Constructive proof that limits preserve inequalities

Question

Suppose that $a_n \leq b_n$ for all $n \in \mathbb{N}$, where $a_n$ and $b_n$ are two convergent sequences. Is there a constructive proof that $\lim_{n \to \infty} a_n \leq \lim_{n \to \infty} b_n$? The proofs that I have seen (e.g. this one) are nonconstructive.

Answer 1

Let $a = \lim a_n, b= \lim b_n$. Choose $\varepsilon >0$ Choose $N$ such that $$n\ge N \Rightarrow |a_n -a| <\frac{\varepsilon}{2}, |b_n -b| <\frac{\varepsilon}{2} $$ Then $$ a \le a_n + \frac{\varepsilon}{2}\le b_n + \frac{\varepsilon}{2} \le b + \varepsilon$$ Since $\varepsilon$ has been chosen arbitrarily, it follows that $a\le b$

Answer 2

I'm assuming by "constructive" you mean not using proof by contradiction.

Let $a = \lim_{n \to \infty} a_n$ and $b = \lim_{n \to \infty} b_n$. Let $\varepsilon > 0$. Then there exists an $N \in \mathbb{N}$ such that $|a_n - a| < \varepsilon/2$ and $|b_n - b| < \varepsilon/2$. Thus,

$$ a- b = a - a_n - (b - b_n) + a_n - b_n < |a - a_n| + |b - b_n| + a_n - b_n < \varepsilon $$ Thus, $a \leq b$. Now, I'm assuming you've proven $a < b + \varepsilon$ for all $\varepsilon > 0$ implies $a \leq b$. But most people prove this fact by contradiction as well. So let's prove it without it.

Let $\eta = a - b$. Then $b + \eta = b + a - b = a$. So $a$ cannot be strictly less than $b + \eta$. This implies $\eta$ cannot be strictly positive. Thus, $\eta \leq 0$, which completes the proof. In the last statement I am assuming the law of the excluded middle, which, depending on your definition, makes the proof nonconstructive.

Answer 3

Given that $a_n\leq b_n$ for each $n$

Suppose $z_n=b_n-a_n\geq 0$

Let $\lim z_n<0$ if it diverges to $-\infty$ there are infintly points when $z_n<0$ which is not the case

In other case we will say if it converges to $l<0$ so we can pick some $\epsilon>0$ such that $l+\epsilon<0$ then $-\epsilon+l<z_n<\epsilon+l<0$ for $n\geq m$ again it is contradiction that $z_n<0$ so Hence

$\lim z_n\geq 0$ $\Rightarrow \lim a_n\leq \lim b_n$