Ok, so I'm having some issues working on these proofs:
a) |xy| = |x| * |y|
b) If |x-y| < c, then |x| < |y| + c
and
c) If |x-y| < E for all E > 0, then x = y.
Any help would be greatly appreciated!
a.) $\displaystyle |xy|=\sqrt{(xy)^2}=\sqrt{x^2y^2}=\sqrt{x^2}\sqrt{y ^2}=|x|\,|y|$
b.) $\displaystyle |x|=|y+x-y|\leq|y|+|x-y|<|y|+c$
c.) Assume $\displaystyle x\neq y$. Then by the properties of a metric, $\displaystyle |x-y|=d>0$. Thus, taking $\displaystyle E=\frac{d}{2}$ gives us an $\displaystyle E>0$ such that $\displaystyle |x-y|>E$, which is a contradiction. So $\displaystyle x=y$. $\displaystyle \square$
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500th Post!
Yes. You could prove (a) by cases (assume $\displaystyle x$ is positive and $\displaystyle y$ is negative, etc.) but it's unnecessary if you know that $\displaystyle |x|=\sqrt{x^2}$.
For the other two proofs, all you need to know is that $\displaystyle |x-y|$ is a metric, and then you can use properties of metrics to complete the proof. No cases required!
b) $\displaystyle \left| x \right| = \left| {x - y + y} \right| \leqslant \left| {x - y} \right| + \left| y \right| < c + \left| y \right|$
c) Suppose that $\displaystyle x\ne y$ then that means $\displaystyle |x-y|>0$.
So let $\displaystyle E=|x-y|>0$ then you have $\displaystyle |x-y|<|x-y|.$
That is a contradiction to $\displaystyle x\ne y.$