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A Dedekind cut is an ordered pair $(A,B)$ where $A$ and $B$ are subsets containing rational numbers such that $A,B \neq \emptyset$, and $A$ doesn't have a greatest element, $A \cup B = \Bbb Q$, and $x\in A$, $y \in B$ $\Bbb \rightarrow$ $x<y$. Let $\Bbb D$ be the set of all Dedekind sets. I need to prove that $f:\Bbb D \rightarrow \Bbb R$, defined as $f(A,B)=\sup A$ is bijective.

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    $\begingroup$ What is your definition of $\Bbb{R}$? $\endgroup$ – Chris Eagle Apr 3 '13 at 14:33
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$f$ is surjective as for $\alpha\in \mathbb R$ we can let $A=\{x\in \mathbb Q\mid x<\alpha\}$, $B=\mathbb Q\setminus A$. Check that $(A,B)\in\mathbb D$ and that $f(A,B)=\alpha$.

To show that $f$ is also injective, note that the converse holds, i.e. if $f(A,B)=\alpha$ then $x<\alpha$ implies $x\in A$ and $x\ge\alpha$ implies $x\notin A$.

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Hint: Use the fact that $\Bbb Q$ is dense in $\Bbb R$.

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  • $\begingroup$ (This is regardless to your definition of $\Bbb R$) $\endgroup$ – Asaf Karagila Apr 3 '13 at 14:38

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