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.
Best Answer
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 $n$.