Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,d)$ be a metric space and consider an increasing sequence $A_n$ of its subsets such that $A = \bigcup_n A_n$ is compact. Can it happen that $A\setminus A_n$ is compact for all finite $n$?

share|improve this question
add comment

2 Answers 2

No. Even if we don't require that the $A_n$'s are increasing.

Assume otherwise. Since $X$ is a metric space, compact sets are closed. Note that $\{A\setminus A_n\mid n\in\Bbb N\}$ has the finite intersection property. Therefore the intersection of all of them must be non-empty. But this is impossible since $\bigcap (A\setminus A_n)=A\setminus\bigcup A_n=\varnothing$.

share|improve this answer
    
You forget the case that $A = A_1 \cup \cdots \cup A_n$ for some $n$. Note that the empty set is compact. –  Omnomnomnom Jul 7 at 16:17
1  
This seems to miss the intention of the question. I expect Ilya to be able and handle the case when the sequence is eventually constant. –  Asaf Karagila Jul 7 at 16:24
    
@AsafKaragila: thanks, though I expected myself to be able to handle also the case in the question, granted that I've applied finite intersection property a couple of times before facing that question. Too much work... –  Ilya Jul 7 at 20:02
add comment

If $A-A_n$ is compact then it is closed and so $A_n$ is a relative open subset of the space $A$. Now we have that $A_n$ is an open cover of $A$ (considered in the space $A$) if this is compact we have $A=A_n$ for some $n$ since the sets are increasing. If the sets are non increasing we still get that $A$ is a finite sum.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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