Kleene wrote:Hi guys,

Thanks for your replies. I attempted to prove it by using some of your suggestions, but I am far from sure it is correct now. Suggestions/corrections are more than welcome! (Please forgive me for my awkward paraphrasing of my LaTeX code.)

Let (epsilon_n) be a sequence in R that converges to 0. Now construct a sequence (x_n) such that x_n is in the closure of B for all n in N and dist(x_n, A) = epsilon_n. Because (epsilon_n) converges to 0, (x_n) converges to 0 and there exists a limit x_n in the closure of B such that dist(x_n, A) = 0. Now it holds that x_n is in the closure of A and x_n is in B because B is closed. A is compact, hence sequentially compact. This means that every sequence in X has a convergent subsequence in A and therefore x_n in A. Conclusion: x_n is in the intersection of A and B which is non-empty.

- B is closed. All points in the closure of B are already contained in B. To choose points in the closure of B is redundant.

- There may not exist an x_n such that dist(x_n,A) = epsilon_n. Consider the possibility of B being a single point. You want to use "<" or "<=".

- Why does the sequence x_n converge to anything? All we know at this point is that it gets closer and closer to the set A. For all you know, each of the x_n is within epsilon_n of a different point of A, and may not converge to any point of A or any point at all. I'm confused at the point where you say the sequence x_n converges, because at that point, you don't use that A is compact at all. Let A be the integers (I know that isn't a compact set) and let be be the set {n+1/n | n \in Z } and let {x_k} = {k+1/k} where k is {1,2,3}... Clearly {x_k} is in B, but does not converge, nor does any subsequence converge. But it fulfills all of your criteria up to that point. Why can't I make a similar argument if A is compact?

- If x_n converged, why is it contained in the closure of A? I mean, if A was a single point, it would be compact, and you can easily construct a sequence x_n in B to converge to that point in A by the premises given, but x_n would not necessarily be in the closure of A, since A would be closed. Why does it matter that x_n is in the closure of A?

- Not every sequence in

X has a convergent subsequence in A. Let X be R, let A be [0,1]. The sequence {42,42,42,42,...} is not contained in A, nor is any of its subsequences. That A is sequentially compact means that every sequence in

A contains a convergent subsequence. But you haven't shown that x_n is in A, so that step makes no sense. And even if you have, that step is completely unnecessary, since it would be obvious from the fact that the sequences are contained in both sets that their intersection is non-empty.