Timeline for Class forcings and elementary embeddings
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2017 at 22:37 | history | edited | Noah Schweber | CC BY-SA 3.0 |
edited body
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 29, 2015 at 21:09 | comment | added | Thomas Benjamin | (cont.) Sorry I didn't respond sooner. | |
| Jul 29, 2015 at 21:08 | comment | added | Thomas Benjamin | (cont.) Namely, the theory keeps everything true in $V$, relativised to a class predicate, and also asserts that there is a $V$-generic filter. [New Paragraph] The language has $\in$, constant symbols for every element of $V$, a predicate for $V$, and constant symbol $G$. The theory asserts: 1. The full elementary diagram of $V$, relativised to predicate. 2. The assertion that $V$ is a transitive proper class. 3. $G$ is a $V$-generic ultrafilter. 4.$ZFC$ holds, and the new universe is $V[G]$." Is this class model a model of set forcing, class forcing, or both? | |
| Jul 29, 2015 at 20:50 | comment | added | Thomas Benjamin | @NoahSchweber: It is a paraphrase of a statement made by Prof. Hamkins in a slide presentation titled "The set-theoretic multiverse: a model-theoretic philosophy of set theory." presented at the Philosophy and Model Theory Conference, Paris, June 2-5, 2010. It is regarding the Naturalist Account of Forcing. The slide reads as follows: "Suppose that $V$ is a universe of set theory and $\mathbb P$ is a notion of forcing. Then there is a class model of the theory expressing what it means to be the corresponding forcing extension. | |
| Jul 28, 2015 at 19:42 | comment | added | Noah Schweber | (But I might be misunderstanding what you're asking - what do you mean by "relativized to a predicate"?) | |
| Jul 28, 2015 at 19:41 | comment | added | Noah Schweber | @ThomasBenjamin Certainly not - for example, consider the trivial forcing extension $V[G]=V$. Truth in $V$ is very much not definable in $V[G]$. :P More generally, I suspect that it is much harder (if possible at all) to build a class- or set-generic extension $V[G]$ in which $Th(V)$ is definable. (Note that there's an apparent proof of impossibility: since the forcing relation is definable in $V$, shouldn't $Th(V[G], G)$ be definable in $V[G]$ from the parameter $G$ if $Th(V)$ is definable in $V[G]$? However, this breaks down since the forcing relation is not uniformly definable.) | |
| Jul 28, 2015 at 17:14 | comment | added | Thomas Benjamin | Does this mean that for certain class forcings, the full elementary diagram of $V$ (relativised to predicate) is not definable in $V[G]$? | |
| Jul 27, 2015 at 3:58 | comment | added | Joel David Hamkins | The same argument shows that your point about restricting the codomain always occurs: if $j:V\to M$ is elementary and $N\subsetneq M$, then $\text{ran}(j)$ is not contained in $N$, since $j(V_\alpha)=V_{j(\alpha)}^M$, which will not be in $N$ for large enough $\alpha$. | |
| Jul 27, 2015 at 3:46 | history | edited | Noah Schweber | CC BY-SA 3.0 |
added 56 characters in body
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| Jul 27, 2015 at 3:32 | comment | added | Joel David Hamkins | If $V$ and $V[G]$ are distinct, then you can never have $\text{ran}(j)\subset V$, since $j(V_\alpha)$ must be equal to $(V[G])_{j(\alpha)}$ by elementarity, and for large enough $\alpha$, this is not contained in $V$. | |
| Jul 27, 2015 at 2:38 | vote | accept | Thomas Benjamin | ||
| Jul 27, 2015 at 2:13 | history | answered | Noah Schweber | CC BY-SA 3.0 |