# Let $A$ be a nonempty subset of $\mathbb{R}$. Prove…

I'm suppose to prove the following...

Let $A$ be a nonempty subset of $\mathbb{R}$. If $\alpha = \sup A$ is finite show that for each $\epsilon > 0$ there is an $a \in A$ such that $\alpha - \epsilon < a\leq \alpha$.

This is what I have so far...

Proof: We claim that $\alpha = \sup A$

Then by definition of a supremum $\alpha$ is the least upper bound of set A

Thus $\forall a \in A$, $a \leq \alpha$

Suppose $\epsilon > 0$

Then $\alpha - \epsilon < \alpha$.

Thus $\alpha - \epsilon < a\leq \alpha$.

I feel like I'm missing some step between my last and second to last step so what am I forgetting? Also is the rest of my proof right?