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 $$-\infty \leq \liminf f_{k \to \infty} a_k \leq \limsup_{k \to
\infty} a_k \leq + \infty,$$ 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 $\liminf f_{k \to \infty} a_k \leq \limsup_{k \to
\infty} a_k$.
Consider $$a = \limsup_{k \to \infty} a_k = \lim_{j \to \infty} \{\sup_{k
\geq j} a_k\},$$ that is to say, $a$ is the largest number amount $a_k$s,
where $k \geq j$; but $$\liminf_{k \to \infty} a_k =\lim_{j \to \infty}
\{\inf_{k \geq j} a_k\}$$ is the smallest number amount $a_k$s under the
same constraint. Hence $a \leq \liminf_{k \to \infty}.$
I am again very not confident with this proof, because the $a_j, a_{j+1},
\dots$ sequence while $j$ approach to infinity.