According to these slides, the axiom $V = \mathrm{Ultimate} \,L$ has the following consequences (p. 55):

  1. It implies the Continuum Hypothesis.

  2. It reduces all questions of set theory to axioms of strong infinity (which are with intrinsic justification).

  3. Provides an axiomatic foundation for set theory which is immune to independence by Cohen's method.

If true, this sounds totally amazing. However, I'm skeptical that (2) is really an accurate description of one of the consequences of $V = \mathrm{Ultimate} \,L.$ In particular, I was under the impression that there's no currently-accepted definition of what makes an axiom a large cardinal axiom.

Question. Does anyone here understand enough of the technical details of Woodin's proposal to be able to comment on the accuracy of the above statements, especially (2)?

  • $\begingroup$ Compare (2) to the last sentence of Woodin's 2010 ICM talk (he adds "arguably"). logic.harvard.edu/EFI_Woodin_StrongAxiomsOfInfinity.pdf $\endgroup$ – literature-searcher Apr 30 at 6:48
  • 5
    $\begingroup$ It is a heuristic claim. Think of $V=L$ first. We know of no statement independent of $V=L$ that is not settled by large cardinal axioms. Woodin's claim is an extrapolation (and recall that the ambient theory is rich in large cardinals). One could argue that the lack of examples may well be a limitation of our methods and nothing else, but Woodin is saying that this is not the case (and he has advanced some arguments justifying this). Again, think of it as a guiding heuristic. If you disagree, it is then an invitation to develop something truly revolutionary. $\endgroup$ – Andrés E. Caicedo Apr 30 at 11:56
  • $\begingroup$ what Ultimate($L$) is supposed to be? $\endgroup$ – Zuhair Al-Johar May 2 at 9:49
  • $\begingroup$ @ZuhairAl-Johar, it's meant to be a variant on $L$ (Godel's constructible universe) that, unlike $L$, is consistent with the existence of very large cardinals. $\endgroup$ – goblin May 2 at 15:04
  • 1
    $\begingroup$ @ZuhairAl-Johar Way, way, way, way above. $\endgroup$ – Noah Schweber May 6 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.