0
$\begingroup$

Let $\{a_n\}$ be a bounded sequence. Prove that if every convergent subsequence of $\{a_n\}$ has limit $L$, then $\lim_{n\rightarrow\infty}a_n = L$.

Proof:

Suppose $(a_i)$ does not converge to $L$. Then for some $\epsilon$, for any $N$ there exists $n>N$ such that $|a_n-L|>\epsilon$. So I can find a subsequence $b_1,b_2,\ldots$ such that $|b_i-L|>\epsilon$ for all $i$. Since $(b_i)$ is bounded, by Bolzano-Weierstrass I can find a convergent subsequence $(c_i)$ of it. Clearly $(c_i)$ cannot have limit $L$, a contradiction.

My problem is, I don't understand why I can find a subsequence $b_1,b_2,\ldots$ such that $|b_i-L|>\epsilon$ for all $i$. Seems very trivial but I just can't get it. Any help would be greatly appreciated!

$\endgroup$
1
$\begingroup$

Let $(a_{n_k}) $ be a convergent subsequence of $(a_n)$ so that $a_{n_k} \to L $. Since $(a_{n_k}) $ is cauchy then for all $\epsilon > 0$, take $N$ such that if $n_k, n_l > N $, then $|a_{n_l} - a_{n_k} | < \epsilon/2 $. In particular, this must hold if $n_l = n $. So, using triangle inequality,

$$|a_n - L| \leq |a_n - a_{n_k} | + | a_{n_k} - L | < \epsilon / 2 + \epsilon / 2 = \epsilon $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.