Comment by Elliot Glazer on Gently changing measure
The claim $\mu_{\omega}^*(A^{\omega})=1$ is right, and is one of the few things in analysis which seems to require $\text{DC}_{\mathbb{R}}$ and not just $\text{CC}_{\mathbb{R}}.$ (I haven't confirmed...
View ArticleComment by Elliot Glazer on What can be the measure of a Vitali set?
Yes. I've added a note on that at the end.
View ArticleComment by Elliot Glazer on How much Dependent Choice is provable in $Z_2$?...
Aren't your two formulations of a theory $T$ having DC equivalent? If DC is "true of $T,$" then $T$ proves $\phi$-DC by substituting $\psi(f, r):= (\forall g \exists s \phi(g, s)) \rightarrow \phi(f,...
View ArticleComment by Elliot Glazer on Long chains of amorphous cardinalities
You’re more likely to get a positive answer here if you use surjective comparison $<^*$ instead.
View ArticleComment by Elliot Glazer on Building the real from Dedekind finite sets
You’re right. I’ve rephrased it in terms of $[A]^{<\omega}.$
View ArticleComment by Elliot Glazer on Existence of finite powerset
You could probably get a counterexample to the original question (with the usual foundation axiom) by starting with a ZFA model with a strongly amorphous set of atoms A and taking all the sets...
View ArticleComment by Elliot Glazer on Does $\mathsf{SVC}^\ast$ exist?
I think it'd be better to use a different symbol then * since that's usually to convert a notion regarding injective cardinalities into surjective. Maybe + since you're strengthening the notion?
View ArticleComment by Elliot Glazer on How to settle the Generalized Continuum...
Even Z (without foundation) is enough to prove the standard equivalences of choice, e.g. AC iff Well-ordering Thm iff cardinal trichotomy iff surjective cardinal trichotomy.
View ArticleComment by Elliot Glazer on Does Well-Ordered Interval Power Set "WOIPS"...
It also goes through in Z (without foundation).
View ArticleComment by Elliot Glazer on Does Well-Ordered Interval Power Set "WOIPS"...
See mathoverflow.net/a/471784/109573
View ArticleComment by Elliot Glazer on Does this rule imply axiom of choice?
Nitpick: the phrase “Scott cardinals” makes your questions unnecessarily reliant on Foundation. It’s fine here to just have $\kappa, \lambda$ be arbitrary sets.
View ArticleComment by Elliot Glazer on Must strange sequences wear Russellian socks?
We’ll justify $3 \rightarrow 1$ by the contrapositive. Fix $s \in \prod A_i.$ Identify $\bigsqcup A_i \setminus \{s_j: j<\omega\}$ with $\{t \in \prod A_i: \exists ! j (s_j \neq t_j)\}.$ The latter...
View ArticleDoes Foundation increase the strength of second-order logic?
Thinking about the recent threads on structural consequences of the Axiom of Foundation (AF) over ZF-AF, I've been trying to find some conservativity result which explains why AF doesn't seem to have...
View ArticleAnswer by Elliot Glazer for Is "There exists an unbounded non-measurable set...
This is more of a long comment than an answer.The "right" notion of an unbounded set being measurable in ZF is less than clear. Suppose $\mathbb{R}$ is a countable union of countable sets. Let $\langle...
View ArticleAnswer by Elliot Glazer for Is "There exists an unbounded non-measurable set...
The answer to your first question is yes, and the answer to your second question is no, under any of the multiple definitions of "measurable" in choiceless contexts.We will prove a theorem relating...
View ArticleAnswer by Elliot Glazer for Can a Vitali set be Lebesgue measurable? (ZF)
The answer to your second question is yes. Let $G=(\mathbb{Q}[\sqrt{2}], +).$ This is a countable abelian group, so $\mathbb{R}/G$ is a hyperfinite Borel equivalence relation. In particular, it embeds...
View ArticleAnswer by Elliot Glazer for Can a Vitali set be Lebesgue measurable? (ZF)
The answer to your first question is no. We will show there is a null Vitali set in Cohen's first model $M.$Recall that in $M,$ there is an infinite (but Dedekind finite) set $A$ of mutually Cohen...
View ArticleAnswer by Elliot Glazer for Example of a $\Pi^2_2$ sentence?
The Suslin hypothesis is $\Pi^2_2,$ and $T = ZFC + GCH + LC$ (LC an arbitrary large cardinal axiom) does not prove it to be equivalent to any $\Sigma^2_2$ sentence. Suppose toward contradiction $T$...
View ArticleAnswer by Elliot Glazer for Can a Vopenka cardinal be supercompact?
If $\kappa$ is almost huge with target $\lambda,$ then $V_{\lambda}$ thinks that $\kappa$ is a supercompact Vopenka cardinal.I'll take for granted the standard facts about almost huge cardinals listed...
View ArticleAnswer by Elliot Glazer for How much choice is necessary to prove this...
Your statement is equivalent to the assertion that there is a function choosing an enumeration of every countable ordinal. From an enumeration of $\alpha,$ you can easily inject it into a countable set...
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