By the definition of $\mathbb{R}$ as the Dedekind complete well ordered field, there should be such a supremum. What is it?
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Can it be $\frac52$? Can it be $\frac32$? – Feb 11 '16 at 15:23
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No, since $2.01$ is smaller. Besides, Dedekind completeness means that it should be in this set. – user285146 Feb 11 '16 at 15:25
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5Dedekind completeness does not mean that it should be in the set. It simply means that it should be in $\mathbb{R}$. – parsiad Feb 11 '16 at 15:25
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Ok, so it's not $2.5$. Can it be $2.01$? – Feb 11 '16 at 15:26
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2$\mathbb{R}$ is Dedekind-complete, ${x : x < 2}$ isn't. – Najib Idrissi Feb 11 '16 at 15:31
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If we call the set $S$ the supremum should be a number $a$ such $x\le a$ for all $x\in S$, and if there exists some $a'$ with $x\le a'$ for all $x\in S$, then $a\le a'$.
Now, if you choose $a<2$ then you have $\frac{a+2}{2}\in S$ and $\frac{a+2}{2}>a$ which means $a$ cannot be the supremum.
Also, if you choose $a>2$ then $2<\frac{2+a}{2}<a$ meaning $x<\frac{2+a}{2}$ for all $x\in S$ but $\frac{2+a}{2}<a$ which means $a$ cannot be the supremum in this case either.
Therefore we must conclude $a=2$.
edit: as has been suggested, there is a clear upper bound of $3$ so our supremum must be at most $3$, meaning our supremum is not infinite either.
Alex Mathers
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Might be worth adding in a short justification that the supremum isn't infinite. – πr8 Feb 11 '16 at 15:52