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Some definitions:

  • A space is hyperconnected if any two nonempty open sets intersect.
  • A space is ultraconnected if any two nonempty closed sets intersect.
  • A space $X$ is biconnected if it is connected and cannot be partitioned into two connected subsets, each with at least two elements.

There is also an equivalent characterization (consequence of the "folklore lemma" here): $X$ is biconnected iff it is connected and there are no disjoint connected subsets, each of size at least $2$. So to show that a connected space $X$ is not biconnected, it is sufficient to exhibit two disjoint connected subsets $A,B$ with $|A|, |B|\ge 2$.

I think the following result should be true:

Proposition: A space that is hyperconnected and ultraconnected with at least $4$ points cannot be biconnected.

The condition $|X|\ge 4$ is to avoid degenerate cases, as a connected space with at most $3$ points is trivially biconnected.

Can anyone provide a proof?

PatrickR
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  • Is it possible that every hyperconnected+ultraconnected space is hereditarily connected? That would give you the result immediately. – Steven Clontz Oct 20 '24 at 16:19
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    @Steven Clontz: No, I guess not: Consider a four-element space, partially ordered as a diamond, with the upper segment topology. – Ulli Oct 20 '24 at 17:17
  • Nicely done Ulli. Generally, every lattice with the upper segment topology will be hyperconnected and ultraconnected, but two incomparable points form a discrete subspace. We should add that counter-example. – Steven Clontz Oct 20 '24 at 17:24
  • BTW, @Steven Clontz: your link to $\pi$-base initiates a search for non hereditarily Lindelöf spaces rather than hereditarily connected spaces. I think the latter property is not included in $\pi$-base. – Ulli Oct 20 '24 at 17:39
  • It is, but only recently: you may need to resync your data at https://topology.pi-base.org/dev – Steven Clontz Oct 20 '24 at 19:44

1 Answers1

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Here is a proof.

Case 1: Assume all subsets $\{a,b\}\subseteq X$ are connected. Since $|X|\ge 4$, we can find two disjoint subsets $\{a,b\}$ and $\{c,d\}$, each of size $2$ and connected. This shows that $X$ is not biconnected.

Case 2: Assume some subset $\{a,b\}$ of size $2$ is not connected. Necessarily $\{a,b\}$ has the discrete topology. So $a$ has an open nbhd $U$ in $X$ not containing $b$, and $b$ has an open nbhd $V$ in $X$ not containing $a$.

Denote by $U^c$ the complement of $U$ in $X$ and similarly for $V^c$. Note that $U^c\ne\emptyset$ and $V^c\ne\emptyset$. The set $U$ is connected, as an open set in a hyperconnected space. The space $U^c$ is connected, as a closed set in an ultraconnected space. Now all that remains to be shown is $|U|\ge 2$ and $|U^c|\ge 2$.

Since $X$ is hyperconnected, $U\cap V\ne\emptyset$. And $U\setminus V\ne\emptyset$ as it contains $a$. Hence $|U|\ge 2$.

Since $X$ is ultraconnected, $U^c\setminus V=U^c\cap V^c\ne\emptyset$. And also $U^c\cap V=V\setminus U\ne\emptyset$ as it contains $b$. Hence $|U^c|\ge 2$.

PatrickR
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