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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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    @JonasMeyer Good point.2011-12-12

2 Answers 2

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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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    Then go over the argument on your own, carefully, one step at a time, and check it. Write it in your own words, try to break it apart as best you can, until you are convinced it is correct. Me telling you isn't going to help with the next problem, and in *this* case, I already essentially gave you everything except the last step; allow me to expect you to take that last step yourself.2011-12-12
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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)