# Equivalent definition of supremum

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

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

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

We want to show that

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

and

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

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

• 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. Commented Dec 4, 2021 at 13:48
• @Anthonny That's subtle, I think. I asked that here math.stackexchange.com/questions/4323732/… Commented Dec 4, 2021 at 14:53