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$a$ and $b$ are fixed real numbers.

Claim: $a < b$ implies $a < b - \varepsilon$ for some $\varepsilon > 0$.

Proof: Utilise the fact that $a \implies b$ is equivalent to $b' \implies a'$.

So this is equivalent to proving $a \ge b - \varepsilon\,$ for all $\varepsilon > 0 \implies a \ge b$.

Now, as $a \ge b - \varepsilon$, take the infimum of both sides.

$$\inf (a) \ge \inf \{b - \varepsilon \mid \varepsilon > 0\}$$

$\inf (a) = a$ and the right-hand side $= b$, hence:

$$a \ge b.$$

Thus, $a \ge b - \varepsilon$ for all $\varepsilon > 0$ implies $a \ge b$. Which then proves our original claim.

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up vote 0 down vote accepted

Soo... looks good, but!:

  1. Rather use different letters for propositions in your 3rd line, at least capitals ($A\implies B$ equivalent to $\lnot B\implies \lnot A$)
  2. Instead of $\inf$ you are considering $\sup$
  3. You are implicitly using $\sup\{b-\varepsilon \mid \varepsilon >0\} = b$, which is true, but... it is equivalent to the given problem. that's why it's not really working.

And the solution is (straight ahead):

If $a<b$, then $b-a>0$, and let $\varepsilon:=\displaystyle\frac{b-a}2$.

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About point 3... Any two true statements are equivalent. – Arthur Oct 10 '12 at 16:41
Ah ok. Yes the infimum was a slip on my part. Ok so I see that epsilon works, thank you! – user64219 Oct 10 '12 at 16:46
Any $\varepsilon\in (0,b-a)$ works. – Berci Oct 10 '12 at 20:04

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