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?
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 $c_1 \geq \liminf a_{n}\geq c_2$.