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.
- If $X$ has the indiscrete topology and $2 \le |X| \le 3$, then $X$ is biconnected, but not $T_0$.
- If $X$ is finite (or, more generally, Alexandroff), the answer is yes, see the post linked above.
- 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.
- In fact, 3. was my motiviation for this question, since, as it is easy to see, a connected space with a dispersion point is $T_0$.
- 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?
- 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?
- Of course, biconnected spaces need not be $T_1$: there are even finite, bicconected spaces of arbitrary size.
Best Answer
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 $T_0$.