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The Maximal Compact Topology is the set $\omega^2\cup\{x,y\}$ topologized by the following basis: points of $\omega^2$ are isolated, $\{x\}\cup\bigcup_{n<\omega}(\omega\setminus f(n))\times\{n\}$ is open for each $f:\omega\to\omega$, and $\{y\}\cup\omega\times(\omega\setminus n)$ is open for each $n<\omega$. Put another way, neighborhoods of $x$ contain all but finitely-many points of each row of $\omega^2$, and neighborhoods of $y$ contain all but finitely-many rows.

Is this space extremally disconnected: every pair of disjoint open sets has disjoint closures? If so, it would be a counterexample to my conjecture at What separation is required to ensure extremally disconnected spaces are sequentially discrete? that all extremally disconnected US spaces are sequentially discrete.

Steven Clontz
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3 Answers3

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No, it isn't: It is well-known, and easy to see, that a space is extremally disconnected, iff the closure of each open set is open again.

Let $U := \omega \times \{0 \}$. Then $U$ is open, $\overline{U} = U \cup \{x\}$, which is not open.

Ulli
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A space is extremally disconnected if and only if the interior of closed sets is closed. The closed subsets in the Maximal Compact Topology are exactly the ones that fall into one of these categories:

  1. subsets that contain $x$ and $y$.

  2. subsets $S$ such that there is some $m\in\omega$ such that $S\subseteq\{x\}\cup(\omega\times\{0,\cdots, m\})$ and $x\in S$.

  3. finite subsets that do not contain $x$ or $y$.

  4. subsets $S$ that contain $y$, do not contain $x$ and such that $S\cap (\omega\times\{m\})$ is finite for all $m$.

You can see that whenever you take a set from $(2)$ its interior will be $S\setminus\{x\}=S\cap(\omega\times\omega)$, which won't be closed if $S$ is infinite.

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To round out these answers, here's how to see that this space is not extremally disconnected using the definition I provided: $\{0\}\times\omega$ and $\{1\}\times\omega$ are disjoint open sets whose closures have $y$ in common.

Steven Clontz
  • 8,773