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first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.

9 votes
Accepted

Can one exhibit an explicit Kuratowski infinite set without invoking Replacement?

The answer is no. Let ZC' be ZFC without replacement and infinity and with the assertion there is a Kuratowski infinite set. We will construct a model $M$ of ZC' such that only hereditarily finite ele …
Elliot Glazer's user avatar
3 votes

Surreals and NSA: some foundational issues

Problem 1: There is a definable proper class saturated real-closed field $\mathbb{R}^*$, defined by a slight modification of your and Shelah's construction, such that there is an $\mathrm{OD}_p$ injec …
Elliot Glazer's user avatar
12 votes
Accepted

Is this set theory equivalent to ZFC?

Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with par …
Elliot Glazer's user avatar
9 votes
Accepted

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 liste …
Elliot Glazer's user avatar
5 votes
Accepted

The difference between Baire 2 and 'effectively Baire 2'

It's provable in ZF that every Baire-2 function is effectively Baire-2. It suffices to prove the following: (ZF) There is an explicit function which maps each Baire-1 function $f: \mathbb{R} \rightarr …
Elliot Glazer's user avatar
6 votes
Accepted

Stronger negation of AC given by rejecting "infinite hat" puzzles

Naturally, the more generalizations of the infinite hats puzzle we consider, the stronger it is to assert that none of them have a paradoxical solution. One of the variants you linked in your question …
Elliot Glazer's user avatar
10 votes
Accepted

Can we make ZF − infinity + “all ordinals are finite” as strong as ZFC?

ZFfin implies every set is finite, and in particular choice holds. Suppose there is an infinite set $X.$ By replacing every element of $\mathcal{P}_{\text{fin}}(X)$ with its cardinality, we see $\omeg …
Elliot Glazer's user avatar
9 votes
Accepted

Must strange sequences wear Russellian socks?

For $n \ge 2$ and $\langle A_i \rangle$ such that $|A_i| = n$ for all $i,$ TFAE: $|\prod A_i| \neq |\mathbb{R}|.$ $\bigsqcup A_i$ is uncountable. $\bigsqcup A_i$ is un-orderable. There are $k \in [2, …
Elliot Glazer's user avatar
9 votes
Accepted

Building the real from Dedekind finite sets

Q1: There is no such partition. Let $\langle X_n \rangle$ be a countable partition of $\mathbb{R}.$ We will construct $n,$ an open interval $I,$ and an injection $g: \omega \rightarrow I \cap X_n$ wit …
Elliot Glazer's user avatar
9 votes
Accepted

The Parity Principle and $\mathbf{C}_2$ (choice for $2$-sets)

Over ZFA, the Parity Principle is strictly weaker. We'll show it follows from Multiple Choice, the assertion that for any family of nonempty sets $\mathcal{F},$ there is $g: \mathcal{F} \rightarrow [\ …
Elliot Glazer's user avatar
7 votes

Global Choice bi-interpretable with Global Wellorder?

The answer to the first question is yes: Global Choice is bi-interpretable with Global Well-Ordering. First, I will prove that a class linear order $(C, <)$ is a well-ordering iff every subset (rather …
Elliot Glazer's user avatar
2 votes

Is "There exists an unbounded non-measurable set but no bounded non-measurable set" consiste...

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 $\lang …
Elliot Glazer's user avatar
11 votes
Accepted

Is $\in$-induction provable in first order Zermelo set theory?

The answer is no. Take the standard model of Z and add in a $\mathbb{Z}$-sequence of objects, each of whose only element is the previous one. I.e., define $M=\bigcup_{n<\omega} \bigcup_{m<\omega}\mat …
Elliot Glazer's user avatar
12 votes
Accepted

Consistency of a strong Fubini type theorem for measure zero sets

ZFC refutes this principle. Let $\kappa=\text{non}(\mathcal{L}),$ i.e. the least cardinality of a set of reals of positive outer measure. Let $X \subset [0,1]$ be such that $|X|=\kappa$ and $\lambda^* …
Elliot Glazer's user avatar
22 votes

What notable theorems cannot be automatically proven without choice using Shoenfield absolut...

$\text{ZF}+ \text{AC}_{\omega}$ is not $\Sigma^1_4$-conservative over ZF and ZF + DC is not $\Sigma^1_4$-conservative over $\text{ZF}+ \text{AC}_{\omega}.$ An example of the former: the sentence $\ome …
Elliot Glazer's user avatar

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