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Robert Furber's user avatar
Robert Furber's user avatar
Robert Furber's user avatar
Robert Furber
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Is $X\times X$ homeomorphic to $X$ for a space of probability measures?
@AlekseiKulikov Agreeing with David Gao, try thinking about the two-point space instead of $S$, where $\mathcal{M}_1(2) \times \mathcal{M}_1(2)$ is a square, but $\mathcal{M}_1(2 + 2)$ is a tetrahedron. In general $\mathcal{M}_1(X) \times \mathcal{M}_1(Y)$ is not a space of probability measures but the state space of the coproduct order-unit space $C(X) \oplus C(Y)$.
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Every tight $\tau$-additive finite measure is Radon
@rfloc Take $Z' = V \cup W \cup Z$. Then $Z' \cap S = Z \cap S$, so $Z' \in \mathcal{V}$, and $V \cup W \subseteq Z'$.
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How many well-orders of reals are there?
It may help you to know that "the number of well orders of $X$, up to isomorphism" is the definition of $|X|^{+}$, and always exists (in ZF). Using the axiom of choice, we can prove that "the next cardinal above $|X|$" exists and is equal to $|X|^{+}$, but without the axiom of choice we cannot prove that there is a "next", so this isn't suitable as a general definition of $|X|^{+}$.
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Nuclear vs Banach spaces: compactness properties
Wikipedia has a bit of a problem in certain places with people copying things from textbooks and papers without understanding. Unfortunately there is also the requirement of "no original research" that means you can't necessarily use your own expertise to fix it.
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Nuclear vs Banach spaces: compactness properties
@JochenWengenroth A necessary and sufficient condition is quasicompleteness, that every closed bounded set is complete.
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CH in non-set theoretic foundations
In general, using a (logically) weaker foundation means there are more possibilities for things to be consistent with it, so we shouldn't expect that a weaker foundation will decide something that is undecidable in a stronger theory. Of course, in another sense logically weaker foundations are "stronger" because it's harder to get an inconsistency out of them and we are less likely to need to change our proofs in the future if an inconsistency were discovered.
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Did Lebesgue like non-measurable set or not?
I think Lebesgue would have had to revise his opinions or arguments from his letter in view of Gödel's construction of $L$, but as far as I know he never commented on it. He didn't live to see the forcing era.
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Did Lebesgue like non-measurable set or not?
A translation of the letter can be found in an appendix of G. H. Moore's Zermelo's Axiom of Choice: Its Origins, Development and Influence, as well as in an appendix to Chapter 1 of Alain Connes Noncommutative Geometry. I should mention that the letter was written in 1905, so around the time of Vitali's paper.
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Did Lebesgue like non-measurable set or not?
The main resource I have on Lebesgue's view on these things is his letter to Borel, part of a discussion that Borel started about Zermelo's paper that used the axiom of choice to prove the well-ordering theorem. Lebesgue says he proved the existence of a Lebesgue measurable set that is not a Borel set in his thesis but he doubts that such a set could be "named". He also says that the existence of infinite Dedekind-finite sets has not been ruled out.
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Did Lebesgue like non-measurable set or not?
I think the trouble comes from the word "wanted". As far as I know, Lebesgue did think that you couldn't get a Lebesgue unmeasurable set without AC, but knew he hadn't proved it. So only a Lebesgue unmeasurable set constructed without AC would contradict his hypothesis, and therefore he "wanted" it as a precondition for changing his prior belief. In any case, it seems nobody predicted the relationship with (strongly) inaccessible cardinals, shown by Solovay and Shelah.
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