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Aug 27, 2014 at 15:24 comment added Emil Jeřábek Note that the last step is nonobvious. For example, let $F$ be an ultraproduct of $\mathbb Q$ over a nonprincipal ultrafilter on $\omega$. Then $F$ is an ordered field, and every Dedekind cut on $\mathbb Q$ can be separated by an element of $F$ (due to its $\aleph_1$-saturation), but there is no embedding of $\mathbb R$ into $F$. In fact, $\mathbb Q$ is relatively algebraically closed in $F$, so $F$ doesn’t even contain a square root of $2$.
Aug 26, 2014 at 18:42 review Late answers
Aug 26, 2014 at 18:56
Aug 26, 2014 at 18:24 review First posts
Aug 26, 2014 at 19:51
Aug 26, 2014 at 18:21 history answered Edward Ross CC BY-SA 3.0