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A bounded seuqence $x_n$ is convergent and converges to {$x$} if and only if every convergent subsequence {$x_{n_k}$} converges to x.

I did

For $\frac{\epsilon}{2}>0$ there exists a natural numberr $M$ such that $\mid x_n - x \mid < \frac{\epsilon}{2}$ for all $n \geq M$.

Let {$x_{n_k}$} be subsequenec of {$x_n$}.

$\mid x_{n_k} - x \mid = \mid x_{n_k} - x_n + x_n - x \mid \leq \mid x_{n_k} - x_n\mid + \mid x_n - x \mid < \mid x_{n_k} - x_n\mid + \frac{\epsilon}{2}$

That's all I did. I am not sure how to make $\mid x_{n_k} - x_n\mid< \frac{\epsilon}{2}$ so I can make $\mid x_{n_k} - x \mid <\epsilon$

user421044
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  • This is a famous problem. See, for instance, http://math.stackexchange.com/questions/776899/if-every-convergent-subsequence-converges-to-a-then-so-does-the-original-boun – Barry Smith Mar 10 '17 at 02:24

1 Answers1

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Theorem 1:

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

(i) If $\lim s_n$ is defined, then $\lim \inf s_n = \lim \sup s_n = \lim s_n$

(ii) If $\lim \inf s_n = \lim \sup s_n$, then $\lim s_n$ is defined, and $\lim \inf s_n = \lim \sup s_n = \lim s_n$

Theorem 2:

Let $(s_n)$ be any sequence of real numbers, and let S denote the set of subsequential limits of $(s_n)$ (the set of limits of all subsequecnes of $(s _n) $).

(i) S is nonempty

(ii) $\sup S = \lim \sup s_n$ and $\inf S = \lim \inf s_n$

Notice that combining the results of Theorem 1 with Theorem 2 imply the statement you're trying to prove. For if $\lim s_n $ if defined, then $\sup S = \inf S$, and therefore the set of subsequential limits has one element, $\lim s_n$ (in other words, all subsequecnes converge to $\lim s_n$). The other side of the if and only if statement is also an apparent consequence of these results.

infinitylord
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