Prove Inequality: |x-x_0|, |y-y_0| & xy-x_0y_0<\epsilon

[objectivesea]

Homework Statement

Prove that if $$|x-x_0| < \textrm{min} \bigg ( \frac{\epsilon}{2|y_0|+1},1 \bigg )$$ and $$|y-y_0| < \frac{\epsilon}{2|x_0|+1}$$ then $$xy-x_0y_0<\epsilon$$

Homework Equations

We can use basic algebra and the following axioms:
For any number $$a$$, one and only one of the following holds:
(i) $$a=0$$
(ii) $$a$$ is in the collection $$P$$
(iii) $$-a$$ is in the collection $$P$$
Note: A number $$n$$ is the collection $$P$$ if and only if $$n>0$$

If $$a$$ and $$b$$ are in $$P$$, then $$a+b$$ is in $$P$$.

If $$a$$ and $$b$$ are in $$P$$, then $$a \cdot b$$ is in P.

We may also use the following consequences of the above axioms:

For any numbers $$a$$and $$b$$, one and only one of the following holds:
(i) $$a=b$$
(ii) $$a < b$$
(iii)$$a > b$$

For any numbers $$a$$, $$b$$, and $$c$$, if $$a<b$$ and $$b<c$$, then $$a<c$$.

For any numbers $$a$$, $$b$$, and $$c$$, if $$a<b$$, then $$a+c<b+c$$.

For any numbers $$a$$, $$b$$, and $$c$$, if $$a<b$$ and $$0<c$$, then $$ac<bc$$.

The Attempt at a Solution

I'm not even sure where to start.

 

[praharmitra]

Are you sure it's not
$$|y-y_0| < \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )$$?

 

[objectivesea]

Are you sure it's not
$$|y-y_0| < \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )$$?

I just double checked. I'm sure.

 

[objectivesea]

I just double checked. I'm sure.

But if $$|y-y_0|<\textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )$$
then $$|y-y_0|< \frac {\epsilon}{2|x_0|+1}$$
Since if $$\textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )=\frac {\epsilon}{2|x_0|+1}$$ then $$|y-y_0|<\frac {\epsilon}{2|x_0|+1}\leq 1$$.
And if $$\textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )=1$$ then $$|y-y_0|<1 \leq \frac {\epsilon}{2|x_0|+1}\$$

either way $$|y-y_0|< \frac {\epsilon}{2|x_0|+1}$$

 

[praharmitra]

No, I actually want the fact that $$|y-y_0|<1$$ in addition to all your above expressions. Now I am wondering, given your first inequality, is it possible to prove that

$$\frac{\epsilon}{2|x_0|+1} <1$$. Do you think you can try to prove that?

If you can do that, then afterwards use the following:

1. $min(a,b) \leq a,~min(a,b)\leq b$

2. $|x-x_0| \geq |x|-|x_0|$

and proceed from there. But first try the first thing. Something I am not able to do.