This question arose in my research of generalized connectedness (see this draft article for the overall idea, but beware that the draft is yet too preliminary and unreadable, however I hope you can understand the overall idea from the draft):

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 call a connector $r$ extendable when

$ \forall X_0, Y_0, X_1, Y_1 \in \mathcal{P} U : (X_1 \cap Y_1 = \emptyset \wedge X_0 r Y_0 \wedge X_1 \supseteq X_0 \wedge Y_1 \supseteq Y_0 \Rightarrow X_1 r Y_1) . $

Below I will require that $r$ is extendable.

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 define connectors $\gamma (r)$ and $\beta (r)$ by the formulas (for every $A, B \in \mathcal{P} U$)

$ A \gamma (r) B \Leftrightarrow \exists X \in \mathcal{P} A, Y \in \mathcal{P} B : X r Y. $

$ A \beta (r) B \Leftrightarrow A \cup B \in \mathrm{CC} (r) . $

Conjecture: $\mathrm{CC} (\gamma (r)) = \mathrm{CC} (\beta (r)) = \mathrm{CC} (r)$.

If it is wrong I want to see the counter-examples with which it fails and additional condition under which it is indeed true.

In the above mentioned draft article I proved that $\mathrm{CC}(\beta(r)) \subseteq \mathrm{CC}(r) \subseteq \mathrm{CC}(\gamma(r))$.

  • 3
    $\begingroup$ You might increase the number of responses by giving some motivation (why generalize connectedness?) or by comparing familiar ideas (how is it a generalization of connectedness?) $\endgroup$ Jul 14, 2010 at 19:05
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    $\begingroup$ I voted to close as "too localized." It really seems way too specific with very little motivation. $\endgroup$ Jul 14, 2010 at 20:22
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    $\begingroup$ mathematics21.org/abel-prize.html $\endgroup$ Jul 14, 2010 at 21:31
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    $\begingroup$ @Harry: on this occasion I'd prefer us to play the ball, and not the man $\endgroup$
    – Yemon Choi
    Jul 14, 2010 at 21:32
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    $\begingroup$ Also, it seems to me you could at least explain your conception of what the relation r represents, since many people will not care to click on the link in order to find out. $\endgroup$
    – Dan Ramras
    Jul 14, 2010 at 22:04

2 Answers 2


First let me note that I agree with many of the comments and my answering the question should not be taken as disagreement. In particular, I find something off-putting about asking such a question on a topic about which you are writing a paper. It would be one thing if this were a mere curiosity (questions of the form "does such and such approach lead anywhere" come up a lot, and can be ok if phrased right). This seems more of the form "I don't feel like checking the details; could you prove my lemmas or give me counterexamples". Nonetheless I had some time and felt like playing around with some random idea which has nothing to do with the conference I'm currently at (on game theory). So I'm answering against my better judgement.

Define the disjointness relation $\delta$ by $A\delta B \Leftrightarrow A\cap B = \emptyset$. Here are some easy things to check:

1) If $r$ is a connector then so is $r\cap\delta$ and $CC(r) = CC(r\cap\delta)$. (Here I'm assuming "connector" means a relation which satisfies the monotonicity property you specified; that wasn't entirely clear)

2) $r\subseteq\gamma(r)$ and $\gamma(r)\cap\delta\subseteq r\cap\delta$, so $\gamma(r)\cap\delta = r\cap\delta$.

3) $CC(\gamma(r)) = CC(\gamma(r)\cap\delta)=CC(r\cap\delta)=CC(r)$

As for the question of $\beta(r)$, it should be noted that as you defined it (or, as I'm assuming you were meaning to define it) $\beta(r)$ is not a connector. Suppose it were. Then $A\in CC(r)$ implies $A\beta(r)\emptyset$. By the definition of a connector, $B\beta(r)\emptyset$ whenever $A\subseteq B$, so $B\in CC(r)$. That is to say, any superset of a connected set is connected. This is usually not true; say, if $r$ defines the usual connectedness of a discrete space. So it's not clear that it makes sense to ask about $CC(\beta(r))$, because as a set it may not have the properties you'd expect for a notion of connectedness.

  • $\begingroup$ To be a connector the monotonicity property (which I call extendability) is not necessary. I edited my question to clarify the difference between connector and extendable connector. $\endgroup$
    – porton
    Jul 15, 2010 at 16:09
  • $\begingroup$ Why $\gamma(r)\cap\delta\subseteq r\cap\delta$? It seems you mistake here. $\endgroup$
    – porton
    Jul 15, 2010 at 16:14
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    $\begingroup$ I was assuming what you are now calling extendability, because you hadn't clarified that they were intended to be separate definitions. So if you hadn't meant to imply that you can ignore my answer. $\endgroup$
    – Noah Stein
    Jul 15, 2010 at 16:20

From the definition of $\beta(r)$ it's obvios that $CC(\beta(r))=CC(r)$.

$CC(\beta(r))=\{A\subseteq U\ |\ \forall X,Y\subset A,\ X\not=\emptyset\not=Y \land X\cup Y=A \land X\cap Y=\emptyset$ $\implies X\cup Y=A\in CC(r)\}=\{A\subseteq U\ | \ A\in CC(r)\}$.

Now assume $A\in CC(\gamma(r))$ and $X,Y\subset A$ non empty such that $X\cup Y=A$ and $X\cap Y=\emptyset$, then $X\gamma(r) Y$. So $\exists N\subset X \land M\subset Y$ such that $NrM$. Now if $r$ is extendable, we have $XrY$. So we can conclude that $A\in CC(r)$.

Is this what you were asking?


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