# 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?

#### Solutions Collecting From Web of "Let $A$ be a nonempty subset of $\mathbb{R}$. Prove…"

First, you don’t need to say that “We claim that $\alpha=\sup A$.” In fact, it was already an assumption. There are many mistakes in the proof you presented. I rather present the following proof.

Proof: Assume that $\alpha=\sup A$. Then $\alpha$ is the Least Upper Bound of $A$. Let $\epsilon>0$. Since $\alpha-\epsilon<\alpha$, then the number $\alpha-\epsilon$ can not be an upper bound of set $A$. Thus, there exists $a\in A$ such that $$a>\alpha-\epsilon.$$ Because $\alpha$ is an upper bound of $A$ and $a\in A$, we get $$a\leq \alpha.$$ Hence, for each $\epsilon>0$, we get
$$\alpha-\epsilon<a\leq \alpha,$$ for some $a\in A$.