# Prove $\liminf (a_n) \leq \limsup (a_n)$

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

$$\liminf (a_n) \leq \lim\sup (a_n)$$

(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|

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

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? Commented Mar 27, 2017 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
Commented Mar 27, 2017 at 11:00

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 ?)
• Hi what is your $x_n$ and $y_n$? Commented Mar 24, 2021 at 14:03