Limit points of $ \{x_n, n\in N\}$ as a subset and limits of all its subsequences In a topological space, let $A$ be the set of limit points of $ \{x_n, n\in N\}$ as a subset, and let $B$ be the set of limits of all subsequences of $ \{x_n, n\in N\}$ as a sequence. Is $A = B$? My intuition says yes. I don't know how to prove or disprove it, although I know both definitions. Thanks! 
 A: No. For $x_n=x$ constant, $A=\emptyset$ and $B=\{x\}$. More generally, $x_n$ can visit countably many isolated points infinitely often, and thus $B$ may be countably infinite while $A$ is empty.
[Edit:]
Example 12.1 from Counterexamples in Topology provides a case where $B$ is empty but $A$ is not. Also, this space is first-countable, so it's a counterexample to the claim in Nate's comment.
Consider an infinite set $X$ with open sets the empty set and all sets containing some point $p$. This space is first-countable, since every point $x$ has an open neighbourhood $\{x,p\}$ that's contained in all open neighbourhoods of $x$.
Take any sequence of distinct points, one of which is $p$. Then each of these points except $p$ is in $A$, since every open set containing any such point also contains $p$. But no subsequence converges to any point, since for any point $x$ the open neighbourhood $\{x,p\}$ contains at most two elements of the sequence.
Thus, the "proof" of Nate's claim that I gave in a comment was wrong. The problem is that I assumed that there's an infinite supply of different points of the sequence that we can pick from the countable neighbourhood basis elements; but in the present case these basis elements are all the same and we can only pick the same point each time, which doesn't result in a subsequence.
