# On the compactness of a certain chain topology [closed]

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.

$$I=[\{\emptyset,A_\lambda,X:A_\lambda\subset X,\lambda\in\Lambda\}]$$

such that $A_\alpha\subset A_\beta$ whenever $\alpha\le\beta$.

It is easy to show that $I$ qualifies as a topology on $X$.

under what condition this chain topology will be compact?

• What do you mean by "this chain topology"? The topology generated by the sets $A_\alpha$? What is the motivation for this question? Jun 16 '12 at 12:38
• The family $I$ has a least nonempty element. Jun 16 '12 at 12:47
• @Edgar...since $\Lambda$ is linearly ordered, there must always exist some $\alpha\in\Lambda$ such that $A_\alpha\subset A_\lambda$ for all $\lambda\in\Lambda$ Jun 16 '12 at 13:32
• Still no motivation; it looks like an exercise from a topology book. The space is compact iff there is a largest open set (excluding $X$ itself). Jun 16 '12 at 13:37
• Yes, I can prove it. This is certainly not a research-level question. Jun 16 '12 at 18:18

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.
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.
• If you don't assume that $I$ is a topology, then it is not correct to say that the topology generated by $I$ is compact if and only if $I$ has a largest proper subset of $X$, since perhaps the proper subsets of $X$ in $I$ union up to something strictly smaller than $X$, but this set does not appear in $I$. In this case, the topology would still be compact, but there wouldn't be a largest proper subset in $I$ (although there would be a largest proper subset in the generated topology). Jun 16 '12 at 17:35