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Steen and Seebach say that: If a topological space $X$ is first countable, sequentially compactness is equivalent to limit point compactness in $X$.

Take $X=\mathbb{N}\times\{0,1\}$, where $\mathbb{N}$ has the discrete topology and $\{0,1\}$ has the indiscrete topology. This space is first countable since the topologies on $\mathbb{N}$ and $\{0,1\}$ are first countable. Thus it should satisfy the statement in the previous paragraph.

However, if one takes a sequence like $\{(n,0)\}$ in $X$, there are no convergent subsequences for limit points like $(1,1)$. In other words, one can't find a subsequence of $\{(n,0)\}$ that converges to $(1,1)$, even though $(1,1)$ is a limit point of the sequence. This seems to violate the statement in the first paragraph.

I don't see where I'm making the error with this example. Any help is appreciated.

Eric Wofsey
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Mapply
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    How is $(1,1)$ a limit point of the sequence? – Eric Wofsey Dec 27 '18 at 00:57
  • Every neighborhood of (1,1) contains an element of the sequence. In fact {(1,0), (1,1)} is the neighborhood base at (1,1)...as far as I understand. – Mapply Dec 27 '18 at 01:05
  • Ah, well, it's a limit point of the set but not of the sequence. Being a limit point of the set is what's relevant though since that's how limit point compactness is defined. So, you are right that the statement is wrong. I can't find such a statement in Steen and Seebach though--can you give a page reference? – Eric Wofsey Dec 27 '18 at 01:17
  • Page 22, paragraph 2 in the Dover edition, in the chapter on Compactness. It is called countable compactness there, but Steen & Seebach's countable compactness is equivalent to what Munkres and others call limit point compactness (pg 19 defn CC3 in Steen & Seebach). – Mapply Dec 27 '18 at 01:51
  • countably compact in S & S is defined on p. 19 of the Dover edition (CC1-CC4) and is correct. Your space is not countable compact but (weakly) limit point compact as defined later. – Henno Brandsma Dec 27 '18 at 07:13
  • S&S say that in first countable spaces: seq. cpt iff countable compact (not limit point compact)! And this is indeed correct. It uses the CC3 equivalent formulation: every sequence has an accumulation point, of countable compactness. – Henno Brandsma Dec 27 '18 at 07:18
  • Yes, I realized this @HennoBrandsma. My example is not a T1 space because it is not Hausdorff. This is the reason why the equivalence of countable compactness and limit point compactness breaks down. Please see my comment below EricWofsey's answer. And, thank you. – Mapply Dec 27 '18 at 14:24

1 Answers1

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You seem to have some confusion about the definitions. Steen and Seebach say that for first-countable spaces, sequential compactness is equivalent to countable compactness, not limit point compactness. You claim that their definition of countable compactness is what others call limit point compactness but this is false. In particular, one of the equivalent conditions for countable compactness they give is:

Every sequence has an accumulation point.

This is not the same as limit point compactness, which means instead that

Every infinite set has a limit point.

Of course, limit point compactness can be checked just on countable sets since every infinite set has a countably infinite subset, but the key distinction is that an accumulation point of a sequence is not the same thing as a limit point of the set of terms of the sequence. In particular, $x$ is an accumulation point of a sequence $(x_n)$ if for each neighborhood $U$ of $x$, there exist infinitely many $n$ such that $x_n\in U$. In contrast, $x$ is a limit point of the set $\{x_n\}$ if for each neighborhood $U$ of $x$, there exists $n$ such that $x_n\neq x$ and $x_n\in U$. In your example, $(1,1)$ is a limit point of the set $\{(n,0):n\in\mathbb{N}\}$ but is not an accumulation point of the sequence $(x_n)$ with $x_n=(n,0)$.

So, your space is indeed first countable, limit point compact, and not sequentially compact. This does not contradict Steen and Seebach's statement though, since your space is not countably compact.

Eric Wofsey
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  • No, I don't think this has anything to do with confusion about the definitions. For instance, the exact statement that is in my post is Proposition 3.19 here https://fenix.tecnico.ulisboa.pt/downloadFile/848204501356648/FTARnotas_3compact.pdf – Mapply Dec 27 '18 at 03:10
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    Proposition 3.19 is just wrong. – Eric Wofsey Dec 27 '18 at 03:13
  • @Mapply 3.19 is contradicted by 3.18 example 3, which is limit point compact, first countable but not sequentially compact. So your document is internally insonsistent. – Henno Brandsma Dec 27 '18 at 07:10
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    @Mapply in short: the document you linked to is wrong and Steen and Seebach are right. The counterexample works against the document formulation but not against that of S & S. The notes misquote S & S, in short. – Henno Brandsma Dec 27 '18 at 07:22
  • Yes, my bad. Steen & Seebach pg 19 last paragraph gives the condition under which limit point compactness and countable compactness are equivalent. Basically needs to be a T1 (or higher) space. In my example, the space is not Hausdorff (thus not T1) and therefore limit point compactness is not equivalent to countable compactness (and sequential compactness). Perhaps @EricWofsey can add this in his answer...that'll make the answer completely self-contained. Thanks everyone. – Mapply Dec 27 '18 at 14:14
  • @Mapply your space is clearly not countably compact by the countable open cover ${n} \times {0,1}$, $n \in \mathbb{N}$, which has no finite subcover. Not being $T_1$ is indeed necessary for such an example. – Henno Brandsma Dec 27 '18 at 17:32