How to prove $\forall \epsilon\in \Bbb{R^+},\exists \delta\in\Bbb{R^+},\forall x\in\Bbb{R^+}, \forall y\in\Bbb{R^+}, |x-y|\le \delta\lor |x^2-y^2|\gt \epsilon$.
My try:
Pick $\delta=\frac{\epsilon}{|x+y|}$. Then either $|x^2-y^2|\gt \epsilon$ or $|x^2-y^2|\le \epsilon$.
Case $1$:$|x^2-y^2|\gt \epsilon$. Then there's nothing to say about it.
Case $2$: $|x^2-y^2|\le \epsilon$. Then $|x+y||x-y|\le \epsilon$. Then $|x-y|\le \frac{\epsilon}{|x+y|}=\delta$
I felt at least part of my proof is wrong. Could someone point out?
The basic problem with your proof is that you are picking a $\delta$ that depends not just on $\epsilon$ but on $x$ and $y$ as well. In other words, you have tacitly changed the order of quantifiers from $\forall \epsilon\exists \delta\forall x\forall y$ to $\forall \epsilon\forall x\forall y\exists \delta$. What you're asked to show is that for a given $e$ there is a $\delta$, depending only on $\epsilon$, for which the given inequalities hold for all $x$ and $y$.
As for a correct proof, note that the equivalence of $P\lor Q$ and $\lnot P\implies Q$ means you can rewrite the conditions $|x-y|\le\delta\lor|x^2-y^2|\lt\epsilon$ as $|x-y|\gt\delta\implies|x^2-y^2|\gt\epsilon$. At this point, it's important to remember that $x$ and $y$ are stipulated to be positive real numbers. This means that $|x-y|\gt\delta$ implies $x\gt\delta$, which in turn implies $|x+y|=x+y\gt\delta$, and this means that $|x^2+y^2|=|x-y||x+y|\gt\delta^2$. Thus it suffices to let $\delta=\sqrt\epsilon$ in order for everything to work out.