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 […]
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 […]
There are integrable functions that are not derivatives: Any function that is continuous except at a single point, where it has a jump discontinuity, is an example. (Derivatives have the intermediate value property.) More interestingly, we can ask whether the existence of an antiderivative ensures integrability. The answer depends on what integral you are co […]
$0^¶$ is the sharp for an inner model with a strong cardinal in the same sense that $0^\dagger$ is the sharp for an inner model with a measurable cardinal. In terms of mice, this is the first mouse containing two overlapping extenders. The effect of this is that, by iterating its top measure throughout the ordinals, you extend the bottom extender in a variet […]
Clearly $\omega^\omega\le(\omega+n)^\omega$. Also, $(\omega+n)^\omega\le (\omega^2)^\omega=\omega^{(2\cdot \omega)}=\omega^\omega$, and the equality follows. If you do not feel comfortable with the move from $(\omega^2)^\omega$ to $\omega^{(2\cdot \omega)}$, simply note the left-hand side is $\omega\cdot\omega\cdot\omega\cdot\dots$, where there are $2\cdot\o […]
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 […]