# 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!

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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.

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There are also examples where $B$ is empty but $A$ is nonempty. However, if the space is first countable, then $A \subset B$. –  Nate Eldredge Nov 24 '11 at 1:37
Thanks, joriki! I often wonder how "a sequence of points $\{x_n, n \in N \}$ in a set $X$ visits a subset $S$ of $X$ infinitely often" can be formulated into a math formula? Does it require some structure on the underlying set $X$? –  Tim Nov 24 '11 at 1:46
@NateEldredge: Thanks! (1) Are there simple examples where $B$ is empty while $A$ is not? (2) Why $A \subseteq B$ when the space is first countable? –  Tim Nov 24 '11 at 1:48
@Tim: On that last question: Given a limit point $a$ in $A$, construct a subsequence by successively choosing points $a_n\ne a$ in $A$ from the neighbourhoods $U_n$ in the countable neighbourhood basis of $a$. –  joriki Nov 24 '11 at 1:59
@Tim: On the first question: No, it doesn't require any structure on $X$; you can formalize it as "for every $n_0\in\mathbb N$ there is $n\ge n_0$ such that $x_n\in S$". –  joriki Nov 24 '11 at 2:03