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

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

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

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

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

  • 5
    $\begingroup$ Did you choose your tags by rolling some dice? $\endgroup$
    – Asaf Karagila
    Commented Mar 27, 2017 at 10:57
  • 3
    $\begingroup$ Please use \liminf and \limsup instead of $\lim_{n\to\infty} inf$ and $\lim_{n\to\infty} sup$, which are absurd. $\endgroup$
    – Did
    Commented Mar 27, 2017 at 11:00

1 Answer 1


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 ?)

  • $\begingroup$ Hi what is your $x_n$ and $y_n$? $\endgroup$ Commented Mar 24, 2021 at 14:03
  • $\begingroup$ @Negrawh well it is defined in the answer. But the bold statement hold for every sequences. $\endgroup$
    – Surb
    Commented Mar 24, 2021 at 19:11

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