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In Counterexamples space 99, the Maximal Compact Topology is asserted to be non-Hausdroff (note 1), KC (compacts are closed, note 3), and first-countable (note 5).

But KC spaces are US, and first countable US spaces are Hausdorff. So where is the error?

Steven Clontz
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  • The page you've linked claims that note 5 of Counterexamples in Topology states that space 99 is NOT first-countable. Can you check? –  Aug 13 '23 at 02:34
  • Thanks for clarifying - the link is correct, but the book actually says the space is first-countable. I'll add a screenshot to my answer. – Steven Clontz Aug 13 '23 at 02:37

1 Answers1

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Despite this claim from the text:

X is second countable since it is first countable and countable.

the space is not first-countable. Any set containing $x$ and containing all but finitely-many points from each row of $\omega^2$ is open. So given a countable collection $U_n$ of neighborhoods of $x$, let's construct an open set that contains none of these.

For each $n<\omega$ choose $m_n<\omega$ with $(m_n,n)\in U_n$ (which must exist, as $U_n$ contains all but finitely-many points on the $n$th row). Then we may consider $$U=\{x\}\cup\{(m,n):n<\omega,m\geq m_n+1\}$$ noting that $U$ is a neighborhood of $x$ but does not contain $(m_n,n)\in U_n$ for each $n<\omega$.

EDIT: It seems this error depends on your edition of Counterexamples. Here is what I find in the copy via Google Books:

X is not first countable since it is first countable and countable.

However this also has a slight error: $y$ does in fact have a local countable basis; only $x$ is a point without a countable basis.

Steven Clontz
  • 5,719