Prove that $\limsup_{n\to\infty}(a_n)\le \sup(a_n)$

Question

Let $a_n$ be a bounded sequence. Prove that $$\limsup_{n\to\infty}(a_n)\le \sup(a_n)$$

So I think that the best way to prove it is assume that $\limsup_{n\to\infty}(a_n)> \sup(a_n)$ and then find element of sequence > $\sup(a_n)$, that is wrong because it's $\sup(a_n)$.

But I don't really understand how to write it correctly in "math" language with $\epsilon$.

Answer 1

Say $\text{lim sup}(a_n) := R \in \mathbb{R}$, and assume $R > \text{sup}(a_n):= L$.

Take $\epsilon >0$ such that $R -L > \epsilon.$

That is, $L + \epsilon < R$. Because $R$ is a limit of a subsequence in $a_n$ there must be some $m \in \mathbb{N}$ such that $L + \epsilon < a_m \leq R$ - contradiction.

Answer 2

$a_n$ bounded;

$b_n=\sup ${ $a_k| k\ge n$ };

Note: $b_0= \sup $ { $a_k| k\ge 0$} $=\sup (a_n)$.

The sequence $b_n$ is monotonically decreasing,

$b_0 \ge b_1 \ge .....\ge b_n$;

Taking the limit:

$b_0 \ge \lim_{ n \rightarrow \infty} b_n$, hence

$\sup (a_n) \ge \lim_{n \rightarrow \infty} \sup (a_n)$,