To prove $a \leq b$

Question

. Given $\lim_{a_n} = a, \lim{b_n} = b$ Given two sequences such that $a_n \leq b_n$ for all $n \in \mathbb{N}$, then $a \leq b$

Assume : $a > b$ Take $$\epsilon = \frac{a-b}{4}$$. so there exists $N_1$ and $N_2$ such that so then $$a_n \in (\frac{3a+4}{4}, a+\epsilon), b_n \in (b_n-\epsilon, \frac{a+3b}{4})$$

Take $N = \operatorname{max} (N_1, N_2)$ so that we have $a_N \geq b_N$. but this is contradiction.

I am not sure if this is correct way to do this. only idea i have is to take disjoint intervels

Answer 1

Response given per the specific request of the OP (i.e. original poster).

Assume $~a > b,~$ and then derive a contradiction.

Let $~\epsilon~$ denote $~\dfrac{a-b}{4} \implies \epsilon > 0.$

Then, per the constraints of the problem:

• $\exists ~N_1 \in \Bbb{Z^+},~$ such that
  for all $~n \geq N_1, ~a_n \in (a - \epsilon, a + \epsilon).$

• $\exists ~N_2 \in \Bbb{Z^+},~$ such that
  for all $~n \geq N_2, ~b_n \in (b - \epsilon, b + \epsilon).$

Set $~n = \max(N_1,N_2).$

This implies that :

• $b_n < b + \epsilon.$

• $a - \epsilon < a_n.$

• Also, per the definition of $~\epsilon,~$ $(b + \epsilon) < (a - \epsilon).$

So, you can conclude that

$$b_n < (b + \epsilon) < (a - \epsilon) < a_n.$$

This contradicts the constraint that for all $~n, ~a_n \leq b_n.$

Answer 2

It suffices to show that $a < b + \epsilon$ for all $\epsilon > 0$. Fix $\epsilon$, and choose $N_1$ and $N_2$ such that $$ n > N_1 \implies |a_n - a| < \frac{\epsilon}{2}, \; n > N_2 \implies |b_n - b| < \frac{\epsilon}{2}. $$ Let $N := \max(N_1, N_2)$. For any $n > N$, we have $|a_n - a| < \frac{\epsilon}{2}$ and $|b_n - b| < \frac{\epsilon}{2}$. In particular, $$ a - \frac{\epsilon}{2} < a_n \leq b_n < b + \frac{\epsilon}{2}, $$ i.e., $a - \frac{\epsilon}{2} < b + \frac{\epsilon}{2}$, hence $a < b + \epsilon$, as claimed.