352: Minus Dedekind Cut Of Dedekind Cut Is Really Dedekind Cut

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A description/proof of that minus Dedekind cut Of Dedekind cut is really Dedekind cut

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of Dedekind cut.
• The reader admits the proposition that for any 2 Dedekind cuts, there is a rational Dedekind cut between them.

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Target Context

• The reader will have a description and a proof of the proposition that for any Dedekind cut, its minus Dedekind cut is really a Dedekind cut.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any Dedekind cut $r$, the minus Dedekind cut, $-r:=\{q\in \mathbb{Q}|\mathrm{\exists }{q}^{\prime }q<q^{\prime }\text{ and }-q^{\prime }\notin r\}$, is really a Dedekind cut.

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2: Proof

Let us prove that $-r\ne \mathrm{\varnothing }$ and $-r\ne \mathbb{Q}$. As $r\ne \mathbb{Q}$, there is a rational number, $q_1\in \mathbb{Q}$, such that $q_1\notin r$. For any $q_2\in \mathbb{Q}$ such that $q_1<q_2$, $q_2\notin r$. $-q_2\in -r$? Is there a $-q_3\in \mathbb{Q}$ such that $-q_2<-q_3$ and $--q_3=q_3\notin r$? Yes, as $-q_3$ can be taken to be $-q_3=-q_1$. As $r\ne \mathrm{\varnothing }$, there is a $q_1\in \mathbb{Q}$ such that $q_1\in r$. $-q_1\notin -r$? For any $-q_2\in \mathbb{Q}$ such that $-q_1<-q_2$, $--q_2=q_2<q_1\in r$, so, $--q_2\in r$. So, yes.

If $q_1\in -r$, for any $q_2\in \mathbb{Q}$ such that $q_2<q_1$, $q_2\in -r$, obviously.

Let us prove that $-r$ has no largest element. Let us suppose that $-r$ had the largest element, $-q_1$. Let us denote the Dedekind cut of $q_1$ by $\widetilde{q_1}$, which would mean that $\widetilde{q_1}=\{q\in \mathbb{Q}|q<q_1\}$. $-q_1\notin -\widetilde{q_1}$, because for any $-q_2\in \mathbb{Q}$ such that $-q_1<-q_2$, $--q_2=q_2<--q_1=q_1$, so, $--q_2\in \widetilde{q_1}$. As $-q_1\in -r$, $r\subset \widetilde{q_1}$, because $--q_2\notin r$ while $--q_2\in \widetilde{q_1}$ for a $q_2$. There would be a rational number, $q_3\in \mathbb{Q}$ such that $r\subset \widetilde{q_3}\subset \widetilde{q_1}$, by the proposition that for any 2 Dedekind cuts, there is a rational Dedekind cut between them. Would be $-q_3\in -r$? Would there be $-q_4\in \mathbb{Q}$ such that $-q_3<-q_4$ and $--q_4=q_4\notin r$? As $r\subset \widetilde{q_3}$, there would be a $q_4\in \mathbb{Q}$ such that $q_4\notin r$ and $q_4\in \widetilde{q_3}$, which would mean that $q_4<q_3$, so, $-q_3<-q_4$. So, yes. But $q_3<q_1$, so, $-q_1<-q_3$, a contradiction against $-q_1$'s being the largest element.

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References

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