2
$\begingroup$

Let $\{a_n\},\{b_n\}$ be given such that $a_n\leq b_n$, then if $\{a_n\}$ converges to $a$ and $\{b_n\}$ converges to $b$, show that this implies $a\leq b$.


Proof:

Let $\epsilon$. Since $\{a_n\}$ converges to $a$ and $\{b_n\}$ converges to $b$, then we have $$\vert a_n-a\vert<\epsilon/2\,\,\,\,\text{for all $n\geq N$}\tag1$$ $$\vert b-b_n\vert<\epsilon/2\,\,\,\,\text{for all $n\geq M$}\tag2$$

Add $(1)$ and $(2)$, we have $\vert a_n-a\vert+\vert b-b_n\vert<\epsilon$

Apply triangle inequality, we have $\vert(a_n-a)+(b-b_n)\vert<\epsilon$ and this is equivalent to $\vert(a_n-b_n)+(b-a)\vert<\epsilon$. Because $a_n\leq b_n$ and let $k\geq\max\{M,N\}$, then we get $$\vert a_k-b_k\vert-\vert b-a\vert\leq\vert(a_k-b_k)+(b-a)\vert\leq\vert a_k-b_k\vert+\vert b-a\vert$$

We know that $\vert a_k-b_k\vert+\vert b-a\vert\geq 0$, so $-\vert a_k-b_k\vert\leq\vert b-a\vert$ and this gives us $0\leq\vert b-a\vert$ and implies $a\leq b$.


I have a feeling that I missed something that I need to show. Can anyone check my proof to see what I missed? Thanks.

$\endgroup$
  • $\begingroup$ You areshowing anything, the absolute value of any complexnumber is always non-negative $\endgroup$ – CIJ Oct 11 '15 at 22:05
  • $\begingroup$ So, since $|2-4|=2\geq 0$ does that mean $4\leq 2$? $\endgroup$ – Thomas Andrews Oct 11 '15 at 22:07
  • $\begingroup$ @ThomasAndrews I see, Thanks. $\endgroup$ – Simple Oct 11 '15 at 22:13
  • 1
    $\begingroup$ My proof here is along the lines of what you're trying to do. $\endgroup$ – Omnomnomnom Oct 11 '15 at 22:16
2
$\begingroup$

You're starting well but going astray in the middle. You can't really reach $a \le b$ by looking at absolute values of their difference.

Start from the end and think geometrically (imagine all of $a_n$, $b_n$, $a$, $b$ on the number line). Suppose the conclusion is wrong and in fact $a > b$. The distance between them is some $\epsilon$. Why is this impossible?

Well, we know that $a_n$ eventually gets very close to $a$, even closer than $\epsilon/2$. And we know that $b_n$ eventually gets very close to $b$, even closer than $\epsilon/2$. Given some $a_k, b_k$ that are both that close to their limits... but they still satisfy $a_k \le b_k$... do you see the contradiction? Draw these four fellows on a line if you don't ($a,b,a_k,b_k$). If you do, now put it into the formalism.

$\endgroup$
  • $\begingroup$ There exists some $b_N$ less than $a_N$, a contradiction $\endgroup$ – Simple Oct 11 '15 at 23:12
1
$\begingroup$

As pointed out in the comments, your proof is not correct. $|b-a|$ is always non-negative, so you haven't really proven anything. You can not conclude that $b \geq a$ based on $|b-a| \geq 0$.

What we can do instead is prove the following:

If $(x_n)$ is a sequence such that $x_n \geq 0$ for all $n$, and $$\lim_{n\to\infty} x_n = x$$ then $x \geq 0$.

The proof is as follows: Suppose that $x<0$. Take $\varepsilon = |x| = -x$. Then for $n$ large enough, we have that $x_n-x \leq |x_n-x| < \varepsilon$, and so $x_n < x + \varepsilon = 0$, which is a contradiction.

Now apply the above result to the sequence $x_n = b_n - a_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.