Search Results

I hold in my hand a printout of Jensen's article from Annals of Mathematical Logic 4 (1972) named as in the title of the question. However the quality of the .pdf I found was very bad and it is very …

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 $3\times A$ …

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 …

A friend of mine and myself (both grad students with a relatively decent set theoretic background) want to venture into the universe of inner models. [pun intended :-)] I would very much like to get …

Let me add that the answer to (3) is negative, if you are willing to assume that inaccessible cardinals are consistent. We say that $\omega_1$ is inaccessible to reals if for every real $x$, $\omega_ …

This is difficult to prove using forcing, for one simple reason. If $M\models\sf ZFC$, and $G$ is an $M$-generic filter (for some forcing notion), then $M[G]\models\sf ZFC$. In other words, the only …

The comments by The User and Joel David Hamkins refer to a previous version of the answer which contained a mistake. The current version is completely disjoint of the previous one, and the comments no …

Yes. The perfect set property will ensure every set of reals is countable or has size continuum. Solovay, R.M., A model of set-theory in which every set of reals is Lebesgue measurable, Ann. Math. …

You might want to look at Grigorieff's paper which shows that every symmetric model is of the form $\mathrm{HOD}(V\cup X)^{V[G]}$, where $X\in V[G]$. In the case of Cohen's model, it is easy to argue …

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 …

If by the Hilbert cube you mean only $[0,1]^\mathbb N$ then the answer is yes. There is such proof, you can find it in Herrlich's The Axiom of Choice as Theorem 3.13. If you mean the general case of …

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 …

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 …

First of all, the axiom of countable choice says that given a countable family of non-empty sets, you can choose from each set simultaneously. If you want to choose from one, then from another, then f …

Let me give an alternative ending to Noah's road map. The splitting point is at symmetric models. After you've understood the basics of forcing well, you can switch to Kanamori's The Higher Infinite. …