Suppose $(a_n)^{n\to\infty}_{n=1}$ is a bounded sequence of real numbers. Prove that: $\liminf a_n \leq\limsup a_n$

Question

Suppose $(a_n)^{n\to\infty}_{n=1}$ is a bounded sequence of real numbers. Prove that:

$$\liminf a_n \leq\limsup a_n$$

This makes sense as $\inf a_n$ is the lowest bound of $a_n$ and $\sup a_n$ is the lowest upper bound of $a_n$ and if $a_n$ converges $\liminf a_n =\limsup a_n = \lim a_n$.

However, I am struggling to prove this. Could I do this by contradiction?

Suppose $$\liminf a_n >\limsup a_n$$

$\liminf a_n$ is the limit of $\inf \{a_k:k\ge n\}$ and is the lowest bound of $a_n$ and $\limsup a_n$ is the limit of $\sup \{a_k:k\ge n\}$ and is the lowest upper bound of $a_n$. By definition, if $\liminf a_n >\limsup a_n$, then $\liminf a_n$ is obviously not the lowest bound. Contradiction.

Is this a way to prove this?

Answer 1

Hint: You could simply note that $\inf(S)\le \sup(S)$ for any non-empty set of real numbers $S$ (make sure you can show this!), and inequalities are preserved under taking limits.

Answer 2

Hint : You can (show and) use the following equivalent definitions of $\liminf$ and $\limsup$ : the $\liminf$ is the smallest limit point, and the $\limsup$ is the greatest limit point.