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?

  • 2
    $\begingroup$ If $a_n$ is bounded below, by very definition the limit inferior is finite. The respective opposite holds for the $b_n$ sequence $\endgroup$
    – FShrike
    Aug 9, 2021 at 17:50
  • $\begingroup$ @FShrike: My guess is that OP wants to show that $\liminf a_n < +\infty$ and $ \limsup b_n > -\infty$. $\endgroup$
    – Martin R
    Aug 9, 2021 at 18:01
  • 3
    $\begingroup$ Be extremely careful with statements like $\infty-\infty$: any argument that relies on such statements is probably a flawed one $\endgroup$
    – FShrike
    Aug 9, 2021 at 18:12
  • $\begingroup$ I think trying by contradiction would be easier $\endgroup$
    – thewatcher
    Aug 9, 2021 at 18:19

1 Answer 1


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


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