# Prove $-\infty \leq \liminf f_{k \to \infty} a_k \leq \limsup_{k \to \infty} a_k \leq + \infty$.

If $\{a_k\}_{k=1}^\infty$ is a sequence of points in $\mathbb{R}^1$, let $b_j = \sup_{k \geq j} a_k$ and $c_j = \inf_{k \geq j} a_k, j = 1,2,\dots$ Prove $$-\infty \leq \liminf_{k \to \infty} a_k \leq \limsup_{k \to \infty} a_k \leq + \infty,$$ where $$\limsup_{k \to \infty} a_k = \lim_{j \to \infty} b_j = \lim_{j \to \infty} \{\sup_{k \geq j} a_k\}$$ $$\liminf_{k \to \infty} a_k = \lim_{j \to \infty} c_j = \lim_{j \to \infty} \{\inf_{k \geq j} a_k\}.$$

Since $-\infty \leq \liminf f_{k \to \infty} a_k$ and $\limsup_{k \to \infty} a_k \leq + \infty$ is assumed by the property of infinity, I am going to show that $\liminf f_{k \to \infty} a_k \leq \limsup_{k \to \infty} a_k$.

According to the definition, $b_j = \sup_{k \geq j} a_k$ so $b_j$ is the supremum (least upper bound) for $a_k$, where $k \geq j$. Similarly, $c_j = \inf_{k \geq j} a_k$ is the infimum (greatest lower bound) for $a_k$. Hence $b_j \geq c_j$. This is true for all $j \to \infty$, therefore,

$$\lim_{j \to \infty} b_j \geq \lim_{j \to \infty} c_j.$$

Hence $$\liminf_{k \to \infty} a_k \leq \limsup_{k \to \infty} a_k.$$

• You seem to have a stranded $f$ in "$\liminf f$".
– Pedro
Aug 21, 2013 at 22:52
• haha, my finger was shaking. Too much AC... :-( Thank you @PeterTamaroff Aug 21, 2013 at 22:56

My favorite way to think about $\limsup$ for instance is that $\limsup_k a_k \leq C$ if and only if for every $\epsilon>0$, the $a_k$ "eventually" fall below $C+\epsilon$. This can be used to prove, but I guess there is something simpler for you.
That being said, your proof is basically right. You are saying that $\inf_{k>n} a_k \leq \sup_{k > n} a_k$, and taking limits of both sides as $n$ tends to $\infty$ preserves the inequality and gives you what you desire.
By the way, you might want to prove this statement rigorously using the definition of inf and sup, you should not say that $a$ is the "largest" satisfying that condition, since that is not quite right (there may be no largest, which is why least upper bound is used instead).
(Whoops, on second thought I should have just left a hint asking you to compare $\inf_{k>n} a_k$ and $\sup_{k>n} a_k$ first.)