Proving $x$ between $\alpha - \epsilon$, if $\alpha$ is supremum Let $S\subseteq\Bbb R$ and $\alpha \in \Bbb R$. If $\alpha = \sup(S)$, then show that for any $\epsilon > 0$, there is some $x \in S$ such that $\alpha - \epsilon < x$.
What I have done :
Since $\alpha$ is the supremum, $x<\alpha$ and $\alpha - \epsilon < \alpha$ I want to show $x$ is in between $\alpha - \epsilon$ and $\alpha$ , but can't and not even sure if it is possible.  
Its been a long I posted a question, so sorry if this seems a total homework question.

Hint: What happens if there is no $x\in S$ such that $\alpha-\varepsilon<x$? (Think about the definition of supremum).
