Timeline for The type of nondefinable elements-2

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Dec 4, 2013 at 15:55	comment	added	Joel David Hamkins		Your question seems to be a moving target. If the definition of $X$ has parameters, it doesn't make sense to use the same definition in $\mathcal{N}$, unless those parameters are also there. And if you don't say something about how the theory of $\mathcal{M}$ and $\mathcal{N}$ are related, then basically anything can happen, since the definition might be: If CH holds, then one thing, but otherwise something else. In this case, $\mathcal{M}$ and $\mathcal{N}$ can think totally different things about $X$.
Dec 4, 2013 at 15:51	comment	added	user38200		I mean that $X$ is definable from ordinals and reals in $\mathcal{M}$, and that a $X$ is definable in $\mathcal{N}$ (where $V=HOD$ in $\mathcal{N}$) by the same formula. And $X$ is nonempty in both models
Dec 4, 2013 at 15:46	comment	added	Joel David Hamkins		I don't understand your question. What do you mean by "$X$ is nonempty in HOD"? And what do you mean when you say that $\mathcal{M}$ is another model of set theory? Are you referring to the HOD of $\mathcal{M}$? And $x\in\mathcal{M}$?
Dec 4, 2013 at 15:43	comment	added	user38200		If we know that $X$ is nonempty in HOD ($X$ infinite), and ask if the type of nondefinable elements of $X$ over the theory of $\mathcal{M}$ (where $\mathcal{M}$ is another model of set theory) is principal, do we have an answer?
Dec 4, 2013 at 15:00	vote	accept	user38200
Dec 4, 2013 at 14:44	history	edited	Joel David Hamkins	CC BY-SA 3.0	deleted 7 characters in body
Dec 4, 2013 at 14:00	history	edited	Joel David Hamkins	CC BY-SA 3.0	added 160 characters in body
Dec 4, 2013 at 13:42	history	edited	Joel David Hamkins	CC BY-SA 3.0	added 160 characters in body
Dec 4, 2013 at 13:28	history	answered	Joel David Hamkins	CC BY-SA 3.0