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This question was motivated by recent posts about biconnected spaces, see here.

A space is called biconnected, if it is connected and the intersection of any two connected subsets $A,B$ with $|A|, |B| \ge 2$ is non-empty. (This is not the original definition, but an equivalent one, which turned out to be more relevant.)

Let $X$ be a biconnected topological space with $|X| \ge 4.$ Is $X$ $T_0$ ?

Notes.

  1. If $X$ has the indiscrete topology and $2 \le |X| \le 3$, then $X$ is biconnected, but not $T_0$.
  2. If $X$ is finite (or, more generally, Alexandroff), the answer is yes, see the post linked above.
  3. Every connected space with a dispersion point (i.e., the subspace without that point is totally disconnected) is biconnected. While the converse is false, both properties seem to be quite close together. For instance, no counterexample in the plane without additional set-theoretic assumptions is known.
  4. In fact, 3. was my motiviation for this question, since, as it is easy to see, a connected space with a dispersion point is $T_0$.
  5. As an answer to the question of the above link, M W proved that there is an $x_0 \in X$, such that $X \setminus \{x_0\}$ is $T_1$. Does this help?
  6. Most often, biconnectness is considered in the realm of $T_2$, or even (separable) metric spaces. Hence, this question is somewhat unusual. Or is there any relevant, already known result?
  7. Of course, biconnected spaces need not be $T_1$: there are even finite, bicconected spaces of arbitrary size.
Ulli
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  • Stab in the dark: is every $T_0$ quotient of a biconnected space biconnected? Then, given a biconnected space, can you double a point and keep biconnected? – Steven Clontz Nov 25 '23 at 17:44
  • I also thought about doubling a point (perhaps even the dispersion point). But this would destroy biconnectness, unless intersections of non-trivial connected subsets in fact have more than 1 element. I'm not familiar with the Knaster-Kuratowski fan. Perhaps it has this property? – Ulli Nov 25 '23 at 17:52
  • Doubling the dispersion point of the Knaster-Kuratowski fan would no longer be biconnected as I believe it would contain two disjoint copies of the Knaster-Kuratowski fan. – Steven Clontz Nov 25 '23 at 22:23
  • Good question. That was going the be my next one. An example for 7. would be a particular point topology or an excluded point topology, which can also be defined for infinite sets. These have a dispersion point and are not $T_1$. – PatrickR Nov 26 '23 at 23:40

2 Answers2

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Motivation for this answer: Ulli suggested the Knaster-Kuratowski fan as a space with a dispersion point that could possibly be doubled without breaking biconnected, but doubling that point would result in these points providing the tips for two disjoint copies of the fan. So any non-$T_0$ space similarly has a pair of topologically indistinguishable points, and assuming biconnected their complement must be disconnected - we'll use this disconnection to disconnect the entire space, contradicting biconnected.

Take a biconnected space $X$ with at least $4$ points; as Ulli points out, M W showed there is some point $x$ such that $X\setminus\{x\}$ is $T_1$. Let $y\in X\setminus\{x\}$. Suppose by way of contradiction that $x,y$ have the same neighborhoods. Then for each $z\in X\setminus\{x,y\}$ pick $V$ open containing $z$ but not $y$ (and thus not $x$). This witnesses that $\{x,y\}$ is closed. It's also connected, so by biconnected we know $X\setminus\{x,y\}$ (which has at least two points) is disconnected, so pick open sets $U,V$ disconnecting it. We may assume both sets miss the closed set $\{x,y\}$ and thus are disjoint in $X$.

I claim $U\cup\{x\}$ and $V\cup\{y\}$ are connected, contradicting the biconnected property.

  • Suppose the open sets $T,W$ disconnect $U\cup\{x\}$ with $x,y\in T$. Then $T\cup V,W\cap U$ are disjoint, nonempty, open, and cover $X$. Thus they disconnect $X$, contradiction.

  • Similarly, suppose the open sets $T,W$ disconnect $V\cup\{y\}$ with $x,y\in T$; it follows that $T\cup U,W\cap V$ disconnects $X$, contradiction.

Since $x,y$ don't have the same neighborhoods, the space is at least $T_0$.

PatrickR
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Steven Clontz
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  • I might be butchering some elementary topology but I think when we say $U\cup{x}$ is not connected, that gives a separation by relatively open disjoint $T$ and $W$, or if we take $T$ and $W$ absolutely open, then we only get disjointness after we intersect with $U\cup{x}$. (The argument still works though, if we do it the latter way, I think.) – M W Nov 25 '23 at 23:58
  • I was operating in the latter fashion. I'll spell out the details; in an draft I neglected to intersect with U (resp. V) which I think would be an error for the concerns you share, but I think it works as written (modulo the need for more clarity). – Steven Clontz Nov 26 '23 at 00:23
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    I would argue as follows: Start with $T, W$ disjoint and open in $U \cup {x}$, $x \in T$, then $W \subset U$, hence $W$ is open in $X$. Then pick $T^\prime$ open in $X$ such that $T = T^\prime \cap (U \cup {x})$, and continue as proposed (with $T^\prime$ instead of $T$). – Ulli Nov 26 '23 at 10:24
  • Great! So, also this is proved now! Thanks a lot! – Ulli Nov 26 '23 at 10:26
  • In bullets at the end, I am still a little confused by the open sets $T$ and $W$ disconnecting $U\cup{x}$ ( Are $T$ and $W$ open in $X$ but not necessarily disjoint? But their intersections with $U\cup{x}$ are disjoint and show the disconnection?...) Maybe it's common language and I need to get used to it. Ulli's latest comment seem clearer. (and the second bullet is basically the same as the first one by symmetry). Great answer otherwise. – PatrickR Nov 27 '23 at 00:09
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    Actually, thinking about it some more and with the right interpretation (that is $T$ and $W$ don't need to be disjoint, etc), Steven's phrasing is a good way to express things and the reasoning in the bullets is easy to follow. I am getting used to it! – PatrickR Nov 27 '23 at 00:21
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(not a direct answer to the question, but a closely related consequence)

Suppose $X$ has a dispersion point $p$ (i.e., $X$ connected such that $X\setminus\{p\}$ is totally disconnected (= all connected components are singletons)), Then $X$ is biconnected. Therefore, from the answer to the question above: $(*)$ if $X$ has a dispersion point and $|X|\ge 4$, it must be $T_0$. And examining all cases for $|X|=3$ shows that such spaces with a dispersion point are also $T_0$. So $(*)$ also holds for all spaces with $|X|\ge 3$.

But the proof in Steven Clontz's answer for biconnected spaces is not so straightforward when taken in combination with the prerequisite result that there is some point $x$ such that $X\setminus\{x\}$ is $T_1$. So I thought it would be instructive to give below a direct proof of $(*)$.

Note: if $X$ has a dispersion point and $|X|=2$, it is not necessarily the case that $X$ is $T_0$. (Counterexample: the indiscrete space on two points)

Proposition: Every space $X$ with a dispersion point and $|X|\ge 3$ is $T_0$.

Proof: Let $p$ be the dispersion point. We have to show that any two distinct points are topologically distinguishable. There are two cases to examine.

First case: Show that $p$ is topologically distinguishable from any other point. Suppose by contradiction that this is not the case. It is easy to check that if a connected topological space has two distinct and topologically indistinguishable points, removing one of the points still leaves a connected space. So in this case, $X\setminus\{p\}$ is connected, with at least two elements since $|X|\ge 3$. But that is impossible, as $X\setminus\{p\}$ is totally disconnected.

Second case: Given distinct points $a,b$ with $a,b\ne p$, show that they are topologically distinguishable. The set $A=\{a,b\}$ is not connected since $X\setminus\{p\}$ is totally disconnected. Hence $A$ is discrete and there is an open set $U\subseteq X$ containing $a$ and not $b$.

PatrickR
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