Suppose A is a closed and bounded subset of R^n. Let {ak} be a sequence in A. Thus, the elements of {ak} are:
(a11,a12,…,a1n),
(a21,a22,…,a2n),
…
…
(ak1,ak2,…,akn),
…
We are not sure if this sequence is finite or infinite, right? The problem asks to prove that every sequence has a convergent subsequence with limit in A.
Now, here is my approach but I am not sure if it is correct. Any suggestions would be very much appreciated.
Let Ai be the set of the ith coordinates of all elements of A with i=1,2,..,n. First, I am trying to prove that Ai is closed. Suppose mi is a limit point/an accumulation point of Ai. Then is it true that the point m=(m1,m2,…,mn) is a limit point/accumulation point of A? If this is true, then m has to be an element of A since A is closed. Then, each mi is an element of Ai. Then Ai is closed. <– for some reasons I feel there is something wrong with my argument here?
Next, I am going to prove that each Ai is bounded. The problem assumes that A is bounded, which implies that there exist p and q such that p =< a =< q with all a in A. Then, for each element ai of Ai, pj =< aij =< qj for j=1,2,..,n and i=1,2,..,n. Hence, each Ai is bounded. <– is this step true?
Now,since each Ai is bounded, by the Bolzano–Weierstrass theorem, each sequence in Ai has a convergent subsequence. Go back to our sequence {ak}. The sequence of the first coordinates of the elements of {ak} has a convergent subsequence {ak1_l1} that converges to d1. Since A1 is closed, d1 is in A1. Now, consider the sequence {ak_l1} in A. Note that {ak_l1} is a subsequence of {ak}. The sequence of the second coordinates of the elements of this sequence has a convergent subsequence that converges to d2 and d2 is in A2…now consider the subsequence {ak_l1_l2} of {ak_l1)…same argument with the third coordinated…d3 is in A3….now we have the subsequence {ak_l1_l2_…._ln}. With the same argument, this subsequence of {ak} has a convergent subsequence with a limit point in A…Hence {ak} has a convergent subsequence with a limit point in A. <– Is there any way we can make this more concise?
Thank you so much for your help. I would be very sincerely grateful.
Best Answer
$A_i$ is closed simply because it is a projection of a compact set, but the compactness is the very thing you're trying to prove, so you can't use it. If $A$ was not bounded, just closed, $A_i$ wouldn't necessarily be closed, or even Borel. Your argument is not correct, as you don't have $m_j$ for $j\neq i$.
The likely simplest proof of the fact you're trying to prove is to notice that if $A$ is closed and bounded, then it is a closed subset of a cube $[-M,M]^n$ for large enough $M$, so you can assume that $A$ is, in fact, a cube. The rest should be straightforward.