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I need help with the following two questions. Let $\alpha$ be a cut. We define $$\alpha^{-1}=\{p \in \mathbb{Q}:\frac{1}{p} \in \alpha^c \text{ and } \frac{1}{p} \text{ is not the least element of } \alpha^c\} \cup \{p \in \mathbb{Q}: p \leq 0\}$$ Also $\alpha$ is called a rational cut if there is $r \in \mathbb{Q}$ such that $\alpha=\{p \in \mathbb{Q}:p<r\}$.

Question 1: Show that $\alpha^{-1}$ is also a cut.

Question 2: Prove that if $\alpha$ is a rational cut then $\alpha^{-1}$ is also so.

(I don't know if I tag this question correctly.)

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1 Answer 1

I suppose your definition of Dedekind cut is: $A$ is a proper subset of $\mathbb Q$ which is downwards closed and does not have largest element.

It seems that you assume $\alpha>0$ (i.e., $\{p\in\mathbb Q; p\le 0\}\subseteq\alpha$), even though you did not mention this in your question.

Then $\alpha^{-1}$:

  • is a proper subset of $\mathbb Q$, since $p\mapsto\frac1p$ is a bijection $\mathbb Q^+ \to \mathbb Q^+$
  • is downwards closed: Suppose $p\in\alpha^{-1}$ and $q<p$. If $p\le 0$, then $q\le 0$ and $q\in\alpha^{-1}$. If $p>0$ then we have $\frac1p\in\alpha^c$, and the inequality $\frac1q>\frac1p$ implies $\frac1q\in\alpha^c$ (since $\alpha^c$ is upwards closed) and it is not the smallest element of $\alpha^c$.
  • does not have largest element. Suppose, by contradict, that $p$ is the largest element of $\alpha^{-1}$. This means $p>0$ and since $p\mapsto\frac1p$ is an order-reversing map on $\mathbb Q^+$, this would imply that $\frac1p$ is the smallest element of $\alpha^c$.

For question 2: I think it should be relatively easy to show that if $\alpha=\{p\in\mathbb Q; p<r\}$ for some $r\in\mathbb Q^+$, then $\alpha^{-1}=\{p\in\mathbb Q; p<\frac 1r\}$.

(By $\mathbb Q^+$ I mean the set $\{q\in\mathbb Q; q>0\}$.)

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