# Homework Help: Proving an equivalence concerning sequences

1. Dec 1, 2011

### RVP91

given closed subsets, A and B, of R^d with A bounded prove the equivalence of:
1) There exists a pair of sequences x_n in A and y_n in B such that |x_n - y_n| -> 0 as n -> infinity
2) A intersection B is non empty.

I have attempted this question but am a bit stuck on proving 1 implies 2 and not sure about my attempt at 2 implies 1.

For 2 implies 1:

Let p be a point of intersection between A and B.
Then let x_n be a sequence in A that tends to p as n -> infinity.
Let y_n be a sequence in B that tends to p as n -> infinity.
Then we have that ||x_n - y_n|| -> 0 as n ->infinity.
Thus we have shown a pair of sequences exists.

For 1 implies 2:
I'm actually clueless as to where to start. I've been told the Bolzano-Weierstrass theorem may help. I know this says that each bounded sequence in R^d has a convergent subsequence. I don't know how to use this though or really where to begin at all.

Any help would be appreciated.

2. Dec 1, 2011

### Dick

x_n is a sequence in A. A is bounded. So x_n must be a bounded sequence in R^d. Now keep going.

3. Dec 1, 2011

### RVP91

x_n is a sequence in A. A is bounded. So x_n must be a bounded sequence in R^d.
Then using B-W we know a bounded sequence has a a convergent subsequence let us call it x_n_k and so we have x_n_k -> Q.
Now ||x_n - y_n|| -> 0 and ||x_n_k - Q|| -> 0 then the subsequence of y_n, y_n_k -> Q since
||y_n_k - Q||-> 0 = ||y_n_k - x_n_k + x_n_k - Q||-> 0 = ||y_n_k - x_n_k|| + ||x_n_k - Q|| -> 0
and we know ||x_n_k - Q|| tends to 0 and so ||y_n_k - x_n_k|| must tend to 0 ans this implies x_n_k = y_n_k and so y_n_k also tends to Q.
This implies the sets E and F have a intersection point and thus E intersection F is non empty.

Is this correct?
Apologies for the fact the proof above is probably rather messy.

Also was my proof for 2 implies 1 correct?

4. Dec 1, 2011

### Dick

Yes, the proof is written pretty sloppily. I'd try and clean it up. But you didn't say why Q is in either A or B! You've got the right idea, though. For 2->1 you can't just say that there is a sequence a_n->p. You'd better show that there is one.

5. Dec 1, 2011

### RVP91

Would the explanation of why Q is in A be because A is closed and bounded and so the sequence must tend to a point in A? And then since Q is in A and we have a sequence in B that converges to Q, the point Q must be in both A and B and hence A intersection B is non-empty? Also where I bought in a subsequence and used y_n_k in the proof would it have worked if I had just used y_n and showed that converges to Q?

How would I show there is a sequence a_n->p? Could you give me a hint please.

Thanks for the help s far and any future help, it is much appreciated!

6. Dec 1, 2011

### Dick

Closed sets contain all of their limit points, don't they? And {p,p,p,p,p,...} is a in A. What does it converge to?

7. Dec 2, 2011

### RVP91

Oh right, so Q must be in A since A is a closed set and so contains it's limit points?

And {p,p,p,p,...} is a constant sequence and so converges to p?

8. Dec 2, 2011

### Dick

Why the question marks?

9. Dec 2, 2011

### RVP91

Just wanted to make sure :)

Thanks for the help