Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this 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
1  
Yes, I can prove it. This is certainly not a research-level question. –  Goldstern Jun 16 '12 at 18:18
show 1 more comment

closed as general reference by Andres Caicedo, Asaf Karagila, Dan Petersen, Andreas Blass, Chris Gerig Jun 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.

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
    
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
    
@ Joel thanx for the answer.can you tell me the areas of topology where such chains could be found? any research paper? Any particular name that could be given to such a topology? –  Kamran Jun 19 '12 at 7:48
    
I have seen them used several times, but only as counterexamples, since a space in which the open sets are linearly ordered by inclusion are unusual in several respects. I am sorry that I don't have any specific reference. –  Joel David Hamkins Jun 19 '12 at 7:59
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.