Timeline for Standard model of ZFC
Current License: CC BY-SA 3.0
21 events
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| Apr 17, 2020 at 16:31 | comment | added | Jori | @JoelDavidHamkins But the arbitrary model you select when you say "take any model of ZFC", which you get from Con(ZFC), is not necessarily transitive. In your sentence "If it does have a transitive model, then it will have an ∈-minimal transitive model of ZFC", the "it" I bold-faced, what model does it refer to? The transitive model, or the original model? Maybe that clarifies it for me. | |
| Feb 5, 2020 at 23:29 | comment | added | mtg | @Jori: this might a little bit off-topic, but if these things about models-in-models are interesting to you, try to prove the following puzzling fact: every model of $ZFC$ contains a model of $ZFC$ as an element (that is, for every $M \models ZFC$ there is a set $N \in M$ such that $M$ believes $N$ is a first-order structure and $N \models ZFC$ (but $M$ might not know about the latter) ;) | |
| Jan 30, 2020 at 16:24 | comment | added | Joel David Hamkins | Yes, I think you've got it. But an $\in$-minimal element of a transitive set is also $\in$-minimal in the ambient universe, so the quibble is not necessary. | |
| Jan 30, 2020 at 13:32 | comment | added | Jori | @Joel The Cantor attic website is malfunctioning, and it exposes some of your data. Also: how do you get to "and this will be a model of ZFC having no transitive models of ZFC."? Did you mean to say that "If it does have a transitive model, then that model will have an $\in$-minimal transitive model of ZFC, and this will be a model of ZFC having no transitive models of ZFC."? Because the model you talk about is not transitive, so need not contain any supposed $\in$-smaller transitive model inside its $\in$-minimal transitive model. Just to check if I understand it correctly (I'm a noob). | |
| Jul 21, 2013 at 19:12 | comment | added | Joel David Hamkins | Yes, that is right. The standard model has the same proofs that $V$ does, and so since $V$ thinks Con(ZFC), it follows that M does also, and so we get Con(ZFC+Con(ZFC)). | |
| Jul 21, 2013 at 19:02 | comment | added | Leonard | Am I right to guess that $ \mathsf{SM} $ implies $ \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})) $ comes from the fact that $ \text{Con}(\mathsf{ZFC}) $ is an absolute $ \mathcal{L}_{\text{Set}} $-formula? This should be the final gap in my understanding. Thanks once again! | |
| Jul 21, 2013 at 18:52 | comment | added | Joel David Hamkins | Both are true, assuming the consistency of ZFC+Con(ZFC), since SM implies Con(ZFC) and also Con(ZFC+Con(ZFC)). | |
| Jul 21, 2013 at 18:46 | comment | added | Leonard | Thank you for your response, Joel. One more thing. Which of the two interpretations is correct: $ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \mathsf{SM} $ or $ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \text{Con}(\mathsf{ZFC} + \mathsf{SM}) $? | |
| Jul 21, 2013 at 12:11 | comment | added | Joel David Hamkins | @Leonard, you can't remove that assumption, since if $\text{ZF}+\text{Con}(\text{ZFC})$ is inconsistent, then $\text{ZF}\vdash\text{Con}(\text{ZFC})\to\text{anything}$. All nonprovability results take place under a consistency assumption, and you shouldn't think that we haven't proved the final non-provability result, since we have proved it under the relevant consistency assumption. Note that any assertion of the form $\text{ZFC}+\text{Con}(\text{ZFC})\not\vdash\psi$ outright implies that $\text{ZFC}+\text{Con}(\text{ZFC})$ is consistent. | |
| Jul 21, 2013 at 9:30 | comment | added | Leonard | The problem is: How can I remove the ‘if $ \mathsf{ZF} + \text{Con}(\mathsf{ZFC}) $ is consistent’ clause so that I get a proof of $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \nvdash \text{Con}(\mathsf{ZFC} + \mathsf{SM})? $$ Thank you for your time! | |
| Jul 21, 2013 at 9:27 | comment | added | Leonard | Proof Suppose instead that $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \vdash \text{Con}(\mathsf{ZFC} + \mathsf{SM}). $$ Then as $$ \mathsf{ZFC} \vdash \text{Con}(\mathsf{ZFC} + \mathsf{SM}) \rightarrow \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})), $$ we would obtain $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \vdash \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})), $$ which contradicts Gödel’s Second Incompleteness Theorem if $ \mathsf{ZF} + \text{Con}(\mathsf{ZFC}) $ is consistent. $ \blacksquare $ | |
| Jul 21, 2013 at 9:17 | comment | added | Leonard | These statements are what I see in most standard texts on set theory, and they somehow seem to mean $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \mathsf{SM}, $$ which I am unable to prove. I had a discussion with Asaf over at Math StackExchange about the proof of $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \text{Con}(\mathsf{ZFC} + \mathsf{SM}), $$ which we can rewrite as $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \nvdash \text{Con}(\mathsf{ZFC} + \mathsf{SM}). $$ | |
| Jul 21, 2013 at 8:56 | comment | added | Leonard | Hi Joel. Do the statements ‘The consistency of $ \mathsf{ZFC} $ does not imply the existence of a standard model of $ \mathsf{ZFC} $’ and ‘The existence of a standard model of $ \mathsf{ZFC} $ is strictly stronger than the consistency of $ \mathsf{ZFC} $’ mean $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \mathsf{SM}, $$ or do they mean $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \text{Con}(\mathsf{ZFC} + \mathsf{SM})? $$ | |
| Mar 17, 2013 at 15:47 | vote | accept | Jaykov Foukzon | ||
| Mar 17, 2013 at 15:46 | vote | accept | Jaykov Foukzon | ||
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| Mar 17, 2013 at 15:46 | vote | accept | Jaykov Foukzon | ||
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| Mar 17, 2013 at 15:46 | vote | accept | Jaykov Foukzon | ||
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| Mar 17, 2013 at 15:46 | vote | accept | Jaykov Foukzon | ||
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| Mar 16, 2013 at 20:38 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 604 characters in body; added 50 characters in body
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| Mar 16, 2013 at 19:58 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Added proof of consistency that there is no transitive model of ZFC
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| Mar 16, 2013 at 19:36 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |