Let $a_n \rightarrow A$ and $b_n \rightarrow B$ as $n \rightarrow \infty$, with $a_n \leq b_n$ for all $n$. Show $A \leq B$. 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?
 A: 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 $n=k$).
A: 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 .
A: 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 .
A: 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 $(b_n - a_n)$ ... 
