Timeline for Cauchy completeness of the real closure

6 events

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Jun 29, 2015 at 9:05	comment	added	Emil Jeřábek		Right. Using these two operators, we can construct from any ordered field $k$ the four fields $\mathcal R(k)$, $\tilde k$, $\mathcal R(\tilde k)$, and $\widetilde{\mathcal R(k)}=\widetilde{\mathcal R(\tilde k)}=\mathcal R(\widetilde{\mathcal R(k)})$. There are natural embeddings $k\to\tilde k\to\mathcal R(\tilde k)$, $k\to\mathcal R(k)\to\mathcal R(\tilde k)$, and $\mathcal R(\tilde k)\to\widetilde{\mathcal R(k)}$, but in general all these may be strict extensions, and there are no other embeddings.
Jun 28, 2015 at 23:38	vote	accept	nombre
Jun 28, 2015 at 23:37	comment	added	nombre		(I had been trying to prove that $\mathcal{R}(\widetilde{k}) = \widetilde{\mathcal{R}(k)}$ (where $\widetilde{.}$ denotes the completion or an ordered field) but I could only get dense embeddings $\mathcal{R}(k) \rightarrow \mathcal{R}(\widetilde{k})$ or $\mathcal{R}(\widetilde{k}) \rightarrow \widetilde{\mathcal{R}(k)}$, that's why I was interested in $\mathcal{R}(\widetilde{k})$ being complete.)
Jun 28, 2015 at 21:58	history	edited	Emil Jeřábek	CC BY-SA 3.0	addendum
Jun 28, 2015 at 21:33	history	edited	Emil Jeřábek	CC BY-SA 3.0	On second thoughts, I’m not sure finite extensions preserve completeness if the value group is not Z. Also include a few links to disambiguate the terminology.
Jun 28, 2015 at 19:55	history	answered	Emil Jeřábek	CC BY-SA 3.0