1. ## sequence, convergence

Let $\displaystyle (x_n)$ be a sequence of real numbers.

(a) Prove that $\displaystyle (x_n)$ converges to a number $\displaystyle A$ iff every subsequence of $\displaystyle (x_n)$ has a subsequence which converges to $\displaystyle A$.
(b) Does $\displaystyle (x_n)$ converge if every subsequence of $\displaystyle (x_n)$ has a subsequence which converges?

2. I figured out part b, but part a is confusing me.

3. Originally Posted by poincare4223
(a) Prove that $\displaystyle (x_n)$ converges to a number $\displaystyle A$ iff every subsequence of $\displaystyle (x_n)$ has a subsequence which converges to $\displaystyle A$.
If $\displaystyle \{x_n\}$ is convergent then all subsequences converge to the same limit. Thus, the forward direction follows.

To prove the backwards direction assume that $\displaystyle \{x_n\}$ [u]did not[/b] converge to $\displaystyle A$. Then it means there is some $\displaystyle \epsilon > 0$ so that for any $\displaystyle N>0$ we have $\displaystyle |x_n - A| \geq \epsilon$ for some $\displaystyle n>N$. Thus, we can form a subsequence, $\displaystyle \{x_{n_k}\}$ with $\displaystyle |x_{n_k} - A|\geq \epsilon$. But this subsequence cannot possible have a subsequence converging to $\displaystyle A$ . Contradiction.