I've been stuck on this problem for over a day, and the answerbook simply says "see chapter 5" for problems 20,21, and 22. But I want to complete the problem without using knowledge given later in the book, so I've been banging my head against the wall trying all sorts of things, but nothing I do seems to lead me anywhere.
The problem is as follows:
Prove that if
$|x – x_0| < min (\frac{\varepsilon}{2(|y_0| + 1)}, 1)$ and $|y – y_0| < min (\frac{\varepsilon}{2(|x_0| + 1)}, 1)$
then
$|xy – x_0 y_0| < \varepsilon$.
Here are some of the things I've been thinking about, I don't know which of these are useful (if any), but they somewhat outline the logic behind my various attempts.
Since at most $|x – x_0| < 1$ and $|y – y_0| < 1$ then it follows that $(|x-x_0|)(y-y_0|) < |x-x_0|$ and $(|x-x_0|)(y-y_0|) < |y-y_0|$
Also, $(|x – x_0|)(|y_0| + 1) < \frac{\varepsilon}{2}$ and $(|y – y_0|)(|x_0| + 1) < \frac{\varepsilon}{2}$ so $(|x – x_0|)(|y_0| + 1) + (|y – y_0|)(|x_0| + 1) < \varepsilon$. and since $|a + b| \leq |a| + |b| < \varepsilon$ I've tried multiplying things out, and then adding them together to see if anything cancels, but I can't make anything meaningful come out of it.
Also since $|a – b| \leq |a| + |b|$ I've also tried subtracting one side from the other, but to no avail.
I was also thinking that since $(|x-x_0|)(y-y_0|) < |x-x_0|$, then I could try something along the lines of $(|x-x_0|)(y-y_0|)(|x_0| + 1) + (|x-x_0|)(y-y_0|)(|y_0| + 1)< \varepsilon$ and various combinations as such, but I just can't seem to get anything meaningful to come out of any of these attempts.
I have a sneaking suspicion that the road to the solution is simpler than I'm making it out to be, but I just can't see it.
Best Answer
By reverse triangle inequality $|x| - |x_0| \leq |x-x_0| < 1$ giving $|x| < 1 + |x_0|$. Now \begin{align*} |xy - x_0y_0| & = |x(y-y_0) + y_0(x-x_0)| \leq |x||y-y_0| + |y_0||x-x_0| \\ & < (1 + |x_0|)\frac{\varepsilon}{2(|x_0| + 1)} + |y_0|\frac{\varepsilon}{2(|y_0| + 1)} \\ & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align*} To my mind the difficult part of this argument is writing down the equality proceeding the string of inequalities and this certainly isn't the only way to do that.