I was reading following text:
For $A \subseteq R$ to show that $\sup A = \alpha$ where $α \in \mathbb{R}$ there are two steps. Firstly you need to show that for all $x \in A$ we have that $x \le \alpha$ (i.e. $α$ is an upper bound). Secondly you need to show that for any $\beta < \alpha$ there exists $a \in A$ with $a > \beta$ (i.e. there is no upper bound smaller than $\alpha$). An equivalent way to approach the second part is to show that for any $\epsilon \gt 0$ there exists $a \in A$ such that $a > \alpha−\epsilon$ (again this shows there is no upper bound smaller than $\alpha$).
So I wondered how to prove the equivalence of aforementioned statements:
$(1) \ \ \ \forall\epsilon\gt0 \ \ \ \ \exists a\in A : a\gt \alpha - \epsilon $
$(2) \ \ \ \forall\beta\lt\alpha \ \ \ \ \exists a\in A : a\gt \beta $
I tried to prove the equivalance by letting $\beta = \alpha - \epsilon$ but it wasn't successful for $(2) \implies (1)$.
$(1)\implies (2)$: For any $\beta <\alpha$ let $\epsilon_{\beta}=\alpha - \beta.$ Then $\epsilon_{\beta}>0$ so by $(1)$ we have $$ \exists a\in A\,(a>\alpha-\epsilon_{\beta}=\beta).$$ $(2)\implies (1)$: For any $\epsilon>0$ let $\beta_{\epsilon}=\alpha -\epsilon.$ Then $\beta_{\epsilon}<\alpha$ so by $(2)$ we have $$ \exists a\in A\,(a>\beta_{\epsilon}=\alpha-\epsilon).$$