MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

The term "continuum" is often used to mean a compact and connected metric space. But it is also used in a broader sense to mean any infinite, complete, separable and connected metric space-which is not necessarily compact. This is the sense in which we use it here. A "continuum" is called "indecomposable" if it is not the union of two of its proper infinite subsets, each of which is itself a "continuum". It is known that a compact "indecomposable continuum" has uncountably many proper infinite subsets that are themselves "continua". If C is a non-compact "indecomposable continuum" and S is the set of all its proper infinite subsets that are themselves "continua", what can be said about the cardinal number of S?

share|cite|improve this question

"Indecomposable continuum" is a bit of a misnomer when you remove compactness. Standard indecomposable continua will not satisfy this condition. Such a continuum is the union of an uncountable pairwise disjoint collection of composants, where each composant is equal to the union of all the proper compact subcontinua containing a given point. If we let $A$ be one of these composants and $B$ be its complement, both $A$ and $B$ will be connected, so we have "decomposed" the space in sense you describe.

In fact, it isn't clear that "indecomposable continua" in the sense you have defined even exist at all. I suspect that they don't.

share|cite|improve this answer

As pointed out by Jeff, the notion you define may not really be what you are after, since indecomposable continua are not 'indecomposable' in your sense. However, we can ask:

Is there a nontrivial connected metric space $X$ such that $X$ cannot be written as the union of two proper connected subsets?

The answer, as Jeff suggested, is no.

Indeed, let $X$ be a nontrivial connected metric space. If $X$ does not have any cut-points, then clearly we can write $$X = (x\setminus\{x_0\}) \cup (X\setminus\{x_1\})$$ for some $x_0\neq x_1$, and are done.

If $X$ does have a cut-point $x_0$, let $A$ and $B$ be open subsets of $X$ such that $$A\cap B = \{x_0\}; \quad A\setminus\{x_0\},B\setminus\{x_0\}\neq\emptyset \quad\text{and}\quad A\cup B = X.$$

We claim that $A$ and $B$ are connected. Indeed, if $U\ni x_0$ is relatively open and closed in $A$, then $U\cup B$ is open and closed in $X$, so we must have $U=A$ (since $X$ is connected).

Regarding your question on the number of proper connected subsets, we can still ask the following question:

If $X$ is any nontrivial connected metric space, what can be said about the cardinality of the set $S$ of proper connected subsets of $X$?

It seems plausible that the set $S$ has at least the cardinality of the continuum, but I wasn't able to find a reference (and haven't thought very deeply about it). Certainly the set $S$ must be infinite.

share|cite|improve this answer

The above analysis is incorrect. $A$ is not necessarily a continuum. And if it is, its complement cannot be. There do exist non compact indecomposable continua. The ones I can construct are not metric. I think, however, there are such continua in any complete metric space that can be represented as the span of an infinite basis of itself, where each element of the basis is a simple curve (ie a locally compact continuum that contains, at most, one non cut point.)

share|cite|improve this answer
You are mistaken in your assessment of Jeff's answer. The composant A of a point is always connected, hence a 'continuum' in the sense of the original poster. Furthermore, the complement of a composant is always connected (see e.g. Nadler, Continuum Theory: An Introduction), hence a 'continuum' in the sense of the original poster. – Lasse Rempe-Gillen Nov 1 '12 at 8:44

A continuum is a closed and connected point set. If a composant of an indecomposable continuum is closed and connected, it's complement cannot be both closed and connected.

Proof: Suppose C is an indecomposable continuum and A is a composant of C. Since A is connected, A plus any Limit point of A is connected. Thus the closure of A is connected.

C reduced by A (C-A) is connected. Otherwise, C-A is the sum of two point sets, X and Y, having no point or limit point in common. The closure of X excludes all points of Y. Since A is connected, X plus A is connected. Sinc the closure of X lies within X plus A, X plus A is closed. By the same argument Y is closed and connected. X is not Y. Therefore X plus A and Y plus A would be two distinct closed and connected proper subsets of C whose sum would be C and thereby contradicting the supposition that C is indecomposable.

A must contain a limit point of C-A, otherwise C is not connected. So C-A is not closed and therefore cannot be a continuum.

Furthermore, suppose p is a point of A that is not a limit point of C-A. The closure of C-A would be a proper sub continuum of C because it would exclude a point of A. A is a proper sub continuum of C. Thus A and it's complement would be propersubcontinua of C whose sum would be C, contradicting the hypothesis that C is indecomposable Thus every point of a composant of an indecomposable continuum is limit point of its complement.

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.