Supremum of a bounded sequence is accumulation point.

Question

Let $(a_n)_{n \in \mathbb{N}}$ be a bounded sequence in $\mathbb{R}$. Denote by $L := \sup\{a_n : n \in \mathbb{N}\}$ the Supremum. Suppose that $L \notin \{a_n : n \in \mathbb{N}\}$. How can I show that $L$ is an accumulation point,i.e., I have to show hat there is a subsequence $(a_{n_{k}})_{k \in \mathbb{N}}$ of $(a_n)_{n \in \mathbb{N}}$ such that $$ \lim\limits_{k \to \infty} a_{n_{k}} = L. $$ In other words $$ \limsup\limits_{n \to \infty} a_n = \sup\{a_n : n \in \mathbb{N}\}. $$ My idea was to give a proof by contradiction. But this seems hopeless. First I have to show that for $\varepsilon > 0$ there exist a $m \in \mathbb{N}$ such that for all $k > m$ we have $$ |a_{n_{k}} - L| < \varepsilon. $$ This would show me that there exists an infinite set of terms of $(a_n)_{n \in \mathbb{N}}$ which satisfy $|a_n - L| < \varepsilon$. How can I prove this? Then I think it is not difficult to construct a subsequence which converges to $L$. I only need help for the first part.

Answer 1

Suppose $$ a_0 = \limsup_{n \to \infty} a_n = \lim_{k \to \infty} \sup_{n \geq k} a_n \neq L.$$ When viewing $\tilde{a}_k = \sup_{n \geq k} a_n$, we find that $\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \tilde{a_n}$. Then, $a_0 < L$ since $\{a_i\}$ is bounded above by $L$ so if we take $\varepsilon = |a_0 - L|$, then there is a natural number $N$ so that we have that $|\tilde{a}_N - a_0| < \varepsilon.$ This then gives $$|\tilde{a}_N - L| = |(L - a_0) - (\tilde{a}_N - a_0)| \geq |a_0 - L| - |\tilde{a}_N - a_0| > 0.$$ That means that $\sup_{k \geq N} a_k < L$. Since $$L = \sup_{n \geq 1} a_n = \max\left(\sup_{k \geq N} a_k, \max_{k \leq N} a_k\right)$$ but $\sup_{k \geq N} a_k \neq L$, it must be the case that $L = \max\{a_1, \dots, a_N\}$ which implies that $L = a_j$ for some $j \leq N$. This proves the claim by contrapositive.