Timeline for Can we have a theory that interpret $\sf ZFC$, proves existence of an infinite set, and yet prove all of its sets being Dedekind finite?
Current License: CC BY-SA 4.0
12 events
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Sep 13 at 15:34 | comment | added | Joel David Hamkins | It seems delicate to interpret ZFC in your theory, since the interpreted model will have Dedekind infinite sets, but to prevent those sets from giving rise to actual Dedekind infinite sets will mean that the interpretation must be using an equivalence relation (i.e. interpreting = nontrivially). | |
Sep 13 at 15:32 | comment | added | Joel David Hamkins | Interpreting ZFC is equivalent to interpreting ZFC+V=L, since L is definable in ZFC and so every interpreted model of ZFC also interprets a model of ZFC+V=L. Your theory doesn't seem to have its own L (unless this is really what you are asking?), so I think the L angle of your second question is irrelevant. | |
Sep 13 at 15:00 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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Sep 13 at 15:00 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, Ah! OK, I'll correct it. | |
Sep 13 at 14:10 | comment | added | Joel David Hamkins | But then you wouldn't be constructing $L$, but interpreting a model. | |
Sep 13 at 12:56 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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Sep 13 at 12:49 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, I thought $\omega$ could be interpreted, even though not provable in the system. | |
Sep 13 at 12:32 | comment | added | Joel David Hamkins | What do you mean by constructing $L$ if you don't have $\omega$? | |
Sep 13 at 12:26 | comment | added | Zuhair Al-Johar | @EmilJeřábek, OK, I've changed the title to better suit the question. | |
Sep 13 at 12:26 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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Sep 13 at 11:50 | comment | added | Emil Jeřábek | The answer to the question in the title, which is quite different from the question in the question, is yes: e.g., take $\mathrm{ZF_{fin}+Con_{ZFC}}$. | |
Sep 13 at 11:26 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |