There exist 3 notions of compactness:

  1. $X$ is compact if any open cover admits a finite subcover;
  2. $X$ is sequentially compact if any sequence in $X$ has a convergent subsequence;
  3. $X$ is limit point compact if any infinite $A\subseteq X$ has at least a limit point.


The three are eequivalent in metric spaces.

Theorem (Eberlein-Šmulian)

The three are equivalent in weak topologies of Banach spaces.

Does there exist a characterisation of spaces where this holds by say some topological property? If not, are there other examples of spaces where the three are equivalent besides metric spaces and weak topologies of Banach spaces? And how would one go about proving whatever the answer to the previous questions is?


1 Answer 1


Some remarks on this, without striving for comprehensiveness:

  • Compactness and sequential compactness do not imply each other in general topological spaces. For example, the set of countable ordinals with the order topology is not compact, but it is sequentially compact, whereas the space $\{0,1\}^{[0,1)}$ with the product topology (each factor space being endowed with the discrete topology) is compact (Tychonoff) but not sequentially so.

  • In any topological space, compactness is equivalent to “net compactness:” that is, the condition that any net have a convergent subnet.

  • Every compact space is countably compact—this means that any countable open cover has a finite subcover (this is different from Lindelöf’s condition, which requires that any arbitrary open cover have a countable subcover).

  • For $T_1$ spaces (that is, in which singletons are closed), countable compactness is equivalent to limit-point compactness as you stated it.

  • Every sequentially compact space is countably compact.

  • If a topological space is countably compact and first-countable, then it is sequentially compact.

  • The previous three claims imply that countable compactness, sequential compactness, and limit-point compactness are all equivalent in first-countable $T_1$ spaces.

  • If a topological space is normal and $T_1$, then it is countably compact if and only if every real-valued continuous function on it is bounded.

For a reference for these claims, see Folland (1999, Section 4.4 and the exercises therein).

  • $\begingroup$ I'm guessing you are missing some condition for the equivalence of compactness and countable compactness from "the previous three claims", since from those I only appear to be able to deduce that countable compactness, limit-point compactness and sequential compactness are equivalent for first-countable $T_1$ spaces, but not to compactness. $\endgroup$
    – MickG
    Commented May 6, 2016 at 6:12
  • 2
    $\begingroup$ @MickG I never stated that compactness is equivalent to the other three conditions in such spaces. Indeed, here is a counterexample: the order topology on the set of countable ordinals is first-countable and $T_1$, so that it is countably compact, sequentially compact, and limit-point compact. However, it is not compact. (See also the first remark.) $\endgroup$
    – triple_sec
    Commented May 6, 2016 at 6:25
  • $\begingroup$ Sorry, must have misread. Specifically, missed a "countable" :). $\endgroup$
    – MickG
    Commented May 6, 2016 at 6:35

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