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Figure 3 on page 20 of Counterexamples in Topology, Steen and Seebach has some assumption? (like $T_1$, first countable) One of the 6 implications shown on the picture is

A sequentially compact space is countably compact.

I'm stuck with the proof and before asking for a direct help, I would like to be sure about the correctness of my interpretation: no assumption.

enter image description here

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    $\begingroup$ As for the proof, try the contrapositive. Assume $X$ isn't countably compact and construct a sequence without convergent subsequence. $\endgroup$ Nov 18 '19 at 20:59
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On page 19 of counterexamples, they mention 4 different equivalent forms of countable compactness (equivalent in all spaces):

  1. Every countable open cover has a finite subcover.
  2. Every infinite subset of $X$ has an $\omega$-accumulation point in $X$. (i.e. $p$ such that every neighbourhood contains infinitely many points of that set.)
  3. Every sequence $(x_n)_n$ in $X$ has an accumulation point in $X$. (i.e. a $p$ such that for every neighbourhood $O$ of $p$ and every $n$ there is some $m \ge n$ such that $x_m \in O$.)
  4. Every countable family of closed subsets of $X$ with empty intersection has a finite subfamily with empty intersection.

So to show space countably compact we can just show any one of these and we're done, depending on what we're given to work with. Other posts on this site show this equivalence and the proof is in the book too.

And it's immediate that $\text{CC}_3$ is implied by sequential compactness, as the limit $p$ of a subsequence (which then exists) is always an accumulation point of the sequence, regardless of extra conditions, just definitions. This justifies their arrow.

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Right: there are no extra assumptions. Take the double pointed countable complement topology (example 21 of their book). It's not even $T_0$, but, for them, it is weakly countably compact.

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