1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hi, thanks for the help, but I am more interested in whether what I have written is a correct/valid proof. $\endgroup$ – Plus Twenty Mar 26 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.