Show that $\liminf a_n$ and $\limsup b_n$ are finite Suppose $(a_n), (b_n)$ are sequences of real numbers where

*

*$a_n \geq b_n$ for all $n$

*$(a_n)$ is bounded below, $(b_n)$ is bounded above.

*$\liminf\, (a_n - b_n) = 0$
With these assumptions, I am trying to show that $\liminf a_n$ and $\limsup b_n$ are finite.
Here's what I have done:
Since $(a_n)$ and $(b_n)$ are bounded below and above, respectively, $\liminf a_n > -\infty$ and $\limsup b_n < \infty$.
Then $\liminf a_n - \limsup b_n$ is not equal to $\infty -\infty$, $-\infty - (-\infty)$, and so $$\liminf\, (a_n - b_n) \geq \liminf a_n - \limsup b_n$$
holds.
Then $0 \geq \liminf a_n - \limsup b_n$, and so $\liminf a_n < \infty$, $\limsup b_n > -\infty$; otherwise, we get $0 \geq \infty$.  $\square$
Is this a valid argument?
 A: You are on the right track. To make things clear, I will go really slow here...
First since $(a_{n})$ is bounded below then there exists $c_{1}\in \mathbb{R}$
such that
$$a_{n}\geq c_1,\quad n\geq 1.$$
Ans since $\inf_{k\geq n} a_{k}\geq \inf_{n\geq 1} a_{n}$ then
$$\inf_{k\geq n} a_{k}\geq c_1,\quad n\geq 1.$$
Therefore
$$\sup_{n\geq 1} \inf_{k\geq n} a_{k}\geq c_1,\quad n\geq 1.$$
This shows that
$$\liminf a_{n} \geq c_1\qquad (1).$$
Similarly there exists $c_{2}\in \mathbb{R}$ such that
$$\limsup b_{n} \leq c_2\qquad (2).$$
Now, since $\liminf (a_n-b_{n})=0$ then
$$0=\sup_{n\geq 1} \inf_{k\geq n}(a_k-b_k)\geq \inf_{k\geq n}(a_k-b_k)
= \inf_{k\geq n}a_k-\sup_{k\geq n}b_k.$$
So
$$\sup_{k\geq n}b_k\geq  \inf_{k\geq n}a_k$$
which implies
$$\inf_{n\geq 1}\sup_{k\geq n}b_k\geq  \inf_{k\geq n}a_k.$$
Thus
$$\inf_{n\geq 1}\sup_{k\geq n}b_k\geq \sup_{n\geq 1} \inf_{k\geq n}a_k$$
that is
$$\limsup b_{n}\leq \liminf a_{n}$$
We finally deduce that
$$0\geq \liminf a_{n}-\limsup b_{n}\geq c_1-\limsup b_{n}$$
Hence
$c_2 \geq \limsup b_{n}\geq c_1$.
Analogously
$c_1 \geq \liminf a_{n}\geq c_2$.
