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I couldn't solve this problem. Any help please?

Let $X$ be a metric space and $E_{i}\subset X$ for $i=1,2,...$ be nonempty compact sets such that $E_{1}\supset E_{2}\supset E_{3}\supset ...$ Prove that $\bigcap_{i=1}^{\infty }E_{i}\neq \varnothing $ and give an example that the claim is not true if the sets $E_{i}$ are not compact.

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I'm pretty sure there's no need for $X$ to be a metric space in this problem. It's just a special case of the Finite Intersection Property. –  Dustan Levenstein Dec 12 '11 at 2:51
    
@Dustan Please write it as an answer so that we can upvote it. –  Srivatsan Dec 12 '11 at 2:56
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@Dustan: In the usual formulation of compactness in terms of the finite intersection property, the sets in question are assumed to be closed. In general topological spaces, compact sets need not be closed. Perhaps you want to consider Hausdorff spaces? –  Jonas Meyer Dec 12 '11 at 3:00
    
@JonasMeyer Good point. –  Dustan Levenstein Dec 12 '11 at 3:03

2 Answers 2

up vote 4 down vote accepted

To give an example where it is not true if the sets are not compact, you can do it easily in $\mathbb{R}$ with sets that "run away to infinity"; that is, sets that get increasingly further away from $0$ with higher $n$.

For the first part, it's pretty easy to argue by contradiction or contrapositive: if $\cap_{i=1}^{\infty}E_i=\emptyset$, then $E_1\subseteq X = X - (\cap_{i=1}^{\infty}E_i) = \cup_{i=1}^{\infty}(X-E_i)$. That means that the sets $X-E_i$ are an open cover (since each $E_i$ is closed) of $E_1$. But $E_1$ is compact, so....

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@ Arturo Magidini Ok, I understand your proof, but I still don't know where you want to go after that in order to reach contradiction? Do you want to say that we can't find a finite subcover for $E1$ –  M.Krov Dec 12 '11 at 4:56
    
If you don't know where I want to go, then I would say you do not understand the argument (at least, you do not see how it ends; to me, "I understand your proof" means you understand it fully, which you are saying you don't). If you have a finite subcover, what does that mean? Think about the complement of that finite union. Then remember your hypothesis about the $E_i$. –  Arturo Magidin Dec 12 '11 at 5:03
    
Magidini: if we have a finite subcover, this means $E_{1} $ is compact. Our hypothesis about each $E_{i} $ is that they are compact. I can't figure out how to use your last hint 'Think about the complement of that finite union'? –  M.Krov Dec 12 '11 at 5:27
    
No; we know there is a finite subcover because $E_1$ is compact. But a finite subcover of this cover is of the form $X-E_{i_1}$, $X-E_{i_2},\ldots,X-E_{i_n}$, $i_1\leq\cdots\leq i_n$. Since $E_{i_1}\supseteq E_{i_2}\supseteq\cdots\supseteq E_{i_n}$, then $X-E_{i_1}\subseteq X-E_{i_2}\subseteq\cdots\subseteq X-E_{i_n}$, so $E_1\subseteq \cup_{j=1}^n(X-E_{i_j}) = X-E_{i_n}$. Is that possible? –  Arturo Magidin Dec 12 '11 at 5:29
    
I think since $E_{in}$ is subset of $E_{1}$, then this is not possible unless $E_{in}$ is empty. Is that right? –  M.Krov Dec 12 '11 at 5:36

Here is a useful proof idea/outline for you.

Suppose, $\bigcap_{i=1}^{\infty}E_i = \emptyset$

Then choose $x_1\in E_1$ for instance. Then we know that $x_1\notin E_k$ fo some $k$, because otherwise the intersection would not be empty. So $x\in E_{k_1}^c$ for some $k_1$. We can do this for each $x_1\in E_1$ so we see that we can find an cover for $E_1$

This open cover is: $\{E_{k_i}^c\}$. Notice, although I wrote it like it is countable it need not be.

But since $E_1$ is compact we can find a finite subcover $E_{k_1}^c, \cdots, E_{k_n}^c$. (why can we do this? i.e. how do we know that we had an open cover?)

But then this means that $E_1 \subset \bigcup_{i=1}^{n}E_{k_i}^c = (\bigcap_{i=1}^n E_{k_i})^c$ (why?)

Which implies that $E_1\cap \bigcap_{i=1}^nE_{k_i} = \emptyset$.

However, by assumption we have that $E_k \supset E_{k+1}$ for all $k$: Use this fact to obtain a contradiction (Hint: Consider the assumption that the sets were non-empty)

Also, for the counterexample part. How about we replace "compact" by either closed or bounded and find counterexamples.

If we replace it with "closed": I suggest considering something which goes of to infinity

and "bounded": I suggest considering something which does not contain its limit points (i.e is not closed)

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