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

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?

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.
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.