What does it mean for something to be strictly less than $\epsilon$ for an arbitrary $\epsilon$?

Question

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

score 17 · Answer 1 · answered May 19 '19 at 20:15

17

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.

answered May 19 '19 at 20:15

J. De Ro

21,438

7

To add to this answer, a common proof technique is to conclude that $a = 0$ from $|a| < \epsilon$ for all $\epsilon > 0$. – Charles Hudgins May 19 '19 at 20:42
Is there a constructive proof of this? I.e. one that doesn't use proof by contradiction or excluded middle? – Michael Bächtold May 20 '19 at 07:56
You can formulate the same proof with contraposition. Is that good enough? – J. De Ro May 20 '19 at 08:09
@EpsilonDelta not sure what you mean. Btw. my question was not directly addressed at you. Maybe I should ask it as a separate question. – Michael Bächtold May 20 '19 at 10:50
@MichaelBächtold You can take $\epsilon \to 0$ in a limit argument, but maybe this also implicitely uses what has to be proven – J. De Ro May 31 '19 at 14:42

What does it mean for something to be strictly less than $\epsilon$ for an arbitrary $\epsilon$?

1 Answers1