Timeline for Non-Archimedean non-standard models for R
Current License: CC BY-SA 3.0
6 events
when toggle format | what | by | license | comment | |
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Apr 22, 2016 at 7:07 | comment | added | Antoine Ducros | And I disagree with ACL: $F(t)$ cannot be real closed. For instance, $t$ will never be a square in $F(t)$. | |
Apr 22, 2016 at 7:00 | comment | added | Antoine Ducros | My construction is perhaps more explicit, and I can make it completely explicit (neither choice nor compactness involved): embed $F(t)$ into $F((s))$ with $s=1/t$ , and set $ K=\bigcup_n F((s^{1/n}))$. Equip $K$ with the ordering extending that of $F$ for which $s$ is positive and smaller than every positive rational number; this ordering extends that of $F(t)$, and makes $K$ a real closed field. Now you may take for $S$ the algebraic closure of $F(t)$ inside $K$. | |
Apr 22, 2016 at 6:11 | comment | added | ACL | Is it so different from the other answer? If you add a constant $t$ to the language with the axioms that it be greater than any integer $x$, then $t$ is transcendental and any model contains $F(t)$ — the simplifying point being that $F(t)$ is already real closed. | |
Apr 21, 2016 at 20:49 | comment | added | Emil Jeřábek | @GeraldEdgar While I agree with the general sentiment, the compactness theorem for countable theories does not need the axiom of choice. | |
Apr 21, 2016 at 20:44 | comment | added | Gerald Edgar | I think it is more satisfying to write down a concrete model like this. Rather than depend on "compactness" which gives you a model out in "Axiom of Choice Land". | |
Apr 21, 2016 at 20:26 | history | answered | Antoine Ducros | CC BY-SA 3.0 |