# if $\{a_n\}$ converges to $a$ and $\{b_n\}$ converges to $b$ where $a_n\leq b_n$, show $a\leq b$

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.

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

• There exists some $b_N$ less than $a_N$, a contradiction – 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 $x_n = b_n - a_n$.