I have glanced over this post but it the answer dealt mainly with $\sf{ZFA}$ and $\sf{ZFC}$. There was no mention of $\sf{NBG}$, $\sf{SP}$ or $\sf{TG}$. So my question is, can the same claim be proved in the set theories that I have mentioned?
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$\begingroup$ What is SF/SP here? $\endgroup$– Asaf Karagila ♦Dec 14, 2015 at 4:24
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$\begingroup$ @AsafKaragila:Scott-Potter Set Theory. $\endgroup$– user170039Dec 14, 2015 at 5:13
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$\begingroup$ Thanks unfortunately that Wikipedia article does not include the axiom of SP, but rather ZU. In any case it seems the axiom of choice is not included there in any reasonable way that you can actually prove something like "every set can be linearly ordered". But since I never dabbled with that set theory, I will leave this to someone else to confirm. $\endgroup$– Asaf Karagila ♦Dec 14, 2015 at 5:19
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Well, it generally depends whether you take the axiom of choice for sets in your coffee. Err... class set theory.
In the case of $\sf NBG$ many axiomatizations do include at least the axiom of choice for sets, and then every set can be well ordered, so in particular totally ordered. If the axiomatization does not include the axiom of choice, or something which can prove it then the answer is negative of course.
In the case of $\sf TG$ things are simpler since Tarski's axiom implies the axiom of choice already. So again every set is well orderable, so every set can be totally ordered.
answered Dec 14, 2015 at 4:29
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$\begingroup$ Is it true in $\sf{NF}$ also? $\endgroup$– user170039Dec 14, 2015 at 5:14
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$\begingroup$ It's unclear how much of the axiom of choice NF can prove, disprove or is it at all consistent. So the question is open, to the best of my knowledge. $\endgroup$– Asaf Karagila ♦Dec 14, 2015 at 5:17
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1$\begingroup$ (To clarify Asaf's comment: the full axiom of choice fails in NF, and as a cute corollary, NF proves the existence of infinite sets. This is a result of Specker.) $\endgroup$ May 24, 2017 at 22:20