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Let $X$ be a compact Hausdorff sequential space. Does $X$ have a dense subset of points which have a countable local base?

Every example that I have seen so far has this property.

user558840
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According to this survey paper (p. 381) Arhangel'skij proved in 1971 (there is only one paper by him from that year in its references, so that might be the paper in question, though the 1970 paper would also interest you, I think, if the title is something to go by) that under CH we have that a compact sequential Hausdorff space has a point of first countability. And also that Malykhin later constructed consistent examples of spaces where this was not the case. I cannot check which paper in the references shows such an example, but with access to a good library one probably could.

So I conclude from that passage that there exists in some model of ZFC a strong counterexample to your conjecture.

Henno Brandsma
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  • Great. That is most helpful. – user558840 Aug 06 '20 at 16:36
  • @user558840 just a google search but I’m glad it helps. – Henno Brandsma Aug 06 '20 at 16:44
  • Of course this gives an answer to this question as well: https://math.stackexchange.com/questions/3778971/is-a-locally-compact-hausdorff-quotient-of-a-locally-compact-sigma-compact-fi A first countable space is sequential, and a quotient of a sequential space is sequential, so under CH any compact neighbourhood of a point of that quotient must have a point of first countability. Thus if the quotient is locally compact, the set of points of first countability is dense. – user558840 Aug 07 '20 at 08:18
  • @user558840 what is "this question"? – Henno Brandsma Aug 07 '20 at 08:19