0
$\begingroup$

This is a sequel to my previous question with the same title. Let $N:=M[G]$ be the full Solovay model. And let $R$ be an archimedean complete real closed field in $M[G]$ (so $R$ has cardinality $2^{\aleph_0}$ in $M[G]$). Let $X$ be a subset of $R$ (or of $R \times R$) which is definable from a countable sequence of ordinals and $R$ (so $R$ is a parameter in the formula defining $X$).

Does is it hold that $X$ has the Baire property in $M[G]$?

Edit: In my case, $R \subset HOD$ and $R \subset K \in HOD$ where $K$ is an algebraically closed field, and the field structure on $R$ is induced by that of $K$. Does the result hold in this case?

$\endgroup$
  • $\begingroup$ We can assume without loss of generality that the underlying set of $R$ is $\omega^\omega$. Is the field structure of $R$ ordinal definable? $\endgroup$ – Asaf Karagila Feb 11 '14 at 12:39
  • $\begingroup$ That is blatantly false. $\endgroup$ – Asaf Karagila Feb 11 '14 at 12:42
  • $\begingroup$ Ah really? Why? $\endgroup$ – user38200 Feb 11 '14 at 12:44
  • $\begingroup$ Err, no, you're right. I was thinking about ordinal definable from a sequence of real numbers. $\endgroup$ – Asaf Karagila Feb 11 '14 at 12:45
  • 2
    $\begingroup$ @AsafKaragila: That’s a good point. And in fact, if we take for $R$ any of these subfields such that $[\mathbb C:R]=2$ (save $\mathbb R$ itself), and for $X$ the intersection of $R$ with the (Euclidean) unit disk in $\mathbb C$, then $X$ is obviously definable from $R$, but it does not have the property of Baire in $R$ (otherwise the isomorphism of $R$ to $\mathbb R$, and thus its extension to an automorphism of $\mathbb C$, would be continuous using a form of Pettis theorem). $\endgroup$ – Emil Jeřábek Feb 11 '14 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.