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Let $A \subset \mathbb{R}, s \in \mathbb{R}$ have the property that $s - \frac{1}{n}$ is not an upper bound for $A$ and $s + \frac{1}{n}$ is an upper bound for $A$, $\forall n \in \mathbb{N}$

Seems a little too easy but calculus comes to mind.

if $s - \frac{1}{n}$ is not an upper bound then $\exists \, x \in A \, s.t. \,\,s+\frac{1}{n} < x$ and $s + ( sup(A) - x ) $ is an upper bound so s is either an upper bound or in A since $s \le sup(A)$ So suppose $s$ is not the supremum

Now, $lim_{n \to \infty} \bigg [ s-\frac{1}{n} < sup(A) \le s+\frac{1}{n} \bigg ] = s < sup(A) \le s \rightarrow\leftarrow$

So $s = sup(A)$

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  • $\begingroup$ Which is your definition of supremum? The least upper bound or wtih $\epsilon$ @oliverjones $\endgroup$
    – DonQuixote
    Sep 26 '15 at 3:05
  • $\begingroup$ What does $\lim_{n\to\infty}[s-\frac1n<\sup A\le s+\frac1n]$ mean? $\endgroup$
    – Jason
    Sep 26 '15 at 3:05
  • $\begingroup$ @DonQuixote the least upper bound. $\endgroup$ Sep 26 '15 at 3:10
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    $\begingroup$ You have a constructive and a non-constructive proof. I hope this will be useful for you :). As @Jason pointed out, your idea was good, but it needs rigor. $\endgroup$
    – DonQuixote
    Sep 26 '15 at 3:21
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Your idea is good but I would formalize it a little more - given you are asking a question about supremums, I am assuming you are taking an introductory analysis course and so rigor is important. As you pointed out, the conditions imply $$s-\frac1n<\sup A\le s+\frac1n$$ for all $n\in\mathbb N$. This implies $|x-\sup A|\le\frac1n$. Fix $\epsilon>0$. Then taking $n>\epsilon^{-1}$, we have $|s-\sup A|\le\frac1n<\epsilon$. Taking infimum, we have $0\le|s-\sup A|\le\inf(0,\infty)=0$.

(By the way, if you have not done so before, proving that $\inf(0,\infty)=0$ is a good exercise!)

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Clearly, $s$ is an upper bound. If not, $\exists a\in A$ such that $s<a$, but $\exists n\in\mathbb{N}$ such that $s+\frac{1}{n}<a$.$\Rightarrow\Leftarrow$. $s+\frac{1}{n}$ is an upper bound.

So, we only need to prove that is the least upper bound. If not, $\exists n\in\mathbb{N}$ such that $s-\frac{1}{n}\geq a\forall a\in A$ $\Rightarrow\Leftarrow$.$s+\frac{1}{n}$ is not an upper bound.

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You need to prove : $s = \text{sup}(A)$, and to achieve this you can prove that: it is not true that : $s > \text{sup}(A)$ or $s < \text{sup}(A)$. Assume $s > \text{sup}(A)$, then choose an $n$ such that $0<\dfrac{1}{n} < s - \text{sup}(A)$, thus : $s- \dfrac{1}{n} > s - ( s - \text{sup}(A)) = \text{sup}(A)$ is an upperbound for $A$ ( its even greater than the least upperbound ), contradiction. Similarly you can argue that $s < \text{sup}(A)$ cannot happen. Thus by trichotomy property of the reals, $s = \text{sup}(A)$.

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