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I've already proved the Cut Property (CP) using the Axiom of Completeness (AoC), but want to now show the converse. Here are my book's definitions:

Cut Property: If $A, B$ are nonempty, disjoint sets with $A \cup B = \mathbb{R}$ and $a < b$ for all $a \in A$ and $b \in B$, then there exists $c \in \mathbb{R}$ such that $x \leq c$ whenever $x \in A$ and $x \geq c$ whenever $x \in B$.

Axiom of Completeness: If $A$ is nonempty and bounded above, then $\sup A$ exists.

My attempt: Let $E$ be nonempty and bounded above. Then there exists some $b$ such that $e < b$ for all $e \in E$... From this point I tried to construct $A$ and $B$ as in the assumption in the Cut Property so that I can invoke it to pinpoint $c = \sup E$. So I tried "expanding" $E$ to go to left to $-\infty$ and construct $B = [b,\infty)$. However, there are possibly gaps between $E$ and $B$ that I don't know how to precisely allocate to $E$ and $B$ without invoking $\sup E$ (which I'm trying to show exists).

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Hint: consider the set of all upper bounds of $E$.

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  • $\begingroup$ I've solved it, thank you. $\endgroup$ – rorty Mar 31 '16 at 18:46

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