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The argument, given to me by Hugh Woodin over a drink in Barcelona in September 2016, is a nice retort to "AC is obviously false since there cannot be a well-ordering of the real numbers". It is argua …

In their book, "Introduction to Cardinal Arithmetic", Holz, Steffens, and Weitz define (on p.71 of the second edition) as follows. Assume that $\kappa$ is an infinite and $\lambda$ is an uncountable …

Not too much that I'm aware of. It comes up in the case of $\omega$, since that would imply that even if Dedekind finite sets exist, they can at least be mapped onto $\omega$ (equivalently, "the power …

It is known for a while now that $\frak p=t$, as a result of Malliaris-Shelah. The original paper draws from model theoretic methods. I've heard rumors that there was a proof which was purely set the …

Yes. This was essentially proved by Honsel and Forti in the 1980s by analysing a model that generalises the Cohen model (essentially, the one Monro used to show it can be consistent for Dedekind finit …

No. Most of the work of Pincus that involved these sort of iterated constructions, where one "pushes the counterexamples out" has never been reworked into modern terms. In my Ph.D. one of the reasons …

The idea is to mimic the permutation models as given in Jech. One can then ask, "Well, in Jech he chooses some set of objects in the full universe, and shows it has a support. But in forcing we don't …

Here are the answers you're looking for: No. No. Yes, no large cardinals needed! (Which explains the previous two answers.) Look no further than John Truss' paper: Truss, John, Models of set theory …

The term you might find in the literature is "weakly Dedekind finite", since a set that maps onto $\omega$ is weakly Dedekind infinite. I'd expect that you'll call these "strongly Dedekind finite". Al …

Let $g$ be the generic subset of $\omega$ added by the Grigorieff forcing, $G(F)$, where $F$ was a free ultrafilter. It is easy to see that for every $A\in F$, $g\cap A$ is non-empty, since if $f$ is …

Nothing is wrong with this version of choice. In $\sf ZF$, and the theories extending it, it is indeed equivalent to Global Choice, exactly by using Scott's trick. You just smooth it out by putting th …

Alex Lubotzky was a parliament member in the late 1990s in Israel. Menachem Magidor was, while being the president of the Hebrew university (a political position in itself), the head of the "preside …

You can find it, amongst other places in my write up: Zornian Functional Analysis or: How I Learned to Stop Worrying and Love the Axiom of Choice. If you need a source to cite, my money is on Handbo …

No, and here is a counterexample. Suppose that $|\Bbb R|<|[\Bbb R]^\omega|$, that is, there are more countable subsets of reals than reals. This is indeed possible, e.g. if all sets of Lebesgue measu …

I don't know if it's lesser known, but it is certainly not on par with some of the other results on this page. Theorem. (Doyle–Conway) Assume $\sf ZF$. If there is a bijection between $3\times A$ …