boombaby
Nov15-08, 01:16 AM
1. The problem statement, all variables and given/known data
Let {si} and {ti} be bounded sequences of real numbers, let E, F, G be the sets of all subsequential limit points of {si}, {ti}, {siti} respectively. Prove that G\subseteq EF= \left\ ef|e\in E,f\in F \right\
2. Relevant equations
3. The attempt at a solution
I have trouble understanding the behavior of the sequence siti. 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?
Let {si} and {ti} be bounded sequences of real numbers, let E, F, G be the sets of all subsequential limit points of {si}, {ti}, {siti} respectively. Prove that G\subseteq EF= \left\ ef|e\in E,f\in F \right\
2. Relevant equations
3. The attempt at a solution
I have trouble understanding the behavior of the sequence siti. 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?