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My textbook gives two different definitions for the supremum, but I've been unable to prove them.

Definition 1
$E \subset \mathbb{R}$ is a set of real numbers.
$\alpha$ :=sup $E$ $\overset {\mathrm{def}} {\Leftrightarrow} $
(1) $\forall x \in E$ , $x \leq \alpha$
(2) If $\gamma < \alpha$ , $\exists x \in E$ : $\gamma < x $

Definition 2
$E \subset \mathbb{R}$ is a set of real numbers.
$\alpha$ :=sup $E$ $\overset {\mathrm{def}} {\Leftrightarrow} $
(1) $\forall x \in E$ , $x \leq \alpha$
(2) $\forall \varepsilon >0 $, $\exists x \in E$ : $\alpha-\varepsilon<x$

Please tell me the proof Definition 1 $\iff$ Definition 2.

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    $\begingroup$ $y<\alpha \iff \exists \varepsilon>0\, s.t.\, y=\alpha-\varepsilon$. $\endgroup$ Commented Dec 2, 2021 at 14:15
  • $\begingroup$ The correspondence is $\gamma\leftrightarrow(\alpha-\varepsilon)$. $\endgroup$
    – Berci
    Commented Dec 2, 2021 at 14:15

1 Answer 1

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We want to show that

$\forall\gamma < \alpha$ , $\exists x \in E$ : $\gamma < x $

and

$\forall \varepsilon >0 $, $\exists x \in E$ : $\alpha-\varepsilon<x$

are equivalent.

Note that

$\gamma < \alpha\iff \alpha-\gamma>0$

and

$\gamma < x \iff \alpha-(\alpha-\gamma)< x$

So, the first proposition is equivalent to

$\forall \alpha-\gamma>0$, $\exists x \in E$ : $\alpha-(\alpha-\gamma)< x$

Now, it is enought to use the change of variable property, in this case $\varepsilon=\alpha-\gamma$. Therefore

$\forall \varepsilon >0 $, $\exists x \in E$ : $\alpha-\varepsilon<x$

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  • $\begingroup$ I'm not really sure about the last step. Could you give some reference about that "rule" that you use? Maybe there are some restrictions, such as that the function $\varepsilon=f(\gamma)$ must be bijective or something like that. $\endgroup$
    – Anthonny
    Commented Dec 4, 2021 at 13:48
  • $\begingroup$ @Anthonny That's subtle, I think. I asked that here math.stackexchange.com/questions/4323732/… $\endgroup$
    – Pathy
    Commented Dec 4, 2021 at 14:53

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