Perhaps a trivial question, but something I never completely understood. If we have shown that $a-b < \epsilon$ for all $\epsilon > 0$, then does that imply that $a-b \le 0$?
I"m interested in it in the context of this question: Proving that $ f: [a,b] \to \Bbb{R} $ is Riemann-integrable using an $ \epsilon $-$ \delta $ definition.
and in page 172 of these notes: http://alpha.math.uga.edu/~pete/2400full.pdf
Yes, it does.
Suppose not. Then $a-b > 0$ and in particular we can take $\epsilon = a-b$. The statement then says that $a-b < a-b$, which is absurd.