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Theorem (I.32, pg 27, Apostol's Calculus Vol.1): Let $\varepsilon \in \mathbf{R}^+$ and let $S \subset \mathbf{R}$. If $\sup S$ exists, then there exists $x\in S$ such that $$x > \sup S -\varepsilon.$$

(Reminder: The real number $B = \sup S$ is the least upper bound of $S$; that is, $B$ is an upper bound for $S$ and no number less than $B$ is an upper bound for $S$.)

Now this is proven very elegantly in 1 line, via contradiction. For if $x\leq \sup S - \varepsilon$ for all $x\in S$ then $\sup S-\varepsilon$ would be an upper bound for $S$ smaller than its least upper bound. I am wondering if this statement could be proven directly, with out arguing by contradiction.

The reason I am interested in a direct proof is because I think I get a better, more permanent understanding of things when I hear their direct proof.

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    $\begingroup$ $sup(S)-\epsilon < sup(S)$. Since $sup(S)$ is the least upper bound, then $sup(S)-\epsilon$ not an upper bound. Hence there is some element of $S$ larger than it. $\endgroup$ – blueInk Mar 20 '18 at 23:47
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Since $\sup S$ is the least upper bound of $S$, then $\sup S-\varepsilon$ is not an upper bound for $S$. Thus, there exists $x\in S$ with $\sup S-\varepsilon< x$.

Notice that this is not really a different argument than the one you stated. A "proof by contradiction" is most often a statement of the contrapositive, i.e. to prove $p\implies q$ you prove $\sim q\implies \sim p$

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For a direct prove:

Let $\epsilon >0$ be given.

Note that $ (\sup S - \epsilon) < \sup S $

Therefore $ (\sup S - \epsilon)$ is not an upper bound of S, because $\sup S$ is the least upper bound of S.

Thus there is an element of S, say $x$ such that $x> \sup S - \epsilon$

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