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!
Limit points of $ \{x_n, n\in N\}$ as a subset and limits of all its subsequences
1 Answers
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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1There are also examples where $B$ is empty but $A$ is nonempty. However, if the space is first countable, then $A \subset B$. – 2011-11-24
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0Thanks, 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 $X$? – 2011-11-24
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0@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? – 2011-11-24
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0@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 $a$. – 2011-11-24
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0@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 $x_n\in S$". – 2011-11-24
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0Thank you so much, joriki! – 2011-11-24
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0@Tim: Sorry, that example contained an error; I'll add it again when I've fixed it. – 2011-11-24
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0@Tim: I've added a (counter)example. It turns out that Nate's claim and my proof of it were actually wrong, as this is an example of a first-countable space where B can be empty while A isn't. – 2011-11-24
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0Oops. But we should have $A \subset B$ whenever the space is first countable and Hausdorff. – 2011-11-24
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0@Nate: Yes; in this case my proof can be made to work since the space being Hausdorff ensures that after picking a point we can throw out that point and all points preceding it in the sequence and still have an open neighbourhood that contains at least one point further along the sequence that we can pick next. – 2011-11-24
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0@joriki: Thanks! If replacing $\{ x_n, n \in \mathbb{N} \}$ with an arbitrary subset $S$, do we still not have any relation between the set $A$ of all its limit points and the set $B$ of limits of all sequences in it? $B$ is also called the sequential closure of $S$ http://en.wikipedia.org/wiki/Sequential_space#Sequential_closure. Can the elements of a sequence in $S$ be the same? (The sequential closure $B$ of $S$ is a subset of the closure of $S$. Here I wonder if $B$ and $A$ has some relation, such as in terms of inclusion.) Feel free to let me know if I better post this as a new question. – 2011-11-24
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0@Tim: I don't understand "Can the elements of a sequence in $S$ be the same?". What might prevent them from being the same? – 2011-11-24
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0If nothing prevents that, then $S \subseteq B$. Since $B$ has all isolated points which are not in $A$, will $A \subseteq B$? – 2011-11-24
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0@Tim: No; a counterexample is given in the [example section](http://en.wikipedia.org/wiki/Sequential_space#Examples) of the article you linked to above. In an uncountable set equipped with the cocountable topology, only eventually constant sequences converge, so $B=S$. On the other hand, every point of the space is a limit point of any uncountable set. Thus, for an uncountable set that isn't the entire space, $A\not\subseteq B$. – 2011-11-24
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0Some of the spaces from that book are listed in wikipedia article [Counterexamples in Topology](http://en.wikipedia.org/wiki/Counterexamples_in_Topology), example 7 is mentioned in [this wikipedia article](http://en.wikipedia.org/wiki/Deleted_integer_topology). – 2011-11-24
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0@MartinSleziak: Thanks! Why was example 7 mentioned? – 2011-11-24
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0Sorry @Tim, my mistake. The google books links in the above post went to example 7, so I thought that it was the space you're talking about. (I should have read the answer more carefully.) Example 12 has a [wikipedia article](http://en.wikipedia.org/wiki/Closed_extension_topology) too. – 2011-11-24