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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 …
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 …
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 …
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 …
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 …
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} …
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 $ …
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 …
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 …
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 $\mathfrak{u} < \mathfrak{s}$ …
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. …
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 …
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 …
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 …