- #1
psie
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- TL;DR
- I'm reading about the construction of the reals from Dedekind cuts in Rudin's PMA. This is the proof of Theorem 1.19 in the appendix of chapter 1. Specifically, I'm on step 4, where the author tries to show ##R## (the real field) has an additive inverse.
We have ##R##, the set of Dedekind cuts which is an ordered set by proper inclusion. Fix an ##\alpha\in R## and let ##\beta## be the set of all ##p## with the following property: $$\text{there exists }r>0\text{ such that }-p-r\notin\alpha.$$
Rudin shows ##\beta\in R## and that ##\alpha+\beta=0^\ast=\{\text{the negative rationals}\}##. When proving ##\supset## in ##\alpha+\beta=0^\ast##, the author picks ##v\in 0^\ast## and puts ##w=-v/2##. Then he claims that ##w>0## and there is an integer ##n## such that ##nw\in\alpha## but ##(n+1)w\notin\alpha##. He says this depends on the archimedean property. How?
I gather we are looking at the set ##\{n:nw\in\alpha\}##, and we want to show it has a maximal element. It's been a long day here, so I don't immediately see how to show this. I think it suffices to show it is nonempty and bounded above. I struggle with both of these. I fail to see the significance of the archimedean property.
A Dedekind cut is any set ##\alpha\subset Q## with the following three properties:
(i) ##\alpha## is not empty and ##\alpha\neq Q##.
(ii) if ##p\in\alpha##, ##q\in Q## and ##q<p##, then ##q\in\alpha##.
(iii) if ##p\in\alpha##, then ##p<r## for some ##r\in\alpha##.
Rudin shows ##\beta\in R## and that ##\alpha+\beta=0^\ast=\{\text{the negative rationals}\}##. When proving ##\supset## in ##\alpha+\beta=0^\ast##, the author picks ##v\in 0^\ast## and puts ##w=-v/2##. Then he claims that ##w>0## and there is an integer ##n## such that ##nw\in\alpha## but ##(n+1)w\notin\alpha##. He says this depends on the archimedean property. How?
I gather we are looking at the set ##\{n:nw\in\alpha\}##, and we want to show it has a maximal element. It's been a long day here, so I don't immediately see how to show this. I think it suffices to show it is nonempty and bounded above. I struggle with both of these. I fail to see the significance of the archimedean property.
A Dedekind cut is any set ##\alpha\subset Q## with the following three properties:
(i) ##\alpha## is not empty and ##\alpha\neq Q##.
(ii) if ##p\in\alpha##, ##q\in Q## and ##q<p##, then ##q\in\alpha##.
(iii) if ##p\in\alpha##, then ##p<r## for some ##r\in\alpha##.