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?

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

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

and this is a contradiction.

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

  • $\begingroup$ Should the $\epsilon$ in the second form be $\epsilon = a - b$? $\endgroup$ – dc3rd Oct 20 '18 at 18:57
  • $\begingroup$ @dc3rd Yes, thank you! $\endgroup$ – Trevor Gunn Oct 20 '18 at 18:58
  • $\begingroup$ 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? $\endgroup$ – dc3rd Oct 20 '18 at 19:01
  • $\begingroup$ @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. $\endgroup$ – Trevor Gunn Oct 20 '18 at 19:06
  • $\begingroup$ @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. $c < a - \varepsilon$. $\endgroup$ – 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 $a-\frac{\epsilon}{2}<a_n$ and $b_n<b+\frac{\epsilon}{2}$. For this specific $n$ we have:

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

So $a<b+\epsilon$. 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 $a\leq b$.


New answer:

Suppose not! If $a\not\le b$ then $b<a$ so $0<a-b$. 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 $a<b$ 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 $a\not<b$

  • $\begingroup$ I've been looking at your solution and I can't seem to see what that would imply. $\endgroup$ – dc3rd Oct 20 '18 at 19:12
  • 1
    $\begingroup$ @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 ($0$). $\endgroup$ – Trevor Gunn Oct 20 '18 at 19:18
  • $\begingroup$ 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 $\endgroup$ – 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.


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