sequence, convergence

**poincare4223** · January 30th 2009, 07:04 AM

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

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

**poincare4223** · January 30th 2009, 09:10 AM

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

**ThePerfectHacker** · January 30th 2009, 10:25 AM

Originally Posted by poincare4223

(a) Prove that $(x_n)$ converges to a number $A$ iff every subsequence of $(x_n)$ has a subsequence which converges to $A$ .

If $\{x_n\}$ is convergent then all subsequences converge to the same limit. Thus, the forward direction follows.

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

. Contradiction.

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