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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

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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.

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    $\begingroup$ To add to this answer, a common proof technique is to conclude that $a = 0$ from $|a| < \epsilon$ for all $\epsilon > 0$. $\endgroup$
    – Charles Hudgins
    May 19, 2019 at 20:42
  • $\begingroup$ Is there a constructive proof of this? I.e. one that doesn't use proof by contradiction or excluded middle? $\endgroup$
    – Michael Bächtold
    May 20, 2019 at 7:56
  • $\begingroup$ You can formulate the same proof with contraposition. Is that good enough? $\endgroup$
    – J. De Ro
    May 20, 2019 at 8:09
  • $\begingroup$ @EpsilonDelta not sure what you mean. Btw. my question was not directly addressed at you. Maybe I should ask it as a separate question. $\endgroup$
    – Michael Bächtold
    May 20, 2019 at 10:50
  • $\begingroup$ @MichaelBächtold You can take $\epsilon \to 0$ in a limit argument, but maybe this also implicitely uses what has to be proven $\endgroup$
    – J. De Ro
    May 31, 2019 at 14:42

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