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Elliot Glazer's user avatar
Elliot Glazer's user avatar
Elliot Glazer's user avatar
Elliot Glazer
  • Member for 7 years, 7 months
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Why is inner model theory evidence for consistency of large cardinals?
@n901 All reasonable formulations of "all sets of reals are Lebesgue measurable" are equivalent in ZF, and prove the $\sigma$-additivity of $\lambda$ (see mathoverflow.net/a/393162/109573). We get a relative consistency proof over ZF by building the Solovay model relative to the Levy collapse of $\omega_{\omega}$ and adapting Solovay's argument to show $\lambda^*$ and $\lambda_*$ agree on subsets of the unit interval.
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Is there an elementary proof of a better result for the finite guessing-box puzzle?
Clarified my probabilistic formalization of "almost always wrong"
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Is there an elementary proof of a better result for the finite guessing-box puzzle?
I expanded the first part. Does the second part address what you were looking for regarding whether the players "almost always fail"?
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Cohen's model yet again
I have a strong interest in this question. Proof strategies for several problems I've worked on are hung up on verifying this conjecture or minor variants of it.
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Sequential continuity and the Axiom of Choice
Yes, that fails for the indicator function of an infinite Dedekind finite set $S:$ globally sequentially usco, sequentially continuous off $S,$ discontinuous at the condensation points of $S$ (which is a perfect set so is not contained in $S$). If you demanded $f$ be in some Baire class AND usco, it would be significantly harder to find a counterexample.
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Can there be a proper class of Dedekind-finite cardinals?
Has anyone checked if every set can be the image of a dually Dedekind-finite set?
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What is the least $\alpha$ such that $L_\alpha$ contains a non-measurable set
Regarding the cases mentioned in the last paragraph: $(2^{\omega})^L$ (as well as any $\omega$-model of $\mathrm{RCA}_0$) is either measure 0, maximally nonmeasurable (inner measure 0 and outer measure 1), or equal to $2^{\omega},$ since it is closed under dyadic translation and from a real $r \not \in A=(2^{\omega})^L,$ we can construct infinitely many disjoint translations of $A.$
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Turning linear ordering into well-ordering
@GabeGoldberg Assume $V=L,$ let $\kappa$ be inaccessible, let $G=\langle G_{\alpha} : \alpha < \kappa \rangle$ be Levy collapse generic. Let $R=\mathbb{R}^{L[G]}$ and define $f: R \rightarrow \kappa$ by setting $f(x)$ to be the least $\alpha$ such that $x \in L[G \restriction \alpha].$ Then $M=L(R, f)$ has $(R, \prec)$ as in the question, and every set of ordinals in $M$ is OD from $f$ and some $G \restriction \alpha,$ but $R \cap HOD^M(f, G \restriction \alpha) = R \cap L[G \restriction \alpha]$ is countable in $M.$ So no subset of $\omega_1$ codes an injection from $\omega_1$ to $R.$
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Must the number of smooth structures be countable or continuum?
Actually, from the equivalence of 4D smooth and PL manifolds, can isomorphism between two 4D smooth manifolds be expressed combinatorially via triangulations $T_i$ of $M_i$? Something like “there are simplices $s \in T_1, t \in T_2$ such that for all $n,$ there is a combinatorial isomorphism from the set of simplices within $n$ adjacencies of $s$ to those of $t$”? I’m out of my depth here geometrically, but that could confirm the equivalence relation to be Borel.
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Must the number of smooth structures be countable or continuum?
Naively this equivalence relation is analytic, because it posits the existence of a diffeomorphism. I don’t think there is a simple trick to proving this conjecture. In the 4D case, it will probably requiring actually overcoming the longstanding problem of finding nontrivial invariants for smooth structures.
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