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If $A$ is compact how can it be shown that any sequence $\{x_n\}$ in $A$ has a limit point in $A$?

I know this is proven in a lot of textbooks but I'm finding this hard to conceptualise.

Stefan Hamcke
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jorgea
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  • By limit point do you mean a point $x$ such that every neighborhood contains infinitely many $x_i$ (not necessarily distinct), or do mean a point $x$ such that each neighborhood $U$ intersects the sequence in a point different from $x?$ – Stefan Hamcke Jun 19 '13 at 16:43

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I assume that $\{x_n\}_{n \in \mathbb{N}}$ is infinite, otherwise the statement is false (if you mean accumulation point); suppose that such a set hasn't got an accumulation point, then it's an infinite discrete closed subset of a compact space, which is impossible. In fact having no accumulation points implies that your subset is closed in $A$ and so it's compact; for each point there 's an open set which contains only a point of your subset (no limit points) and so that open cover has no finite subcover. (I still have the doubt that you mean cluster point)

Edoardo Lanari
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  • Yes I believe it should be cluster point, though the notes I am following have it written as above. – jorgea Jun 19 '13 at 13:59
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    Sequentially closed is not the same as closed. – André Caldas Jun 19 '13 at 14:16
  • I meant no accumulation points, since the OP somehow confused those two terms – Edoardo Lanari Jun 19 '13 at 14:21
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    Sorry, I made the same confusion myself... – André Caldas Jun 19 '13 at 14:27
  • I think in the space $(\mathbb R_{\ge0},\tau)$, where the topology is generated by all intervals $[0,x),\ x\ge0$, the set $\mathbb N$ is an infinite sequence without a cluster point, but it is not closed as there are adherence points outside of $\mathbb N$. So in a non-$T_1$ space a sequence without an accumulation point isn't automatically closed. – Stefan Hamcke Jun 19 '13 at 14:44
  • Correct me if I'm wrong, but is OP asking that every compact space is sequentially compact? If so, that is false. See: http://math.stackexchange.com/questions/44907/whats-going-on-with-compact-implies-sequentially-compact – Dylan Yott Jun 19 '13 at 15:12
  • @StefanH: I repeat myself: I meant accumulation point, not cluster point for the sequence. – Edoardo Lanari Jun 19 '13 at 15:20
  • I thought accumulation point is the same as cluster point. That's what Wikipedia says http://en.wikipedia.org/wiki/Accumulation_point. Then what do mean by accumulation point? A limit point? – Stefan Hamcke Jun 19 '13 at 15:26
  • A cluster point for a sequence is a point (not necessarily in the image of that sequence) such that chosen one of its neighborood you get infinitely many naturals $n$ for which $x_n$ is in that neighborood. An accumulation point is a point such that each of its neighborhood intersects the set minus that point. – Edoardo Lanari Jun 19 '13 at 15:31
  • Isn't that called a limit point? See the same wikipedia-page. – Stefan Hamcke Jun 19 '13 at 15:32
  • It's a bit ambiguous in my opinion: a cluster point doesn't need to be an acc point (think of a constant sequence) – Edoardo Lanari Jun 19 '13 at 15:38
  • No, a cluster point and an accumulation point are equal by definition, except you define an accumulation point to have infinitely many points in each of its neighborhoods, but then it is called a $\omega$-accumulation point. – Stefan Hamcke Jun 19 '13 at 16:38
  • Tell me where I make that infinite request – Edoardo Lanari Jun 19 '13 at 16:47
  • Please, use "@Stefan" in your comment, else I don't get notified about your comment. I translated word for word from German, that's why I missed a word, I meant "... except if you define ...". Never mind, I now notice your comment from earlier where you wrote how you define it. But I think the notion limit point is more common for what you refer to as accumulation point. Also the OP wanted to know if it has a cluster point, not a limit point. What you proved is that the sequence has a limit point, which is a cluster point only if the space is $T_1$. – Stefan Hamcke Jun 20 '13 at 08:55
  • Does the reverse hold? If every sequence in a space X has at least one cluster point, is the space X necessarily compact? – Michael Oct 08 '22 at 12:58
  • By the way, there's a flaw in your proof. It should be 'for each point there's an open set which contains only finite points of your subset (no limit points)'. – Michael Oct 08 '22 at 12:59