# 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?

-

## closed as general reference by Andres Caicedo, Asaf Karagila, Dan Petersen, Andreas Blass, Chris GerigJun 16 '12 at 20:22

This question is too basic; it can be definitively and permanently answered by a single link to a standard internet reference source designed specifically to find that type of information. If this question can be reworded to fit the rules in the help center, please edit the question.

What do you mean by "this chain topology"? The topology generated by the sets $A_\alpha$? What is the motivation for this question? –  Goldstern Jun 16 '12 at 12:38
The family $I$ has a least nonempty element. –  Gerald Edgar 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 $\lambda\in\Lambda$ –  Kamran 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). –  Goldstern Jun 16 '12 at 13:37
Yes, I can prove it. This is certainly not a research-level question. –  Goldstern 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). –  Joel David Hamkins Jun 16 '12 at 17:35