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?
Best Answer
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 $a\in A$.