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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.

15 votes
2 answers
2k views

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

This is a follow-up to this question. We say that a set $A \subseteq \mathbb{R}$ is bounded if there exists a finite interval $(a,b)$ such that $A \subseteq (a,b)$. Working in $\mathsf{ZFC}$, the exis …
Clement Yung's user avatar
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8 votes
1 answer
231 views

Is $\operatorname{non}(\mathcal{M}) < \mathfrak{a}$ consistent?

Let $\operatorname{non}(\mathcal{M})$ be the least cardinality of a non-meagre subset of the reals. Let $\mathfrak{a}$ be the least cardinality of an infinite maximal almost disjoint family (i.e. $\ma …
Clement Yung's user avatar
  • 1,442
7 votes
1 answer
381 views

Models with fixed cardinality of non-Lebesgue measurable sets

In the usual $\mathsf{ZFC}$, we know that there are $2^\mathfrak{c}$ many subsets of $\mathbb{R}$ that are not (Lebesgue) measurable. On the other hand, the Solovay model also provides us a model of $ …
Clement Yung's user avatar
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6 votes
2 answers
476 views

Properties of Jech's hierarchy of stationary sets (Exercise 8.13, 8.14 of Jech)

I must first preface that while this is indeed a question on an exercise, I believe this is advanced enough for MathOverflow. Let $\kappa$ be a regular uncountable cardinal. Recall that the notion of …
Clement Yung's user avatar
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5 votes
1 answer
306 views

Locating generic filters in the Lévy collapse

Let $\operatorname{Col}(\omega,<\kappa)$ denote the Lévy collapse of an inaccessible cardinal $\kappa$. A variant of the Factor Lemma is as follows: Lemma. Suppose that $\kappa$ is an inaccessible ca …
Clement Yung's user avatar
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5 votes
1 answer
227 views

Minimum number of dense sets to make a filter an ultrafilter

$\newcommand{\U}{\mathcal{U}}$ $\newcommand{\F}{\mathcal{F}}$ $\newcommand{\D}{\mathcal{D}}$ $\newcommand{\C}{\mathcal{C}}$ For any infinite $X \subseteq \omega$, we define: $$ \D_X := \{Y \in [\omega …
Clement Yung's user avatar
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5 votes
1 answer
428 views

Uniform strategy on Kastanas' game

I think my question applies to most games, but for the sake of concreteness, I shall consider one specific game in this question. We consider the game posed by Ilias Kastanas in his paper On the Ramse …
Clement Yung's user avatar
  • 1,442
3 votes
1 answer
364 views

What does it really mean for a polynomial to be solvable in $\mathbb{Z}$ iff $\mathsf{ZFC}$ ...

I was reading this answer, which says that: In his Master's Thesis, Merlin Carl has computed a polynomial that is solvable in the integers iff ZFC is inconsistent. A joint paper with his advisor Bor …
Clement Yung's user avatar
  • 1,442
3 votes
1 answer
244 views

A submodel of set theory with all reals which every set is analytic

This is a continuation of my previous question. Recall that a subset $A \subseteq {}^\omega\omega$ is analytic if it is the continuous image of the Baire space. I would like to know if there exist two …
Clement Yung's user avatar
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3 votes
2 answers
471 views

A Baire subset of reals that is not Suslin measurable

EDIT: The definition of a Suslin measurable set I wrote here is incorrect. It should be that $\mathcal{S}$ contains the field (or algebra) of open subsets of ${}^\omega\omega$ (or, in other words, it …
Clement Yung's user avatar
  • 1,442
3 votes
2 answers
267 views

Is the set of $\kappa$-complete ultrafilters closed in $\beta X$?

Given an arbitrary set $X$, let $\beta X$ be the set of all ultrafilters over $X$. Consider endowing $\beta X$ with a topology consisting of the following open sets: $$ \{\mathcal{U} \in \beta X : A \ …
Clement Yung's user avatar
  • 1,442
2 votes
0 answers
104 views

Does every ordered-union coideal contain an ordered-union ultrafilter?

$\newcommand{\FU}{\operatorname{FU}}$ $\newcommand{\H}{\mathcal{H}}$ Recall that an ordered-union ultrafilter is an ultrafilter on $\omega$ with a base of sets of the form $\FU(A)$. Here, $A = \{a_0,a …
Clement Yung's user avatar
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