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When reading through the past answer https://math.stackexchange.com/a/2536917/577288, I found a statement

$\forall \epsilon >0, s\leq t + \epsilon$, then $s\leq t$

But, I couldn't understand why this holds. Intuitively, $\epsilon$ can be very small but it never cannot be 0...

orematasaburou
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2 Answers2

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This is a proof by contradiction. Suppose $s>t$. Then let $\epsilon = \frac {s-t} {2} > 0$. But then $s<t+\epsilon=t+\frac {s-t} {2} = t - \frac {t} {2} + \frac {s} {2} = \frac {t+s} {2} < \frac {2s} {2} = s $. But $s<s$ is a contradiction, and so we are done.

(This technique and result are important to fully understand as they'll be used all throughout analysis. The result is also a good tool to use when you want to prove two things are equal, and the method is a good illustration of how to do a $\delta-\epsilon$ proof).

Xela
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    Xela's answer is excellent, but if you want a bit more intuitive way to think about it, $s$ and $t$ have fixed values, but $\epsilon$ is allowed to change values and move around, so long as it stays positive. So if $s$ and $t$ had any distance between them , $\epsilon$ would be able to wiggle in between them and produce a contradiction. Statements that assert that something is true $\forall \epsilon \gt 0$ are really powerful! – JonathanZ Feb 15 '23 at 04:00
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If $s > t$, then you could take $\epsilon$ small enough such that $t+\epsilon < s$, which is a contradiction with your assumption, and hence necessarily $s \leq t$.

Gibbs
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