I received a letter on Friday from Bob Kustra, the president of BSU.
“I am pleased to inform you that you are being promoted to the rank of Associate Professor and being awarded tenure.”
A kind of library
Andrés E. Caicedo
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Mathoverflow- Answer by Andrés E. Caicedo for Is the Riemann Hypothesis equivalent to a $\Pi_1$ sentence? July 14, 2010Yes. This is a consequence of the Davis-Matiyasevich-Putnam-Robinson work on Hilbert's 10th problem, and some standard number theory. A number of papers have details of the $\Pi^0_1$ sentence. To begin with, take a look at the relevant paper in Mathematical developments arising from Hilbert's problems (Proc. Sympos. Pure Math., Northern Illinois Un […]Andrés E. Caicedo
- Two elementary inequalities for real-valued polynomials March 4, 2016I am looking for references discussing two inequalities that come up in the study of the dynamics of Newton's method on real-valued polynomials (in one variable). The inequalities are fairly different, but it seems to make sense to ask about both of them in the same post. Most of the details below are fairly elementary, they are mostly included for comp […]Andrés E. Caicedo
- A translation of the Cantor set contained in the irrationals February 22, 2014Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Is the Martin's axiom number $\mathfrak m$ regular January 14, 2016Not necessarily. That $\mathfrak m$ is consistently singular is proved in MR0947850 (89m:03045) Kunen, Kenneth. Where $\mathsf{MA}$ first fails. J. Symbolic Logic 53(2), (1988), 429–433. There, Ken shows that $\mathfrak{m}$ can be singular of cofinality $\omega_1$. (Both links above are behind paywalls.)Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Are there any non-linear solutions of Cauchy's equation ($f(x+y)=f(x)+f(y)$) without assuming the Axiom of Choice? March 6, 2011Ignas: It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here. […]Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Is the Riemann Hypothesis equivalent to a $\Pi_1$ sentence? July 14, 2010
- Math.StackExchange
- Answer by Andrés E. Caicedo for Injection - Bijection function Question September 7, 2016There are several nice proofs of the result. The most intuitive I'm aware of is the following: We may as well assume that $A$ and $B$ are disjoint. Consider the directed graph whose set of vertices is $A\cup B$, in which you add an edge from $a$ to $b$ precisely if $a\in A$, $b\in B$, and $f(a)=b$, or $a\in B$, $b\in A$, and $g(a)=b$. Now consider the c […]Andrés E. Caicedo
- Answer by Andrés E. Caicedo for What does it mean to say that a forcing "collapses cardinals"? June 4, 2014A forcing collapses cardinals iff (by definition) some cardinal of the ground model is no longer a cardinal in the forcing extension. Naturally, this means that there is some $\kappa$ in the ground model whose cardinality in the extension is strictly smaller than $\kappa$ (e.g., let $\kappa$ be the first cardinal that witnesses the definition above). Note th […]Andrés E. Caicedo
- Answer by Andrés E. Caicedo for How does HOD differ from L? August 22, 2016$\mathrm{HOD}$ always contains $L$ because any inner model contains $L$, by absoluteness. How easy it is to exhibit a difference really depends on your background. For instance, $0^\sharp$, if it exists, is a real that always belongs to $\mathrm{HOD}$ but is not in $L$. If you are not too comfortable with large cardinals, but know forcing, you may enjoy prov […]Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Why is $0^\sharp$ not definable in $ZFC$? August 10, 2016The classical definition of $0^\sharp$ is as (the set of Gödel numbers of) a theory, namely, the unique Ehrenfeucht-Mostowski blueprint satisfying certain properties (coding indiscernibility). This is a perfectly good definition formalizable in $\mathsf{ZFC}$, but $\mathsf{ZFC}$ or even mild extensions of $\mathsf{ZFC}$ are not enough to prove that there are […]Andrés E. Caicedo
- Answer by Andrés E. Caicedo for What's meaning of the inverted Greek letter iota “ι” in Principia Mathematica I* 14? August 8, 2016This is the descriptor operator. $(\iota x)\varphi x $ is the unique $x $ with the property specified by $\varphi $ (should it be the case that, indeed, there is precisely one such $x $). The Wikipedia entry on Principia has a very decent explanation of their notation.Andrés E. Caicedo
- Answer by Andrés E. Caicedo for Injection - Bijection function Question September 7, 2016
Congratulations! Here’s to a transfinity of good research in the future!
Thanks!
It is a little late, but congratulations!
Not late at all. Many thanks!
Congratulations!!!
Thank you, John!
Congratulations, Andres!
Many thanks, Joel!