existence of the supremum.

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One proof is by bisection:

Let $(a_n,b_n)$ be a pair of element of the set and upper bound. Set $c_n=(a_n+b_n)/2$ their midpoint. Either $c_n$ is an upper bound, then $(a_{n+1},b_{n+1})=(a_n,c_n)$. Or there is a point $a_{n+1}\ge c_n$ in the set, then $b_{n+1}=b_n$.

Use that this sequence of pairs provides a sequence of nested intervals or use that the sequence $(b_n)$ can be shown to be Cauchy.

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Updated on August 01, 2022

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  • idm
    idm over 1 year

    How can I show that every upper-bounded set $A\subset \mathbb R, A\neq \emptyset$ has a supremum ? I tried by contradiction, but it isn't conclusive. I also tried to prove that $$\inf\{M\mid \forall x\in A , M\geq x\}$$ exist, but it looks equivalent than to show that $\sup A$ exist.