Let $a_n \rightarrow A$ and $b_n \rightarrow B$ as $n \rightarrow \infty$, with $a_n \leq b_n$ for all $n$. Show $A \leq B$.

Question

This is my proof:

We have the sequence $a_n$ is convergent. Consider some arbitrary $b_k$. We have for all $n$ that $a_n \leq b_k$. By the Limit Location theorem, we have $\lim_{n\rightarrow \infty} a_n \leq b_k$, i.e. $A \leq b_k$.

Since $b_k$ is arbitrary, we can generalize this inequality to all $b_n$: namely, for all $n$, $b_n \geq A$. By the Limit Location Theorem, $\lim_{n \rightarrow \infty} b_n \geq A$, i.e. $B \geq A$. QED.

Is this correct?

Answer 1

No: we don't know that $a_n\leq b_k$ for all $n$. We only know that $a_n\leq b_n$ (in other words, that $a_n\leq b_k$ when $n=k$).

Answer 2

We may assume $A , B$ are real numbers .

Now , $$\forall n\in{\bf N}~,~~ b_{n}-a_{n}\geq0 \Longrightarrow\lim_{n\rightarrow\infty}(b_{n}-a_{n})\ge0.$$

But $$B-A=\lim_{n\rightarrow\infty}b_{n}-\lim_{n\rightarrow\infty}a_{n}\color{red}{=}\lim_{n\rightarrow\infty}(b_{n}-a_{n})\ge0$$,where the red line holds by the convergence of the two sequences $a_{n},b_{n}~,$ that is $\lim_{n\rightarrow\infty}a_{n}=A$ amd $\lim_{n\rightarrow\infty}b_{n}=B$ are both exist .

Answer 3

Hint by Paramanand:

Assume $A>B.$

1) For $\epsilon _1>0$ there is a $n_1$ such that for $n \ge n_1$:

$|a_n -A| \lt \epsilon_1 .$

2)For $\epsilon_2 >0$ there is a $n_2$ such that for $n \ge n_2:$

$|b_n-B| \lt \epsilon_2.$

Choose $\epsilon := \epsilon_1=\epsilon_2 = (A-B)/4$,

and $N = \max(n_1,n_2).$

For $n\ge N :$

$|a_n-A| \lt \epsilon$, and $|b_n-B| \lt \epsilon.$

Or:

$-\epsilon \lt a_n -A\lt \epsilon$, and

$-\epsilon \lt b_n -B \lt \epsilon.$

Putting together:

$A-\epsilon \lt a_n$, and $b_n \lt B + \epsilon.$

With $\epsilon =(A-B)/4:$

$a_n \gt A -(A-B)/4 =$

$3/4 A + B/4$ , and

$b_n \lt B+(A-B)/4 = $

$3/4 B + A/4,$ or finally

$ a_n \gt b_n$ , a contradiction .

Answer 4

It suffices to show that if $a_n \to a$ and $\forall n, \ a_n ≥ 0$ then $a ≥0$.

$\forall n, \ a_n ≥ 0$ i.e. $\forall n, \ a_n = |a_n|$ ; so by uniqueness of limit, $a = |a| ≥ 0$

Then consider $(b_n - a_n)$ ...