0
$\begingroup$

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?

  • 0
    You are forgetting that part of the definition of sup is that if $b < sup A then B is *not* an upper bound. Your line alpha - epsilon < a < alpha doesn't make sense. Let A = (0,1). sup A =1. So a <= 1 for all a in A. Let epsilon =.01. Then 1-.01=.99 < 1 but .99 < a < 1 is not true for all a in A. – 2017-02-24
  • 0
    The first line is not appropriate and the last line needs justification. Please see my answer below for the correction and formal proof. – 2017-02-24

1 Answers 1

1

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