Subsequencial limit set of a product sequence SiTi

[boombaby]

Homework Statement

Let {s_i} and {t_i} be bounded sequences of real numbers, let E, F, G be the sets of all subsequential limit points of {s_i}, {t_i}, {s_it_i} respectively. Prove that G\subseteq EF= \left\ ef|e\in E,f\in F \right\

Homework Equations

The Attempt at a Solution

I have trouble understanding the behavior of the sequence s_it_i. For this question, it seems that the right thing to do is to choose any x in G, show that x is in EF. Here's what I've done:
There is a subsequence s_{\alpha _{i}}t_{\alpha_{i}} that converges to x. Now, consider the sequences {s_{\alpha _{i}}} and {t_{\alpha _{i}}}. Being a sequence in the compact set (i.e. [-M,M] in R), some subsequence of {s_{\alpha _{i}}} converges to a point a, let A be the set of these subscripts. And some subsequence of {t_{\alpha _{i}}} converges to a point b, Let B be the set of these subscripts.
If A\cap B has infinitely many elements, then x=ab\in EF. But what if A\cap B is empty? I've no idea why this cannot happen. Any hint would be greatly appreciated! Is there any different way to prove it?

 

[Office_Shredder]

Don't work with both sequences at the same time.

s_{a_i}t_{a_i} converges to x. Then s_{a_i} has a convergent subsequence s_{a_{i_j}}. What does the sub-subsequence t_{a_{i_j}} do? Remember, you know s_{a_{i_j}}t_{a_{i_j}} converges to x also

 

[boombaby]

yea, I think I get it now.
if s_i (denotes the sub-subsequence...) ->a. a=0 is a trivial case. If a!=0, then the corresponding t_i converges to b such that ab=x, since
|a*t_i-ab|<=|a*t_i-s_i*t_i|+|s_i*t_i-x|<=t_i*|s_i-a|+|s_i*t_i-x|<=Me+e
Thanks a lot!