## Does connectedness follow?

(It is an other question than my previous question, despite it starts with the same discussion about connectors.)

This question arose in my research of generalized connectedness (see this draft article for the overall idea):

Let $U$ is a set, $r$ is a binary relation on $\mathcal{P} U$. I call $r$ a connector.

Informal note: The relation $r$ of two sets $A$ and $B$ represents that $A$ and $B$ are in some sense "near" or "touch". For example $r$ may be a proximity.

I will define the set $\mathrm{CC} (r)$ of connected subsets of $U$ by the formula

$\mathrm{CC} (r) = \lbrace A \in \mathcal{P} U | \forall X, Y \in \mathcal{P} A \setminus \lbrace \emptyset \rbrace : (X \cup Y = A \wedge X \cap Y = \emptyset \Rightarrow X r Y) \rbrace .$

As I mentioned above, $r$ may be a proximity and in this case $\mathrm{CC}(r)$ is proximal connectedness, that is a set a $A$ is connected iff every partition of the set a $A$ consists of two near sets.

As an other important example $ArB$ may mean that the topological closure (given some topological space) of the set $A$ in the subspace generated by the set $A\cup B$ intersects $B$ or the closure of $B$ intersect $A$. This is equivalent to the classic definition of connectedness of a set on topological space, because it happens if and only if $A$ and $B$ are not both open-closed on $A\cup B$.

There are other examples of connectedness following this scheme: graph connectedness, digraph strong connectedness, uniform connectedness, etc. (see my draft article)

I will call a link a ternary relation $\tau\in U\times U\times\mathcal{P}U$. (The link $\tau(a,b,A)$ may mean for example that there are some path from a point $a$ to a point $b$ through a set $A$.) I will call a link $\tau$ increasing when $A\subseteq B\implies \tau_A\subseteq\tau_B$.

I will define connector $Q(\tau)$ by the formula $\forall X,Y\in\mathcal{P}U:(X Q(\tau) Y\iff \exists x\in X,y\in Y:\tau(x,y,X\cup Y))$.

Let $\tau$ is an increasing link.

Prove or give a counter example: $A\in\mathrm{CC}(Q(\tau)) \implies \forall x,y\in A:\tau(x,y,A)$.

Does it hold if we remove the requirement to be increasing?

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You should probably tell us what $\tau_A$ means. – Mariano Suárez-Alvarez Jul 23 2010 at 21:23
I think that $\tau_A(a,b) =\tau (a,b,A)$, for what I understand. Isn't it right? – dan232 Aug 9 2011 at 23:57
@dan232: You are right. – porton Aug 10 2011 at 11:06