A question about indecomposable continua. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:23:04Z http://mathoverflow.net/feeds/question/37411 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37411/a-question-about-indecomposable-continua A question about indecomposable continua. Garabed Gulbenkian 2010-09-01T19:15:49Z 2013-05-01T05:57:13Z <p>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?</p> http://mathoverflow.net/questions/37411/a-question-about-indecomposable-continua/42091#42091 Answer by Jeff Norden for A question about indecomposable continua. Jeff Norden 2010-10-14T00:06:01Z 2010-10-14T00:06:01Z <p>"Indecomposable continuum" is a bit of a misnomer when you remove compactness. Standard indecomposable continua will <i>not</i> 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. <p> 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.</p> http://mathoverflow.net/questions/37411/a-question-about-indecomposable-continua/87012#87012 Answer by Joel Finegold for A question about indecomposable continua. Joel Finegold 2012-01-30T07:10:52Z 2012-01-30T07:10:52Z <p>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.)</p> http://mathoverflow.net/questions/37411/a-question-about-indecomposable-continua/111145#111145 Answer by Lasse Rempe-Gillen for A question about indecomposable continua. Lasse Rempe-Gillen 2012-11-01T12:01:50Z 2012-11-01T12:01:50Z <p>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:</p> <p>Is there a nontrivial connected metric space $X$ such that $X$ cannot be written as the union of two proper connected subsets?</p> <p>The answer, as Jeff suggested, is <strong>no</strong>.</p> <p>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.</p> <p>If $X$ does have a cut-point $x_0$, let $A$ and $B$ be open subsets of $X$ such that $$A\cap B = {x_0}; \quad A\setminus{x_0},B\setminus{x_0}\neq\emptyset \quad\text{and}\quad A\cup B = X.$$</p> <p>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).</p> <p>Regarding your question on the number of proper connected subsets, we can still ask the following question:</p> <p>If $X$ is any nontrivial connected metric space, what can be said about the cardinality of the set $S$ of proper connected subsets of $X$?</p> <p>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.</p> http://mathoverflow.net/questions/37411/a-question-about-indecomposable-continua/129284#129284 Answer by Joel Finegold for A question about indecomposable continua. Joel Finegold 2013-05-01T05:57:13Z 2013-05-01T05:57:13Z <p>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.</p> <p>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. </p> <p>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.</p> <p>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.</p> <p>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. </p>