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Professor of Philosophy, Baylor University


Jan
15
revised Maximal isometrically invariant finitely additive extension of Lebesgue measure in dimension $\ge 3$?
added 254 characters in body
Jan
15
comment Maximal isometrically invariant finitely additive extension of Lebesgue measure in dimension $\ge 3$?
Yup! It is obvious. Thanks!
Jan
15
asked Maximal isometrically invariant finitely additive extension of Lebesgue measure in dimension $\ge 3$?
Jan
13
comment Does a nonlinear additive function on R imply a Hamel basis of R?
If one could show that there is a bounding function $g:\mathbb R\to\mathbb R$ such that for any finite-dimensional subspace $S$ of $\mathbb R$ (considered as a vector space over $\mathbb Q$) containing $1$ and $\pi$ (say), there is an additive function $f:S\to\mathbb R$ with $f(1)=1$ and $f(\pi)=0$ and $|f| \le g$ everywhere on $S$, then one could indeed use BPI--or just Hahn-Banach!--to prove that there is such a function on all $\mathbb R$. This would provide an alternate route to showing that Hahn-Banach implies non-measurable sets. But I don't see how to prove there is such a $g$.
Jan
11
comment Does a nonlinear additive function on R imply a Hamel basis of R?
Goldstern: The Consequences of the Axiom of Choice search lists the question of the implication from BPI to a discontinuous additive function as unknown. I would think that if it weren't hard to clean up your completeness argument, it would be known.
Jan
11
comment Does a nonlinear additive function on R imply a Hamel basis of R?
@Goldstern: This is a naive question. In your BPI argument, how do you ensure that $f$ is real-valued (rather than having values in some hyperreal field)?
Dec
30
comment Totally right preorderable groups
OK, I see it now. You must have had "$h\in G$" instead of "$h\in H$". I know that you're talking of a left-invariant total preorder, but nonetheless as you correctly said "The preorder is actually right-$H$-invariant" (but not necessarily right $G$-invariant).
Dec
30
comment Totally right preorderable groups
But since $\le$ is right $H$-invariant, if $g\ne g'$ and $h\in H$, then $gh \equiv g'h$ iff $g \equiv g'$. Thus the minimal $h$ will always be the same (i.e., the $\prec$-minimal member of $H$). Perhaps you have a typo, though, and meant $h\in G$ instead of $h\in H$? I was assuming $H=\{ g : 1 \le g \le 1 \}$, as earlier in your post.
Dec
29
comment Totally right preorderable groups
I am having some trouble following your last direct proof. By your definition, we have $g\le' g'$ iff for all $h\in H$ we have $gh\le g'h$ or there is an $h'\prec h$ such that not: $gh'\le g'h'\le gh'$. But as you showed, $\le$ is right $H$-invariant, so the first disjunct is just equivalent to $g\le g'$. Thus $\le'$ is weaker than $\le$ and so $\{ g : 1 \le g \le 1 \}\subseteq N$. But you claim the opposite inclusion (and surely in general they aren't equal). What am I missing?
Dec
29
revised Two kinds of invariance of full conditional probabilities
fix minor mistake
Dec
23
comment Full conditional probabilities and versions of AC?
Stone representation is equivalent to BPI, so that doesn't help. (There is a version of Stone that holds in ZF--use a filter instead of an ultrafilter in the proof and get a weaker conclusion--but I don't know how to use that for the purposes needed here.)
Dec
22
comment Full conditional probabilities and versions of AC?
Equivalent versions: (1) every filter (collection $F$ of subsets of $X$ such that $A,B\in F$ implies $A\cap B\in F$ and $A\in F$ and $A\subseteq B\subseteq X$ implies $B\in F$) extends to an ultrafilter (maximal filter); (2) every ideal $I$ in a boolean algebra $B$ ($a\in I, b\in B$ implies $a\wedge b\in I$, and $a,b\in I$ implies $a\vee b\in I$) is contained in a prime ideal ($a\wedge b\in I$ implies $a\in I$ or $b\in I$). Form 14 here: consequences.emich.edu/CONSEQ.HTM
Dec
22
revised Full conditional probabilities and versions of AC?
Amusing note on amorphous sets.
Dec
19
comment Probabilities in a riddle involving axiom of choice
But the opponent can win by foreseeing what which value of i we're going to choose and which choice of representatives we'll make. I suppose we would ban foresight of $i$?
Dec
19
comment Probabilities in a riddle involving axiom of choice
What we have then is this: For each fixed opponent strategy, if $i$ is chosen uniformly independently of that strategy (where the "independently" here isn't in the probabilistic sense), we win with probability at least $(n-1)/n$. That's right. But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy".
Dec
18
comment Probabilities in a riddle involving axiom of choice
I was assuming that "independently" has the meaning it does in probability theory ($P(AB)=P(A)P(B)$ and generalizations for $\sigma$-fields). But that does require a probabilistic description of the opponent's choice. Of course, one could mean "independently" here in some non-mathematical causal sense. (And there may be philosophical reason for doing this: fitelson.org/doi.pdf ) Still, mixing the probabilistic with nonprobabilistic concepts might lead to some difficulties, though.
Dec
17
comment Probabilities in a riddle involving axiom of choice
In the probabilistic variant, I don't see that you can win against any strategy of the opponent. If we are making no probabilistic assumptions whatsoever, then in particular we are not assuming that our choice of index $i$ is independent of the opponent's choice of numbers.
Dec
16
revised Full conditional probabilities and versions of AC?
FCU implies order extension principle
Dec
16
revised Strength of claims about extensions of partial preorders and orders to linear ones
indicate connection with conditional probabilities and maybe with Hahn-Banach
Dec
12
revised Probabilities in a riddle involving axiom of choice
added some more fun stuff