Comment by Elliot Glazer on Does $2^{\aleph_0}\rightarrow [\aleph_1]^2_3$...
Shelah's writing is hard for me to follow but this new preprint appears to claim no large cardinal strength is needed to realize this partition relation: arxiv.org/pdf/2601.02923
View ArticleComment by Elliot Glazer on Why doesn't Bukovský's classification of forcing...
@TimothyChow That seems reasonable. Perhaps one concrete scenario that would refute the "uniqueness of forcing" thesis is if there is a sentence $\phi$ such that ZFC is mutually interpretable with ZFC...
View ArticleComment by Elliot Glazer on Are we stuck with Lean?
Clarification: the more familiar theory ZFC+$\{n \text{ inaccessibles}: n<\omega\}$ matches idealized Lean on arithmetic. PFIRS has the stronger property that PFIRS proves an arbitrary sentence iff...
View ArticleAnswer by Elliot Glazer for Cofinality without choice: can this coarse...
We get an example if PSP +$\omega_1, \omega_2 \in \text{Reg} + \Theta \ge \omega_3$ holds, e.g. under determinacy (I don’t know if this hypothesis has any consistency strength beyond $\omega_1$ being...
View ArticleAnswer by Elliot Glazer for Is there a double covering of the plane with...
There is no such covering, and in general, the only $m$-fold coverings of the plane with finitely many slopes are the $m$-slope coverings.Let $\mathcal{C}$ be an $m$-fold covering with lines of $n$...
View ArticleAnswer by Elliot Glazer for Generically separating Hartogs numbers
No, suppose $M \models ``X$ is an o-amorphous set$"$, i.e. $X$ is infinite and there is a linear order $(X, <)$ such that every subset of $X$ is a finite union of intervals. Then $M$ has an...
View ArticleComment by Elliot Glazer on Generically separating Hartogs numbers
@Gro-Tsen $\aleph(\mathcal{P}^2(A))=\aleph(A)$ is consistent: mathoverflow.net/a/98381/109573 ... but $\aleph(\mathcal{P}^3(A)) > \aleph(\mathcal{P}(A))$ is provable: mathoverflow.net/a/500818/109573
View ArticleAnswer by Elliot Glazer for Definable witness to $\mathcal{P}(X)\not\preceq X$
Let $\langle x_{\gamma}: \gamma < \alpha \rangle$ be the standard enumeration of $\mathrm{OD}_f \cap X,$ and for $S \subset \alpha,$ define $x_S$ to be $\{x_{\gamma}: \gamma \in S\}.$ As Emil...
View ArticleAnswer by Elliot Glazer for Can this extension of ZC evade having distinct...
We will show that ZC + Iteration (more commonly denoted ZCR, i.e. ZC + Ranks) is solid, and therefore tight. See [1] to review concepts of interpretation such as identity preservation and dimension,...
View ArticleAnswer by Elliot Glazer for Does Foundation increase the strength of...
Some terminological notes: In the time since asking this question, I have rephrased "fundamentally $\Pi_2$" to "structural $\Pi_2$" and now use the equivalent characterization of "a sentence asserting...
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 Can ZFC be interpreted in this theory of finitely...
Yes and you don't need choice, just do the construction you linked verbatim. For showing $U \models$ Replacement, let $F$ be a $U$-definable map from some $X$ to $U$-ordinals $\alpha$ such that $U...
View ArticleAnswer by Elliot Glazer for How to interpret ZU in a finitely ranked infinite...
This theory interprets Z (and thus ZU). Let $U$ be the class of all pointed well-founded extensional digraphs.Define an equivalence relation $(G_0, p_0) \equiv (G_1, p_1)$ if there is $H$ and...
View ArticleAnswer by Elliot Glazer for Is ultrafilter lemma over N sufficient for...
Very little is known here. It hasn't even been ruled out that a single nonprincipal ultrafilter on $\omega$ implies the Banach-Tarski paradox. Paul Larson recently told me this is one of the motivating...
View ArticleAnswer by Elliot Glazer for Cauchy completeness in ordered families of...
This isn't even provable in ZF. In fact, ZF proves that the principle "weak completeness = strong completeness" (*) is equivalent to countable choice ($\mathrm{AC}_{\omega}$).$\mathrm{AC}_{\omega}...
View ArticleAnswer by Elliot Glazer for Without AC can there be a function with linearly...
Yes this is a ZF theorem. Precisely, ZF proves TFAE for all sets $X:$$|X| \in [1, 3].$There is $f: \mathcal{P}(X) \rightarrow X$ such that each fiber is a $\subset$-chain.$\mathcal{P}(X)$ is an...
View ArticleAnswer by Elliot Glazer for Is it possible to formulate the axiom of choice...
The Gabay-O'Connor Theorem (GO), which is that there is a winning strategy for arbitrary sets of players and colors and $\kappa=\omega,$ is strictly weaker than the Axiom of Choice over ZF. GO holds in...
View ArticleAnswer by Elliot Glazer for Is 0# still unique in ZFC without powerset?
Suppose $r_0$ and $r_1$ each have the $0^{\#}$ property, with respective Silver indiscernible classes $J_0$ and $J_1.$ Then $J_0 \cap J_1$ is club below $\kappa=\omega_1^{L[r_0, r_1]},$ even if...
View ArticleAnswer by Elliot Glazer for (AC) and existence of basis of $\{0,1\}^X$ for...
The statement (S) does not imply $\mathrm{AC}_{\omega}(\mathbb{R})$ (which is much weaker than PP) or the statement (S+) that every $\mathbb{F}_2$-vector space has a basis. We will show that (S) and...
View ArticleAnswer by Elliot Glazer for How do the finite partitioning principle compare...
Here's a separation over ZFA.In the Russell's socks model $M,$ there is a Dedekind finite (in particular, non-idemmultiple) set $X$ which is a countable union of pairs.$M$ is a model of Multiple...
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