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 $a\leq b$.


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.

  • $\begingroup$ 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
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    $\begingroup$ My proof here is along the lines of what you're trying to do. $\endgroup$ – Omnomnomnom Oct 11 '15 at 22:16

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.

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

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 $x_n = b_n - a_n$.


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