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I wanted to check if this proof works. Any comment would be appreciated!

Proof. Note that $\sup(A)$ is the smallest number such that $a \leq \sup(A)$ for every $a \in A$

Now it is true that $a \leq 1$ for all $a \in A$. Therefore $\sup(A) \leq 1$.

Now suppose $\sup(A) = b < 1$. Then $b \in A$

Let $\epsilon >0$ and b = $1 - \epsilon$. Then 1 - $\frac{\epsilon}{2}$ > b

So b can not be an upper bound of A since it can't be true that a $\le b$ for all a $\in A$

Hence we only have that $\sup(A) = 1$.

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  • 1
    $\begingroup$ "...but then A has no maximum..." ?? $\endgroup$ – DonAntonio May 13 '16 at 21:10
  • $\begingroup$ @Joanpemo because 1 is not included in A, whatever that b is, wouldn't there be any number that is slightly greater than b? That was what I was thinking.. $\endgroup$ – Meer May 13 '16 at 21:19
  • $\begingroup$ That's something you have to prove, yet that wasn't my point but the word "then" there. $\endgroup$ – DonAntonio May 13 '16 at 21:21
  • $\begingroup$ @Joanpemo wording is wrong? $\endgroup$ – Meer May 13 '16 at 21:26
  • $\begingroup$ The logic is wrong! $\endgroup$ – The Chaz 2.0 May 13 '16 at 22:03
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Take any $\;\epsilon>0\;$, (we can assume $\;\epsilon<1\;$ , otherwise the argument is trivial), and take $\;a:=1-\frac\epsilon2\;$ , then $\;a\in A\;$ and

$$1-\epsilon<a<1\,,\;\;\text{and since this is true for any}\;\;\epsilon >0$$

we get that $\;1=\sup A\;$

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