Proof $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \sup a_n = \lim_{n \to \infty} \inf a_n$

Question

Let $(a_n)_{n \in \mathbb{N}}$ be a bounded set. I know that $(a_n)_{n \in \mathbb{N}}$ converges exactly when

$$\lim_{n \to \infty} \sup a_n = \lim_{n \to \infty} \inf a_n$$

But how can one prove that in this case the following is true:

$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \sup a_n = \lim_{n \to \infty} \inf a_n$$

Answer 1

$\lim (a_n)$ is defined.

Suppose $\lim(a_n)=\infty$. Let $M$ be a positive real number. Then there exists $M$ s.t. $$n>N\implies a_n>M$$

Then $u_N= \inf\{a_n:n>N\}\ge M$. It follows that for $m>N\implies u_m\ge M$ which means $u_N$ satisfies $\lim u_N=\infty$, then $\lim \inf a_n=\infty$

Likewise $\lim\sup a_n=\infty$. And the proof for $-\infty$ is similar.

Then, suppose $\lim a_n=a,a\in\mathbb{R}$, There exists $N$ s.t. $n>N\implies|a_n-a|<\epsilon\implies a_n<a+\epsilon\implies u_N=\sup\{a_n:n>N\}\le a+\epsilon$

$m>N\implies u_m\le a+\epsilon\implies \lim\sup a_n=\lim u_N\le a+\epsilon$

Since this is true for all $\epsilon>0$, we conclude that $\lim \sup a_n\le a$

Similarly, one can prove $\lim\inf a_n\ge a$.

Since $\lim\inf a_n\le\lim\sup a_n$, we infer that $\lim\inf a_n=\lim a_n=\lim\sup a_n$.

Answer 2

Basic idea: Assume that the limit exists and show that $$\liminf_{n\to\infty}a_n\leq \lim_{n\to \infty}a_n\leq \limsup_{n\to \infty}a_n = \liminf_{n\to\infty}a_n$$The general properties of $\leq$ will handle the rest.