I am looking at the relationship between the existence of a dispersion point and the biconnected property. Some definitions:
- A space $X$ is biconnected if it is connected and is not the union of two disjoint connected subsets, each with at least two points.
- The point $p\in X$ is a dispersion point for $X$ if $X$ is connected and the subspace $X\setminus \{p\}$ is totally disconnected (where totally disconnected = all connected components are singletons).
When the space $X$ has a very small number of elements, the definitions degenerate into trivial cases:
To witness that a connected space is not biconnected, one needs at least four elements (two elements in each subset). So any connected space with at most three elements is biconnected.
Also, a space with less than two elements is trivially totally disconnected. So if $X$ has a single point, that point is trivially a dispersion point. And if $X$ is connected with exactly 2 points, each of its points is a dispersion point.
For size 3, there are examples of $X$ connected (hence biconnected) and without dispersion point. For example, if the corresponding specialization preorder is a chain, there is no dispersion point ($X\setminus\{p\}$ is connected for every $p$). Explicitly, take $X=\{a,b,c\}$ with topology $\{\emptyset,\{a\},\{a,b\},X\}$.
Now some results. First a standard one, which does not depend on the cardinality of $X$.
Proposition 1: If $X$ has a dispersion point, then it is biconnected.
Proof: If $X=A\cup B$ with $A$ and $B$ disjoint and connected of size at least $2$ and if the dispersion point $p\in A$ for example, then $B$ would be a connected subset of $X\setminus\{p\}$ of size at least $2$. That is not possible since $X\setminus\{p\}$ is totally disconnected.
The converse implication is not true. The classic example of biconnected space without dispersion point is "Miller's biconnected set" from Counterexamples in topology, example #131 (taken from E. Miller, Concerning biconnected sets Fundamenta Mathematicae 29 (1937), 123-135). Another (trivial) such example with three elements was given above.
But I think there could be a valid converse for most finite spaces.
Proposition 2: If $X$ is biconnected and finite with at least four points, then it has a dispersion point.
Can this be proved?
Best Answer
Here is a proof:
$X$ is biconnected, finite, $n := |X| \ge 4$.
Let $\le$ be the specialization preorder on $X$, i.e. $x \le y \Leftrightarrow x \in \overline{\{y\}}$.
every chain $C$ in $X$ has at most 2 elements.
[Assume $C = \{a, b, c\} $ are pairwise distinct, $a \le b \le c$. Then $C = \overline{\{c\}}$: "$\subset$" is clear. "$\supset$": Let $x \in \overline{\{c\}}$ with $x\ne c$. Then $\{x,c\}$ and $\{a,b\}$ are connected, hence $x=a$ or $x=b$ (see below). Analogously, $C = \{x \in X: a <= x\}$. Hence C is closed and open, hence $C= X$. Contradiction.]
X is T0, i.e. $\le$ is a partial order.
[Assume not, then there are distinct $a, b \in X$, such that $a \le b \le a$. By 1. we have $\overline{\{a\}} = \overline{\{b\}} = \{a, b\}$, and for $x \in X$: if $a \in \overline{\{x\}}$, then $x \in \{a, b\}$. Hence for $x \in X \setminus \{a, b\}$: $\overline{\{a\}} \cap \overline{\{x\}} = \emptyset$ and $\overline{\{x\}} = \{x\}$, since X is biconnected. Since X is finite, we have $X \setminus \{a, b\}$ is closed. It follows $X = \{a, b\}$. Contradiction.]
$X$ has a dispersion point.
[Let $m = max\{|A|: A \text{ is an antichain in X} \}$. By Dilworth's theorem there exists a partition $X = \bigcup_{i=1}^m C_i$ of $X$ into chains. By 1., $|C_i| <= 2$ for each i, and, again by biconnectness, there is at most one $C_i$ with two elements. Hence, $ n = |X| = \sum _{i=1}^m |C_i| <= 2 + (m-1) = m+1$.
Thus, there is an antichain $A$ of size $n-1$. It follows, that $A$ is discrete, hence the unique element in $X \setminus A$ is dispersion point.]
Addendum
Thanks to @M W for reminding me to explicitly mention the following fact and a proof thereof:
If $X$ is biconnected, then for all connected subsets $A, B$ of $X$, each with more than one point, one has $A \cap B \neq \emptyset$.
For a proof see for instance here ("strongly SG").