June | 2015 | A kind of library

Update

June 29, 2015

So, we moved to Ann Arbor.

We visited the Hands-On Museum (thanks to Zach and Diana, and congratulations!),

and attended the Math Reviews Summer picnic.

I started today.

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3 Comments | Life | Tagged: Francisco, Isabel | Permalink
Posted by andrescaicedo

You are currently browsing the A kind of library blog archives for June, 2015.
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- Answer by Andrés E. Caicedo for Question about Jech's proof of V = L implies GCH June 3, 2019
  You assume $\omega_\alpha\subseteq M$ and $X\in M$ so that $X$ belongs to the transitive collapse of $M$ (because if $\pi$ is the collapsing map, $\pi(X)=\pi[X]=X$. You assume $|M|=\aleph_\alpha$ so that the transitive collapse of $M$ has size $\aleph_\alpha$. Since you also have that this transitive collapse is of the form $L_\beta$ for some $\beta$, it fol […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Smallest $\beta$ such that it is provable that $2^{\aleph_\beta} > 2^{\aleph_0}$ January 16, 2019
  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 […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Arguments against large cardinals October 29, 2010
  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 […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Nonessential use of large cardinals December 10, 2010
  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 […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for What definitions were crucial to further understanding? December 3, 2018
  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 […]
  Andrés E. Caicedo
Math.StackExchange
- Answer by Andrés E. Caicedo for What is the consistency strength of generalized failure of the continuum hypothesis. April 14, 2019
  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 […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Is "integrability" equivalent to "having antiderivative"? March 9, 2014
  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 […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Meaning of $0^{¶}$ in set theory May 2, 2019
  $0^¶$ 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 […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for ¿Is, for $1\le n May 1, 2019
  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 […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Does $2^{O(n)} $ mean $O(2^n)$? If not, what does $2^{O(n)} $ mean? April 24, 2019
  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 […]
  Andrés E. Caicedo
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