# Limit of sequence implying the limit of convergent subsequences question

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!

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