Relationship between dispersion point and biconnected property

Question

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).

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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 $\{\emptyset,\{a\},\{a,b\},X\}$.

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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?

Answer 1

Here is a proof:

$X$ is biconnected, finite, $n := |X| \ge 4$.
Let $\le$ be the specialization preorder on $X$, i.e. $x \le y \Leftrightarrow x \in \overline{\{y\}}$.

1. 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.]

2. 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.]

3. $X$ 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").

Answer 2

Here's a purely topological answer using the useful characterization noted by Ulli: biconnected implies for each pair of connected subsets $A,B$ with $|A|,|B|\geq 2$, $A\cap B\not=\emptyset$.

Take $X$ biconnected with $|X|\geq 4$. We begin by noting $cl\{x\}\setminus\{x\}$ is $T_1$ for each $x\in X$; equivalently, there do not exist distinct points $x,y,z$ with $y\in cl\{x\}$ and $z\in cl\{y\}$. If not, choose $y,z\in cl\{x\}\setminus\{x\}$ with $z\in cl\{y\}\setminus\{y\}$. Note that every subset of $\{x,y,z\}$ is connected. Since $X$ is connected and $X\setminus\{x,y,z\}$ is non-empty, $\{x,y,z\}$ is not clopen. It follows that either there is a point of $X\setminus\{x,y,z\}$ which is a limit of some point of $\{x,y,z\}$, or vice versa. But either way, this would witness two disjoint connected sets of size two, a contadiction.

Since $X$ is connected and finite, it is not discrete, and there is some $x\in X$ with $y\in cl\{x\}\setminus \{x\}$; note $\{x,y\}$ is connected. It follows that $X\setminus\{x,y\}$ is $T_1$: if not we'd have distinct $w,z\in X\setminus\{x,y\}$ with $w\in cl\{z\}$, and $\{x,y\},\{w,z\}$ would be disjoint connected sets, a contradiction.

Take each $z\in X\setminus\{x,y\}$.

Suppose $\{z\}$ is open; $\{z\}$ is not closed, so either $x\in cl\{z\}$ or $y\in cl\{z\}$. But if $x\in cl\{z\}$ we'd have a contradiction with $y\in cl\{x\}$, so $y\in cl\{z\}$.

Then if $\{z\}$ is not open, every open neighborhood intersects either $x$ or $y$, that is, $z\in cl\{x\}$ or $z\in cl\{y\}$. But if $z\in cl\{y\}$ we'd have a contradiction with $y\in cl\{x\}$, so $z\in cl\{x\}$.

Finally, these scenarios must be the same for all $w,z\in X\setminus\{x,y\}$. If not, take $y\in cl\{z\}$ and $w\in cl\{x\}$; then $\{y,z\},\{w,x\}$ are disjoint connected sets, a contradiction.

If $z\in cl\{x\}$ for all $z\in X\setminus\{x,y\}$, then $X$ has the particular point topology with particular point $x$, and $x$ is a dispersion point. And if $y\in cl\{z\}$ for all $z\in X\setminus \{x,y\}$, then $X$ has the excluded point topology with excluded point $y$, and $y$ is a dispersion point.

Answer 3

Combining Ulli and Steven Clontz's approaches, we can actually prove a more general statement: If $X$, not necessarily finite, is biconnected with $|X|\geq 4$, then for some point $x_0\in X$, $X\backslash \{x_0\}$ is $T_1$. From this, the original statement follows immediately, as a finite $T_1$ space is discrete, hence totally disconnected.

In what follows we use the equivalent characterization of biconnectivity, that $X$ is connected, and two disjoint subsets of $X$ of size at least two cannot both be connected (as pointed out in the post linked by Ulli, and also in the original paper on the subject by Knaster and Kuratowski).

Step 1

We begin with a slight modification to Ulli's argument. Let $\preccurlyeq$ denote the specialization preorder $x\preccurlyeq y \iff x\in \overline{\{y\}}$. We also sometimes say "$x$ is connected to $y$" to mean "$\{x,y\}$ is connected", or equivalently, $x$ and $y$ are comparable in this preorder.

We first argue we can never have $x\preccurlyeq y \preccurlyeq z$ for distinct $x,y,z\in X$.

Supposing otherwise, we have $\overline{\{z\}}\supseteq\{x,y,z\}$. Since $X$ is biconnected, we cannot have a fourth $w\in\overline{\{z\}}$, as then $\{z,w\}$ and $\{x,y\}$ are disjoint and connected. Thus $A=\{x,y,z\}$ is closed.

But then, again by biconnectivity, since $\{y,z\}$ is connected, we have $X\backslash \{y,z\}=B\cup C$ for disjoint (relatively) closed nonempty sets $B$ and $C\ni x$. Since $x\notin B$, we must have $\overline{B}\cap A=\emptyset$, therefore $B$ is actually closed in $X$, and since $C$ is closed relative to $X\backslash\{y,z\}$ and $A$ is closed, we have $A\cup C$ is closed, so that $A\cup C$ and $B$ are a separation of $X$, contradicting connectivity.

Step 2

We now argue more or less as in Steven Clontz's answer. If $X$ is $T_1$ already there is nothing to prove, so otherwise let $x\neq y\in X$ with $\{x,y\}$ connected. By biconnectivity, $X\backslash \{x,y\}$ is totally disconnected. Then for each $z\in X\backslash \{x,y\}$, $z$ cannot be connected to any point other than $x$ or $y$, and may only be connected to one of these, as otherwise we have a chain of size $3$, contradicting the previous result.

Moreover, if $z_1\neq z_2$ are connected to $x$ and $y$ respectively, then $\{x,z_1\}$ and $\{y,z_2\}$ are disjoint and connected, contradicting biconnectivity, so we must have every $z\in X\backslash \{x,y\}$ connected to (WLOG) $x$ and no other point, or not connected to any other point at all. It follows that $y$ is also not connected to any point other than $x$, as there are no allowable points left.

Therefore $X\backslash \{x\}$ is $T_1$.

Remark

To upgrade this result to a dispersion point, we don't actually need $X$ to be finite - it suffices to assume that $X$ is Alexandrov (arbitrary intersections of open sets are open), as in such a space $T_1$ is equivalent to discrete. In this case, most of the arguments from the other two answers carry over as well, particularly that $X$ is $T_0$, and has either a particular point topology or an excluded point topology, with the particular or excluded point coinciding with the dispersion point.