Timeline for Taking a proper class as a model for Set Theory
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Mar 13, 2018 at 23:30 | comment | added | Noah Schweber | By contrast, of course, every model of second-order ZFC is well-founded, but that's a very different thing. (Similarly, second-order Peano arithmetic has no nonstandard models, but first-order Peano arithmetic has lots by the compactness theorem.) | |
| Mar 13, 2018 at 23:29 | comment | added | Noah Schweber | @JustinaColmena Your most recent comment again makes the mistake of conflating second-order ZFC and (first-order) ZFC. By the compactness theorem, there is no way to rule out ill-founded models in a first-order way. The axiom of regularity prevents the existence of descending $\in$-sequences which are definable inside the model, but this is only "internal well-foundedness." The situation here is identical to that of nonstandard models of arithmetic. | |
| Mar 13, 2018 at 21:19 | comment | added | Andreas Blass | Unfortunately, the latest comment, claiming that the membership relation of a model of ZFC has to be well-founded, is another error. I agree, though, with @JohannesHahn that the post doesn't need to be deleted. I've seen the same errors elsewhere, so it may be useful for some readers to see them here along with the comments and downvotes, which (I think) suffice to make it clear that they are errors. | |
| Mar 13, 2018 at 21:01 | comment | added | Justina Colmena | @JohannesHahn if $\epsilon$ is not a well-founded relation, then a model $U,\epsilon$ cannot be a model of $ZFC$, because the axioms of $ZFC$ require $\in$ to be well-founded. | |
| Mar 13, 2018 at 20:42 | comment | added | Noah Schweber | (Sorry to keep spamming comments, but further on the universe issue, this post may be helpful.) | |
| Mar 13, 2018 at 20:38 | comment | added | Asaf Karagila♦ | @JohannesHahn: How is a post which only introduces mistakes, where the person posting insists being correct is useful for anything? | |
| Mar 13, 2018 at 20:37 | comment | added | Johannes Hahn | @AsafKaragila There is no button for "Not a great post, but doesn't necessarily need to be deleted". That's what downvotes are for as I understand the system. | |
| Mar 13, 2018 at 20:30 | comment | added | Noah Schweber | (In my first comment above, it should say "$L_\alpha^M$" rather than "$L_\alpha$," of course (the latter doesn't really make sense, although it's clear what's meant). A bit too late for me to edit it, unfortunately.) | |
| Mar 13, 2018 at 20:24 | comment | added | Noah Schweber | Meanwhile, your claim "If a model U of ZFC exists within the universe V of ZFC, then its cardinality is "inaccessible" with respect to the universe V" is false as observed above. It's not even true for models of the form $V_\alpha$ - not every worldly cardinal is inaccessible (assuming they exist). I believe you're thinking of universes in the sense of category theory, but those (as Asaf says) are really about second-order ZFC in a precise (e.g. in a universe, powersets need to be computed correctly). | |
| Mar 13, 2018 at 20:21 | comment | added | Noah Schweber | A proof of the fact Johannes mentioned: let $M$ be a model of ZFC + Con(ZFC) + "There is a standard model of ZFC," let $\alpha$ be the thing $M$ thinks is the least height of a standard model of ZFC, and consider $L_\alpha$ ... | |
| Mar 13, 2018 at 20:21 | comment | added | Asaf Karagila♦ | @JohannesHahn: How does this post "Looks OK"? | |
| Mar 13, 2018 at 19:49 | comment | added | Johannes Hahn | A model $(U,\epsilon)$ is isomorphic to one of the form $(U',\in)$ (a "standard model") iff $\epsilon$ is a well-founded relation. This is not necessarily satisfied for an arbitrary model. (And that's true even though the model believes itself to be well-founded. It can be mistaken about that!) Even worse: if ZF(C) is consistent, then $Con(ZFC)+$"There is no standard model" is also consistent, i.e. it is perfectly possible that, while there are models of ZFC, all of them are ill-founded. | |
| Mar 13, 2018 at 19:43 | review | Low quality posts | |||
| Mar 13, 2018 at 20:03 | |||||
| Mar 13, 2018 at 19:39 | comment | added | Asaf Karagila♦ | Again? When did I confuse things unnecessarily before? And no, you are absolutely wrong. But it seems impossible to get you to admit that, so I'm going to stop trying. | |
| Mar 13, 2018 at 19:30 | comment | added | Justina Colmena | "Second-order ZFC" -- No. You are confusing things unnecessarily again. Of course another relation may be used, but if there is a model then it may be represented by the standard relation $\in$ on a set $U$, (where $U\in V$ and $U\subsetneqq V$.) | |
| Mar 13, 2018 at 18:40 | comment | added | Asaf Karagila♦ | This does not add to the answers already appearing, with the exception that it adds mistakes (e.g. inaccessibility is only obtained when one restricts to second-order ZFC; being a model of ZFC need not be using the real $\in$ of $V$, but rather it can be using some arbitrary binary relation on the set $U$, etc.) | |
| Mar 13, 2018 at 18:06 | review | First posts | |||
| Mar 13, 2018 at 18:07 | |||||
| Mar 13, 2018 at 18:04 | history | answered | Justina Colmena | CC BY-SA 3.0 |