# 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 $$\alpha-\varepsilon? (Think about the definition of supremum).
We want to show the existence of such $$x$$.
Suppose it doesn't exist, then we have $$\alpha - \epsilon$$ being an upper bound of $$S$$ which contradicts that $$\alpha$$ is the least upper bound.