# Prove $\lim_{n\to\infty} inf (a_n) \leq \lim_{n\to\infty} sup (a_n)$

I've been working on this for a bit and haven't got far :/

So the full question is:

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

$\lim_{n\to\infty} inf (a_n) \leq \lim_{n\to\infty} sup (a_n)$

(Hint: apply the result of the previous question to a difference of two sequences.)

So the previous question was basically: $(a_n)_{n=1}^{\infty}$ is a convergent sequence and $a_n \in [0, \infty) \forall n$. Prove the limit lies in $[0, \infty)$.

How I solved it was fixed $\epsilon > 0$ then $\exists$ $N \in \mathbb{N}$ $\forall$ $n\geq N$ I let the limit = a $\therefore$ $|a_n - a| < \epsilon$ Since $a_n$ took finitely many values $\implies$ $M_1 = max{|a_1|,...,|a_{N-1}|}$, let $M_2 = |a| + \epsilon$ Since it converges it's bounded by some M $\implies$ $(a_n)^{\infty}_{n=1}$ within $[0, M]$ Since $[0, M]$ is closed, it contained its limit points $\implies$ $\lim_{n\to\infty} a_n \in [0,M] \subset [0, \infty)$.

So far I've said;

$(a_n)_{n=1}^{\infty}$ $\in$ $[\mathbb{R}]$

$M_1 =$ inf{$a_n$}, $M_2 =$ sup{$a_n$} (From previous question)

$\therefore$ sup{$a_n$} = $|a_n|$ + $\epsilon$

inf{$a_n$} = max{$a_1, a_2, ..., a_{N-1}$}

$|a_n|<M_1 \implies |a_n| <$inf{$a_n$}

$|a_N| < |a| + \epsilon$ $\implies$ $|a_N|<$ inf{$a_n$}

That's all I have not sure if it actually means anything but yeah as I said any help would GREATLY be appreciated. Thanks ;)

• Did you choose your tags by rolling some dice? – Asaf Karagila Mar 27 '17 at 10:57
• Please use \liminf and \limsup instead of $\lim_{n\to\infty} inf$ and $\lim_{n\to\infty} sup$, which are absurd. – Did Mar 27 '17 at 11:00

## 1 Answer

If you know that $x_n\leq y_n$ for all $n$ implies that $\lim_{n\to \infty }x_n\leq \lim_{n\to \infty }y_n$ :

Let $x_n=\inf_{k\geq n}x_k$ and $y_n=\sup_{k\geq n}x_k$. In particular, $x_n\leq y_n$ for all $n$, and thus $$\lim_{n\to \infty }x_n\leq \lim_{n\to \infty }y_n.$$ The claim follow.

If you don't know that $x_n\leq y_n$ for all $n$ implies that $\lim_{n\to \infty }x_n\leq \lim_{n\to \infty }y_n$, prove it !

Hint : It's enough to prove that $z_n\geq 0$ for all $n$ implies that $\lim_{n\to \infty }z_n\geq 0$ (why ?)