# Given that $\{a_n\}$ is a sequence of $\mathbb{R}$ s.t. $a_n \leq b_n$ and $a_n \rightarrow a$ and $b_n \rightarrow b$ then $a \leq b$

Given that $$\{a_n\}$$ is a sequence of $$\mathbb{R}$$ s.t. $$a_n \leq b_n$$ and $$a_n \rightarrow a$$ and $$b_n \rightarrow b$$ then $$a \leq b$$

Proof Verification:

I feel I'm one step away, but can't find the right $$\epsilon$$

let $$\epsilon > 0$$, and let $$b < a$$. Given the convergence of $$a_n$$ and $$b_n$$ this means:

$$\forall \ \epsilon >0,\ \exists \ N_1\in \mathbb{N} \ s.t.\ \forall \ n \geq N_1 \ |a_n - a| < \epsilon$$ AND $$\forall \ \epsilon >0,\ \exists \ N_2\in \mathbb{N} \ s.t.\ \forall \ n \geq N_2 \ |b_n - b| < \epsilon$$

Consider $$\epsilon = \frac{a-b}{2}$$. Then:

$$b < \frac{a}{2} + \frac{b}{2} < a_n < \frac{3a}{2} - \frac{b}{2}$$ AND $$\frac{3b}{2} - \frac{a}{2} < b_n < \frac{a}{2} + \frac{b}{2} < a$$

I could feel it in my soul that I have to subtract equation-2 from equation-1 and somehow manage to get a zero on one side which would show $$b_n < a_n$$ which would create a contradiction. But when I tried this all I got was: $$b-a < b_n - a_n < b-a$$ This doesn't help me because all it establishes is that $$a_n \rightarrow a$$ and $$b_n \rightarrow b$$. Which I already assumed.......What is the $$\epsilon$$ or the inequality I should try to get?

• Maybe you mean to prove $a \leq b$? The statement is false otherwise. – Hugo Oct 20 '18 at 18:34
• Not quite true. It should be "...then $a\leq b$." Not a strict inequality. – Mankind Oct 20 '18 at 18:34
• fixed @Hugo.... – dc3rd Oct 20 '18 at 18:37

If you want to prove an inequality in real analysis there are two main ways of doing that.

First, we could show that $$a \le b + \varepsilon$$ for every $$\varepsilon > 0$$.

The quantity $$b + \varepsilon$$ should remind you of the interval $$(b - \varepsilon, b + \varepsilon)$$. We know that $$b_n \to b$$ means that for every $$n \ge$$ some $$N$$, we have $$b_n \in (b - \varepsilon, b + \varepsilon)$$. So in particular, $$b_n \le b + \varepsilon$$ and therefore $$a_n \le b_n \le b + \varepsilon$$ (this is for $$n \ge N$$).

This is a step in the right direction because we have replaced $$a_n \le b_n$$ with $$a_n \le c$$ where $$c = b + \varepsilon$$ is constant which, at least in principle, is simpler.

Second, we could argue by contradiction, as you have done.

Suppose that $$a > b$$, which one should think of as $$a = b + \varepsilon$$ where $$\varepsilon = a - b > 0$$ and proceed as before. We know that for $$n \ge N$$ we have $$b_n \in (b - \varepsilon, b + \varepsilon)$$ so $$b_n \le b + \varepsilon = a$$ and hence $$a_n \le b_n \le a$$.

You will notice that this isn't particularly helpful. So we try again. What happens if we make $$\varepsilon$$ smaller, say to $$\varepsilon = \frac{a - b}{2}$$? Well, then we have

$$a_n \le b_n \le b + \varepsilon = b + \frac{a - b}{2} \tag{*}$$

Again we have something of the form $$a_n \le$$ some constant. So if we appeal to the theorem that says that if $$a_n \le c$$ (a constant) and $$a_n \to a$$ then $$a \le c$$ we have from $$(*)$$:

$$a \le b + \frac{a - b}{2} < b + (a - b) = a,$$

As an exercise, you might like to finish the first method as well.

• Should the $\epsilon$ in the second form be $\epsilon = a - b$? – dc3rd Oct 20 '18 at 18:57
• @dc3rd Yes, thank you! – Trevor Gunn Oct 20 '18 at 18:58
• Question....what if we don't have this theorem to appeal to? I may be mistaken, but doesn't that theorem follow from the one I am proving here or is it the other way around? – dc3rd Oct 20 '18 at 19:01
• @dc3rd It does follow but what I am suggesting is that it can be used to prove this as well. Because if the sequence $b_n$ is constant then you are only working with one sequence and it becomes just that much easier. For instance see what Mark has written. You need to have an $N_1$ for $a_n$ and an $N_2$ for $b_n$ and you have to realize to take the same $\varepsilon$ for both. If you can eliminate one of the sequences from the picture, it doesn't necessarily immediately lead to a solution but the hope is that it becomes an easier problem to think about. – Trevor Gunn Oct 20 '18 at 19:06
• @dc3rd As a hint for the $a_n \le c$ theorem. Consider $c < a$ and take an interval $(a - \varepsilon, a + \varepsilon)$ that lies entirely above $c$. I.e. $c < a - \varepsilon$. – Trevor Gunn Oct 20 '18 at 19:07

First of all, the statement as it was written at the beginning is not true because you can take $$a_n$$ and $$b_n$$ be the same sequence as a counterexample. Perhaps you mean $$a\leq b$$, then I can give an easy proof. Let $$\epsilon>0$$. There exists a large enough $$n$$ which satisfies both $$a-\frac{\epsilon}{2} and $$b_n. For this specific $$n$$ we have:

$$a

So $$a. This is true for any $$\epsilon>0$$. But if we assume $$a>b$$ then we get it isn't true for $$\epsilon=\frac{a-b}{2}$$ which is a contradiction. Hence $$a\leq b$$.

Suppose not! If $$a\not\le b$$ then $$b so $$0. Let $$c=a-b$$, then $$c>0$$. Then $$c=\lim_{n\to \infty} a_n - \lim_{n\to \infty} b_n = \lim_{n\to \infty} (a_n-b_n) = \lim_{n\to \infty} c_n$$

Now by the definition of "limit", it follows that for

for all $$\varepsilon>0$$, there exists $$N$$ such that for all $$n\ge N$$, we have $$|c_n-c|<\varepsilon$$

## Why does this fail?

Well... $$c>0$$ is always positive and $$c_n$$ is always negative. Can you finish it from there?

Original answer (for $$a case):

Let $$a_n=b_n=1$$ (for every $$n$$) then $$a_n\le b_n$$ and $$\lim_{n\to \infty} a_n=a=1$$ and $$\lim_{n\to \infty} b_n=b=1$$ but $$a\not

• I've been looking at your solution and I can't seem to see what that would imply. – dc3rd Oct 20 '18 at 19:12
• @dc3rd The idea is that if $c > 0$ then there is a gap between it and $0$ and if $c_n \le 0$ there is no way for it to cross this gap in the limit. Another case of when you have a sequence ($c_n$) that is less than a constant ($0$). – Trevor Gunn Oct 20 '18 at 19:18
• I thought about it more ..... and while my answer is correct and this comment (of Trevor) is correct.... it's not very "satisfying" because it is almost nothing more than a shift in perspective rather than a proof. One may not be satisfied by saying "a limit of negatives is never positive"... after all, a limit of rational numbers may be irrational! in any case, maybe my answer helps with intuition whereas Trevor's answer is just the technically correct answer – Squirtle Oct 20 '18 at 19:23

From what you have done follows than for $$n \geqslant N_1, N_2$$:

$$b_n < \frac{a}{2} + \frac{b}{2} < a_n$$

so $$b_n < a_n$$, which is a contradiction.