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 Can we have an infinite sequence of decreasing cardinality all terms of which have equal sized power sets? August 12, 2018
  This is consistent, at least under a rather tame large cardinal assumption. (One can also produce examples by manipulating Dedekind finite sets, but Asaf's answer addresses this. The answer here works even in the context of $\mathsf{DC}$.) For instance, see MR3612001. Conley, Clinton T.; Miller, Benjamin D. Measure reducibility of countable Borel equiva […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Are there known examples of sets whose power set is equal in size to power set of larger sets only in absence of choice? August 11, 2018
  Let me add some comments to Asaf's answer. We are looking for situations with $|X|
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Is $Z_3+{\rm AD}$ equiconsistent with $\text{ZFC+AD}^{L(\mathbb{R})}$ May 10, 2018
  The only reference I know for precisely these matters is the handbook chapter MR2768702. Koellner, Peter; Woodin, W. Hugh. Large cardinals from determinacy. In Handbook of set theory. Vols. 1, 2, 3, 1951–2119, Springer, Dordrecht, 2010. (Particularly, section 7.) For closely related topics, see also the work of Yong Cheng (and of Cheng and Schindler) on Harr […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Does constructing non-measurable sets require the axiom of choice? October 14, 2010
  As other answers point out, yes, one needs choice. The popular/natural examples of models of ZF+DC where all sets of reals are measurable are models of determinacy, and Solovay's model. They are related in deep ways, actually, through large cardinals. (Under enough large cardinals, $L({\mathbb R})$ of $V$ is a model of determinacy and (something stronge […]
  Andrés E. Caicedo
- Relative Densities September 12, 2013
  Throughout the question, we only consider primes of the form $3k+1$. A reference for cubic reciprocity is Ireland & Rosen's A Classical Introduction to Modern Number Theory. How can I count the relative density of those $p$ (of the form $3k+1$) such that the equation $2=3x^3$ has no solutions modulo $p$? Really, even pointers on how to say anything […]
  Andrés E. Caicedo
Math.StackExchange
- Answer by Andrés E. Caicedo for Clarification about Infinite Sums in Jech's Set Theory 3rd Edition August 12, 2018
  Using the regularity of $\kappa$ and the fact that the infinite cardinal $\lambda$ is less than $\kappa$, Jech notes that $$ \kappa^\lambda=\bigcup_{\alpha
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Formal definition of "hereditarily" August 7, 2018
  A set $x$ is hereditarily $\varphi$ if and only if every member of its transitive closure (including $x$ itself) has property $\varphi$. For instance, being hereditarily finite means not just that $x$ is finite, but also every member of $x$ is finite, and every member of every member of $x$, and so on. (Note that in the absence of foundation, this is a bit p […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Collection of measurable sets closed under countable unions implies existence of a set with maximal measure July 12, 2018
  Let $s$ be the supremum of the $\mu$-measures of members of $\mathcal G$. By definition of supremum, for each $n$, there is $G_n\in\mathcal G$ with $\mu(G_n)>s-1/n$. Letting $G=\bigcup_n G_n$, then $G\in \mathcal G$ since $\mathcal G$ is closed under countable unions, and $\mu(G)=s$, since it is at least $\sup_n\mu(G_n)$ but it is at most $s$ (by definiti […]
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Set theory ordinals problem Prove that if $b$ is a limit ordinal and $a June 27, 2018
  The result you are trying to prove is false. For example, if $a=\omega+1$ and $b=\omega+\omega$, then $a+b=\omega\cdot 3>b$. Here is what is true: first, the key result you should establish (by induction) is that An ordinal $\alpha>0$ has the property that for all $\beta
  Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Are there any new axioms being developed? June 5, 2018
  Very briefly: Yes, there are several programs being developed that can be understood as pursuing new axioms for set theory. For the question itself of whether pursuing new axioms is a reasonably line of inquiry, see the following (in particular, the paper by John Steel): MR1814122 (2002a:03007). Feferman, Solomon; Friedman, Harvey M.; Maddy, Penelope; Steel, […]
  Andrés E. Caicedo
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