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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.

5 votes
Accepted

A property of the Frechet filter and every ultrafilter

The filters that you're looking for don't exist. Every filter with your property, other than the Frechet filter, maps to an ultrafilter via a finite-to-one map. Let's call the filters with your proper …
Will Brian's user avatar
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8 votes
Accepted

Gaps in cardinalities of MAD families

Yes, this is consistent. Suppose we force to add $\kappa$ mutually generic Cohen reals to a model of $\mathsf{CH}$, where $\kappa$ is some cardinal with uncountable cofinality. In the extension, there …
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7 votes

Is there nontrivial structure to forcing axioms?

This paper seems to be the sort of thing you're looking for: J. H. Barnett, "Weak variants of Martin's Axiom," Fundamenta Mathematicae 141 (1992), pp. 61-73. The abstract is pretty short and to …
Will Brian's user avatar
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9 votes

Uniqueness results that follow from CH

Here's one about my favorite topological space, the Stone–Čech remainder of the natural numbers, denoted $\mathbb N^*$ or $\omega^*$. The characterization is due to Parovicenko. Assuming CH, $\mathbb …
6 votes

Characterizing "bounded" distributivity in terms of dense open sets

This is just a comment on the question (but too long for a comment box), not an answer. Let $\kappa$ and $\lambda$ be cardinals. A complete Boolean algebra $\mathbb B$ is called $(\kappa,\lambda)$-di …
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14 votes
Accepted

Can the cardinal $2^{\aleph_0}$ be order-embedded in ${\cal P}(\omega)/(\text{fin})$?

Yes! The fact that this is consistent is originally due to Laver. Later, Baumgartner, Frankiewicz, and Zbierski strengthened Laver's result to the following theorem: Theorem: Is it consistent that $\m …
Will Brian's user avatar
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12 votes
Accepted

"Towers" on singular cardinals with countable cofinality

For $\lambda > 2^{\aleph_0}$, there is no such sequence. Suppose $\lambda > 2^{\aleph_0}$. Because $2^{\aleph_0}$ cannot have countable cofinality, there is some $\kappa < \lambda$ with $2^{\aleph_0} …
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27 votes
Accepted

Does the axiom of choice follow from the statement "Every simple undirected graph is either ...

(S) is a theorem of ZF. Proof: Let $G$ be a graph, and let $v$ be a vertex of $G$. Define $$P_v = \{w \,:\, \text{there is a path from } v \text{ to } w\}.$$ If $P_v$ is the vertex set of $G$, then $ …
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10 votes

Forcing notions adding minimal reals

Prikry-Silver forcing adds a minimal real: that is, if $g$ is Prikry-Silver generic over $V$ and $h$ is any real in $V[g] \setminus V$, then $V[g] = V[h]$. Interestingly, while Prikry-Silver forcing i …
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3 votes
Accepted

Minimal cardinality of non-bipartite sub-family of $[\omega]^\omega$

The cardinal $\mathfrak{nb}$ is equal to the reaping number $\mathfrak{r}$. An unsplit family is a collection $\mathcal R$ of infinite subsets of $\omega$ such that there is no set $D \subseteq \omega …
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11 votes
Accepted

Can we force $\mathfrak{r}<\mathfrak{s}$?

The inequality $\mathfrak{r} \leq \mathfrak{u}$ is provable in ZFC (because every base for an ultrafilter is a reaping family). Blass and Shelah proved the consistency of $\mathfrak{u} < \mathfrak{s}$ …
Will Brian's user avatar
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5 votes

Strategic vs. tactical closure

This post is an addition to the argument posted by bof last night. In bof's post, it is proved that posets having a special kind of dense subset cannot provide a counterexample for Monroe's question. …
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29 votes

Is it possible to define higher cardinal arithmetics

Let me begin with an observation: $2 \!\uparrow\uparrow\! n$ is equal to the number of sets of rank at most $n$. [If you followed the combinatorics tag here and have forgotten what the r …
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3 votes

The "strong" measure number

Great question! This is not a complete answer, but hopefully it gets the ball rolling . . . Theorem: $\mathfrak s_{-}\geq \mathrm{cov}(\mathcal M)$, where $\mathrm{cov}(\mathcal M)$ is the smalles …
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14 votes
Accepted

On the Large Cardinal Strength of Normal Moore Space Conjecture

If $\text{NMSC}$ is consistent, then so is $\text{NMSC}+\text{"there are no strongly inaccessible cardinals"}$. This is because if $V \models \text{NMSC}$, then $V_\kappa \models \text{NMSC}$ for an …
Will Brian's user avatar
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