Sequence has a convergent subsequence in R^n 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.
 A: $A_i$ 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.
A: First, it seems that proving $A_i$ is closed is unnecessary and most likely false. In the end all you need is to get the subsequential limit of the whole sequence. note by completeness of real numbers that each coordinate sequence has to converge to something but since you assumed $A$ itself was closed you know such a limit must be in $A$.
You do need to prove that they are bounded, but you wrote something a little funny. Try using the definition of bounded that uses magnitudes, I.e.  $|x|\leq M$ for all $x$ in $A$. Can you relate the magnitude with the individual coefficients?
