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Let $S\subseteq\Bbb R$ and $\alpha \in \Bbb R$. If $\alpha = \sup(S)$, then show that for any $\epsilon > 0$, there is some $x \in S$ such that $\alpha - \epsilon < x$.

What I have done :

Since $\alpha$ is the supremum, $x<\alpha$ and $\alpha - \epsilon < \alpha$ I want to show $x$ is in between $\alpha - \epsilon$ and $\alpha$ , but can't and not even sure if it is possible.

Its been a long I posted a question, so sorry if this seems a total homework question.

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Hint: What happens if there is no $x\in S$ such that $\alpha-\varepsilon<x$? (Think about the definition of supremum).

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We want to show the existence of such $x$.

Suppose it doesn't exist, then we have $\alpha - \epsilon$ being an upper bound of $S$ which contradicts that $\alpha$ is the least upper bound.

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  • $\begingroup$ Got it, Thanks. $\endgroup$ – Pushkar Soni Oct 1 '18 at 1:57

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