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I am trying to prove the following.....

Let $A$ be a nonempty bounded subset of $\mathbb{R}$ and let $\alpha = \sup A$. If $\alpha \notin A$, prove that for every $\varepsilon >0$, the interval $\left ( \alpha -\varepsilon, \alpha \right )$ contains infinitely many points of $A$.

However, I don't really understand how to prove it. This is what I have so far... Let $\emptyset \neq A \subseteq \mathbb{R}$, $\alpha = \sup A $ and $\alpha \notin A$ be given. Assume $(\alpha - \varepsilon, \alpha) \cap A = {a_1, a_2...a_n}$. Since $(\alpha - \epsilon, \alpha) \cap A$ is a finite set then it has a min and a max. Thus there exists a $\varepsilon$ so that $\alpha - \varepsilon $ is in the set.

I'm not really sure to go from here or if I'm even on the right track. Could someone explain/show how to prove this.

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2 Answers 2

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You are on the right track. Since $(\alpha - \epsilon, \alpha) \cap A$ is a finite set, it has a maximum, say $a_1$. Then $a_1$ is an upper bound for $A$ that is strictly smaller than $\alpha=\sup A$ which is a contradiction.


More details, since apparently I am causing confusion...

$A$ is a nonempty set with supremum $\alpha \notin A$. To prove that $(\alpha-\epsilon,\alpha)$ contains infinitely many points of $A$, suppose for sake of contradiction $(\alpha-\epsilon,\alpha)\cap A$ is finite. Then it has some maximum element $a$. All other elements of $A$ must be in the interval $(-\infty, \alpha-\epsilon]$ since $\sup A=\alpha$ and $\alpha \notin A$, so all elements of $A$ are $\le a$. However, $a<\alpha$, which contradicts the fact that $\alpha$ is the least upper bound for $A$.

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There is $x_1 \in (\alpha - \epsilon, \alpha) \cap A $, since $\alpha - \epsilon$ is not an upper bound of $A$ and $ \alpha \notin A$.

Since $x_1$ is not an upper bound of $A$ and $ \alpha \notin A$, there is

$x_2 \in (x_1, \alpha) \cap A $.

Now it should be clear how to proceed to get a sequence $(x_n)$ in $(\alpha - \epsilon, \alpha) \cap A $ with

$x_1 < x_2 < x_3 < ....$

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