# 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?

• If $a_n$ is bounded below, by very definition the limit inferior is finite. The respective opposite holds for the $b_n$ sequence Aug 9, 2021 at 17:50
• @FShrike: My guess is that OP wants to show that $\liminf a_n < +\infty$ and $\limsup b_n > -\infty$. Aug 9, 2021 at 18:01
• Be extremely careful with statements like $\infty-\infty$: any argument that relies on such statements is probably a flawed one Aug 9, 2021 at 18:12
• I think trying by contradiction would be easier Aug 9, 2021 at 18:19

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