Mathematical Reviews will have an open house at the upcoming AMS meeting here in Ann Arbor. There will be some snacks, at the very least. The event will take place Saturday October 20, 12:30-2:00 pm at 416 Fourth Street.

For some of the history of the building and a few pictures, see here.

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Perhaps the following may clarify the comments: for any ordinal $\delta$, there is a Boolean-valued extension of the universe of sets where $2^{\aleph_0}>\aleph_\delta$ holds. If you rather talk of models than Boolean-valued extensions, what this says is that we can force while preserving all ordinals, and in fact all initial ordinals, and make the contin […]

I do not know of any active set theorists who think large cardinals are inconsistent. At least, within the realm of cardinals we have seriously studied. [Reinhardt suggested an ultimate axiom of the form "there is a non-trivial elementary embedding $j:V\to V$". Though some serious set theorists found it of possible interest immediately following it […]

There is a fantastic (and not too well-known) result of Shelah stating that $L({\mathcal P}(\lambda))$ is a model of choice whenever $\lambda$ is a singular strong limit of uncountable cofinality. This is a consequence of a more general theorem that can be found in 4.6/6.7 of "Set Theory without choice: not everything on cofinality is possible", Ar […]

In set theory, definitely the notion of a Woodin cardinal. First, it is not an entirely straightforward notion to guess. Significant large cardinals were up to that point defined as critical points of certain elementary embeddings. This is not the case here: Woodin cardinals need not be measurable. If $\kappa$ is Woodin, then $V_\kappa$ is a model of set the […]

The first example that came to mind was MR0270881 (42 #5764) van der Waerden, B. L. How the proof of Baudet's conjecture was found. 1971 Studies in Pure Mathematics (Presented to Richard Rado) pp. 251–260 Academic Press, London. There, van der Waerden describes some of the history as well as his proof of his well-known theorem. Another example: MR224589 […]

A function $f:\mathbb N\to\mathbb R$ is $2^{O(n)}$ if and only if there is a constant $C$ such that for all $n$ large enough we have $f(n)\le 2^{Cn}$. We can think of the $O$ notation as decribing a family of functions. So, $2^{O(n)}$ would be the family of functions satisfying the requirements just indicated. In contrast, a function $f$ is $O(2^n)$ if and o […]

An interesting example of a different kind is any model where all sets of reals have the Baire property. In any such set the quotient of $\mathbb R$ by the Vitali equivalence relation is not linearly orderable. See here for a sketch. Examples of such models are Solovay's model where all sets of reals are Lebesgue measurable, or natural models of the axi […]

The precise consistency strength of the global failure of the generalized continuum hypothesis is somewhat technical to state. As far as I know, it has not been published, but I think we have a decent understanding of what the correct statement should be. The most relevant paper towards this result is MR2224051 (2007d:03082). Gitik, Moti Merimovich, Carmi. P […]

P=NP is an arithmetic statement: we can code the relevant deterministic Turing machines by numbers in a fairly explicit recursive way (which also explicitly involves codes for polynomial upper bounds), and then the equality between both classes can be discussed by discussing numerical properties of the indices involved in the coding, and using a specific NP- […]

Update: The problem has been solved. See below for the original answer, with the state of the art in 2013. In 2017, Ł. Grabowski, A. Máthé and O. Pikhurko showed in Measurable circle squaring, Ann. of Math. (2) 185 (2017), no. 2, 671–710, MR3612006, that Tarski's problem can be solved using pieces that are both Lebesgue and Baire measurable. Their proof […]