MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let (δ;U) is a proximity space.

I will call a set A connected iff for every partition {X,Y} of the set A holds X δ Y.

I will call connected component a maximal connected set.

Is this true: U is partitioned into connected components?

share|cite|improve this question
I mistakenly wrote "connected sets" instead of "connected components". Now this error in the question is fixed. – porton Jun 14 '10 at 23:57
The notion of "proximity space" is not widely known in math. You might want to explain what a proximity space is in the body of your question. – André Henriques Sep 7 '10 at 13:57
See for a short account of proximity space. – Willie Wong Oct 5 '10 at 13:26

Say that $x\sim y$ for $x,y \in U$ if $\exists C\subset U$ such that $C$ is (proximally) connected and $x,y\in C$.

$\sim$ is reflexive since $x\in\{x\}$. It is symmetric by definition. That it is transitive follows from the fact that if $C,D$ are connected sets, and $C\cap D\neq \emptyset$, then $C\cup D$ is connected. (This follows because, for any partition $W,Z$ of $C\cup D$, we have $Z = C\setminus W \cup D \setminus W$, and that $(C\setminus W) \delta (W\cap C)$, ditto $D$, and $(W\cap C) \delta (W \cap (C\cap D)) \delta (W \cap D)$.)

So in any case, proximity induces a partition of $U$ since $\sim$ is an equivalence relationship.

If suffices to check that (a) the equivalence classes are connected and (b) each class is maximal.

Let $x_0 \in U$ be arbitrary, take its equivalence class $[x_0]$. By definition there exists some connected set $C_0\subset U$ such that $\{ y\sim x_0\} \subset C_0$. Conditions (a) and (b) then follows from the fact that for any $D\supset [x_0]$ we must have $D\setminus [x_0] = \emptyset$, which follows simply by observing that by definition there cannot by $y_0 \sim x_0$ while not belonging to $[x_0]$.

share|cite|improve this answer

It looks like a space is proximally connected iff every proximally continusous function from it to a discrete space is constant. So a family of proximally connected sets that all have a point in common, will have a proximally connected union, and the connected components of the points will partition the space.

In fact, we can say more than that - for any two distinct components X and Y, you'll have X and Y not close to each other, as otherwise $X \cup Y$ would still be proximally connected. (A proximally continuous function to a discrete space would be constant on X, constant on Y, and the two constants would be close and hence equal.) If there's only finitely many components that means the original space would decompose as the disjoint union of the components (as a proximity space); I don't actually know proximity spaces at all so I have no idea whether that still holds if there are infinitely many.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.