A description/proof of that minus Dedekind cut Of Dedekind cut is really Dedekind cut
Topics
About: set
The table of contents of this article
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.
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.
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.
Main Body
1: Description
For any Dedekind cut \(r\), the minus Dedekind cut, \(-r:= \{q \in \mathbb{Q}\vert \exists q' q \lt q' \text{ and } -q' \notin r\}\), is really a Dedekind cut.
2: Proof
Let us prove that \(-r \neq \emptyset\) and \(-r \neq \mathbb{Q}\). As \(r \neq \mathbb{Q}\), 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 \(q_1 \lt q_2\), \(q_2 \notin r\). \(- q_2 \in - r\)? Is there a \(- q_3 \in \mathbb{Q}\) such that \(- q_2 \lt - q_3\) and \(- - q_3 = q_3 \notin r\)? Yes, as \(- q_3\) can be taken to be \(- q_3 = - q_1\). As \(r \neq \emptyset\), 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 \(- q_1 \lt - q_2\), \(- - q_2 = q_2 \lt q_1 \in r\), so, \(- - q_2 \in r\). So, yes.
If \(q_1 \in - r\), for any \(q_2 \in \mathbb{Q}\) such that \(q_2 \lt q_1\), \(q_2 \in - r\), 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 \(\tilde{q_1}\), which would mean that \(\tilde{q_1} = \{q \in \mathbb{Q}\vert q \lt q_1\}\). \(- q_1 \notin - \tilde{q_1}\), because for any \(- q_2 \in \mathbb{Q}\) such that \(- q_1 \lt - q_2\), \(- - q_2 = q_2 \lt - - q_1 = q_1\), so, \(- - q_2 \in \tilde{q_1}\). As \(- q_1 \in - r\), \(r \subset \tilde{q_1}\), because \(- - q_2 \notin r\) while \(- - q_2 \in \tilde{q_1}\) for a \(q_2\). There would be a rational number, \(q_3 \in \mathbb{Q}\) such that \(r \subset \tilde{q_3} \subset \tilde{q_1}\), 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 \(- q_3 \lt - q_4\) and \(- - q_4 = q_4 \notin r\)? As \(r \subset \tilde{q_3}\), there would be a \(q_4 \in \mathbb{Q}\) such that \(q_4 \notin r\) and \(q_4 \in \tilde{q_3}\), which would mean that \(q_4 \lt q_3\), so, \(- q_3 \lt - q_4\). So, yes. But \(q_3 \lt q_1\), so, \(- q_1 \lt - q_3\), a contradiction against \(- q_1\)'s being the largest element.
