most recent 30 from http://mathoverflow.net 2013-05-19T13:10:04Z http://mathoverflow.net/feeds/question/99783 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99783/on-the-compactness-of-a-certain-chain-topology Kamran 2012-06-16T12:26:23Z 2012-06-16T14:20:09Z

<p>Let $X$ be a non-empty set and $I$ a collection of some nested subsets of $X$ indexed by a linearly ordered set $(\Lambda,\le)$ such that $I$ always contains the void set $\emptyset$ and the whole set $X$, i.e.</p> <p>$$I=[{\emptyset,A_\lambda,X:A_\lambda\subset X,\lambda\in\Lambda}]$$</p> <p>such that $A_\alpha\subset A_\beta$ whenever $\alpha\le\beta$.</p> <p>It is easy to show that $I$ qualifies as a topology on $X$.</p> <p>under what condition this chain topology will be compact?</p>

http://mathoverflow.net/questions/99783/on-the-compactness-of-a-certain-chain-topology/99792#99792 Joel David Hamkins 2012-06-16T14:20:09Z 2012-06-16T14:20:09Z

<p>Your claim that those hypotheses ensure that $I$ is a topology is not correct. What you have is a family of subsets of $X$ that is linearly ordered by $\subset$ and includes the empty set and $X$ itself, and not every such family is a topology. For example, consider the family of intervals in the real line of the form $(-q,q)$, for $q\in\mathbb{Q}$, plus the empty set and all of $\mathbb{R}$. These intervals are nested in the sense you describe, but they do not form a topology, since this family is not closed under arbitrary unions.</p> <p>Meanwhile, if you have an actual topology that consists of a family of sets that is linearly ordered by $\subset$, then this topology is compact if and only if it contains a largest proper subset of $X$. If it does have such a set, then every open cover must contain the whole set $X$, since the union of all smaller sets does not cover the space. Conversely, if it does not have such a set, then the union of all the proper subsets of $X$ is $X$ itself, and so this will be an open cover with no finite subcover. </p>