Answer by Abolfazl Tarizadeh for On the universal property of the completion...
The universal property of the completion of an ordered field as stated in Cohn's book (I mean, P.M. Cohn, Basic Algebra, Theorem 8.7.1) is true if we replace the assumption "each dense order embedding...
View ArticleAnswer by Emil Jeřábek for On the universal property of the completion of an...
François G. Dorais has already described in detail the construction of completion using cuts, but let me add another perspective that might look more algebraically-minded.The following are equivalent...
View ArticleAnswer by François G. Dorais for On the universal property of the completion...
Sine Emil seems to be rather busy, here is an outline of the Dedekind-style approach to completing an ordered field.$\newcommand{\cut}[1]{\langle#1\rangle}$A good cut in $K$ is a pair $\cut{L,R}$ such...
View ArticleAnswer by Pete L. Clark for On the universal property of the completion of an...
I wrote up the material on sequential completion of an ordered field. It appears in Chapter 16 of these notes. My perspective was that of Francois Dorais's answer: namely, the correction of Cohn's...
View ArticleAnswer by Niels J. Diepeveen for On the universal property of the completion...
Note: This answer has been made redundant by the latest version of theother answer. I am leaving it in place only to save the discussions attached to it.You seem to have fully answered almost all of...
View ArticleAnswer by François G. Dorais for On the universal property of the completion...
The statement can be corrected by adding one word:Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$...
View ArticleOn the universal property of the completion of an ordered field
I have been trying to write up some notes on completion of ordered fields, ideally in the general case (i.e., not just completing $\mathbb{Q}$ to get $\mathbb{R}$ but considering the completion via...
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