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forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.

3 votes
2 answers
294 views

Hyperfinite set containing the reals, with specified upper bound on internal cardinality?

Is this true? For any hyperfinite $n$ that isn't finite, there is a hyperfinite set $A$ such that $\mathbb R \subset A$ and $|A|\le n$ (that's the crucial part, of course)? Intuitively it seems righ …
Alexander Pruss's user avatar
9 votes
2 answers
865 views

Is it consistent with ZFC that some translation-invariant extension of Lebesgue measure assi...

It is consistent with ZFC (but not ZFC+CH, of course) that there is a subset $A$ of nonzero outer Lebesgue measure that has cardinality less than $c$. There will then be an extension of Lebesgue meas …
Alexander Pruss's user avatar
3 votes

Is it consistent with ZFC that some translation-invariant extension of Lebesgue measure assi...

For what it's worth, here is a slightly more elementary proof. If there is a translation-invariant measure on $\mathbb R$ assigning a non-zero value to some set of cardinality less than $\frak c$ and …
Alexander Pruss's user avatar
9 votes
1 answer
2k views

A set of positive measure with cardinality less than that of the continuum?

Is it consistent with ZFC that there is a subset of $[0,1]$ whose cardinality is less than that of the continuum but which has positive Lebesgue measure? Obviously not given CH. And, given ZFC, ther …
Alexander Pruss's user avatar
1 vote

Consequences of ZF+"all subsets of reals are Lebesgue measurable"

One can multiply examples, of course. For the amusement of anyone interested, here are three more consequences of there being no nonmeasurable sets (also of the Banach-Tarski decomposition failing). …
Alexander Pruss's user avatar
20 votes
0 answers
449 views

Hahn-Banach and the "Axiom of Probabilistic Choice"

Stipulate that the Axiom of Probabilistic Choice (APC) says that for every collection $\{ A_i : i \in I \}$ of non-empty sets, there is a function on $I$ that assigns to $i$ a finitely-additive probab …
Alexander Pruss's user avatar
10 votes

Unique existence and the axiom of choice

In A definable nonstandard model of the reals, Kanovei and Shelah surprisingly managed to prove the existence of a ZFC-definable (i.e., specifiable via an explicit ZFC construction) nonstandard model …
Alexander Pruss's user avatar
3 votes

Strength of some claims about finitely additive measures on infinite sets?

A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC. (2)+Countable Choice for Finite Sets (CC(fin)) implies that every uncountable set has a non-principal (finite …
Alexander Pruss's user avatar
6 votes
4 answers
609 views

Strength of some claims about finitely additive measures on infinite sets?

Assume ZF. Consider the claim: (1) For any infinite set $\Omega$, there is a finitely additive probability measure $\mu:2^\Omega\to[0,1]$ with $\mu(A) = 0$ whenever $|A|<|\Omega|$. Then (1) is impl …
Alexander Pruss's user avatar
10 votes
0 answers
755 views

Full conditional probabilities and versions of AC?

A probability is a finitely additive measure on a boolean algebra with total measure $1$. A function $P:\scr B \times (\scr B - \{ 0 \})$ is a full conditional probability on $\scr B$ (for a boolean …