actually use the following theorem twice: every subsequence of converges to ?
Is this correct?
Prove that if is a sequence such that every subsequence of has a further subsequence converging to , then .
Let be a strictly increasing function. Suppose that is a subsequence of . Then is a subsequence of . If , then there is an such that implies that .
From here, I have to show that .
Now what? . Could I use the Squeeze theorem?
Let be a sequence such that for any subsequence of we can find a subsequence which is convergent to zero. Assume that does not converge to zero, then it means for some there are some such that . The sequence is a subsequence of but it does not have any subsequence which is convergent to zero since all its terms are in absolute value. This is a contradiction. Thus, must be convergent to zero.
This is Mine 11th Post!!!