. Perhaps there is something trivial about this notion that I am overlooking.

Answer 1

A sequence in a complete metric space whose closed balls are totally bounded, $Y$, is semiconvergent if and only if it has at most one limit point in $Y$.

Proof:

Suppose $(x_n)$ has two limit points in $Y$, then choose a bounded open set containing both. Choose the subsequence of $(x_n)$ that lies in this open set. Now we have a bounded subsequence that necessarily still has two limit points, hence is not convergent.

For the converse, it suffices to show that a bounded sequence, $(x_n)$, with a unique limit point, $x$, is convergent. Since $(x_n)$ is bounded, it is contained in a closed ball, which is compact by total boundedness of closed balls and completeness of $Y$. Call this closed ball $K$. Then if $(x_n)$ doesn't converge to the unique limit point $x$, there is $\epsilon > 0$ such that $(x_n)$ has infinitely many terms not contained in the open ball $U=B_\epsilon(x)$. Then let $(y_n)$ be the subsequence of $(x_n)$ contained in $K\setminus U$, which is a closed and hence compact subset of $K$. Since compactness implies sequential compactness for metric spaces, $(y_n)$ has a limit point in $K\setminus U$, and thus so does $(x_n)$. Contradiction.

Note:

In particular, this applies to $Y=\Bbb{R}^n$ for any $n$.

Answer 2

As jgon's answer showed, a sequence $(x_n)$ in $\mathbb{R}$ is semiconvergent iff it has at most one limit point. Since you seem to be interested in what a semiconvergent sequence can "look like", let me point out that there is a very concrete description you can get from this. Namely, a sequence $(x_n)$ is semiconvergent iff there is a partition of $\mathbb{N}$ into two sets $A$ and $B$ such that the subsequence $(x_n)_{n\in A}$ converges and the subsequence $(x_n)_{n\in B}$ satisfies $|x_n|\to \infty$. (If $A$ or $B$ is finite, these conditions are taken to hold vacuously.) In other words, $(x_n)$ is obtained by splicing together a convergent sequence with a sequence that goes to (unsigned) $\infty$.

The proof is simple. If $(x_n)$ has no limit points, then it has no bounded subsequences, so $|x_n|\to \infty$ and we can take $A=\emptyset$ and $B=\mathbb{N}$. Otherwise, suppose $x\in\mathbb{R}$ is the unique limit point of $(x_n)$. Let $U$ be some bounded neighborhood of $x$ and let $A=\{n:x_n\in U\}$. Then $(x_n)_{n\in A}$ is a bounded sequence whose only limit point is $x$, so it converges to $x$. If $|x_n|\not\to\infty$ on $B$, then there is a bounded subsequence of $(x_n)_{n\in B}$. But by definition of $B$, this subsequence cannot converge to $x$, and so its limit would be a limit point of $(x_n)$ besides $x$. Thus $|x_n|\to\infty$ on $B$.

Conversely, if such $A$ and $B$ exist, any bounded subsequence of $(x_n)$ must have all but finitely many of its terms in $A$, and thus must converge to the limit of $A$.

As in jgon's answer, all of this generalizes to any metric space in which closed balls are compact, with "$|x_n|\to\infty$" taken to mean that $x_n$ is eventually outside any bounded set. More generally, a similar argument shows that if $(x_n)$ is a sequence in any compact Hausdorff space with only finitely many accumulation points, then $(x_n)$ is obtained by splicing together finitely many sequences converging to each of those accumulation points. (In the metric space case above, the compact Hausdorff space is the one-point compactification of your original space.)