Intereting Posts

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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 Verification - Every sequence in $\Bbb R$ contains a monotone sub-sequence
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- Proof that a sequence of continuous functions $(f_n)$ cannot converge pointwise to $1_\mathbb{Q}$ on $$
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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?

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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 $a\in A$.

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