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Results for "mathfrak b" tagged with set-theory
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forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.
3
votes
1
answer
158
views
Is there any class of ideals for which $\mathfrak{b}(\mathcal I)\neq\mathfrak{b}$?
Let $\mathfrak{b}(\mathcal{I})$ be the minimum cardinality of a $\leq_\mathcal{I}$-unbounded subset of $\mathbb{N}^\mathbb{N}$. … Is there any class of ideals for which $\mathfrak{b}(\mathcal I)\neq\mathfrak{b}$? …
- 505
2
votes
0
answers
81
views
A convex version of the small uncountable cardinal $\mathfrak b$
It is clear that $\omega_1\le \mathfrak b'\le\mathfrak b$. Problem 1 asks whether $\mathfrak b'=\mathfrak b$?
Problem 2. Is it consistent that $\omega_1<\mathfrak b'$? … In particular, is $\mathfrak b'=\mathfrak c$ under Martin's Axiom? …
- 41.8k
6
votes
1
answer
415
views
$\text{cov}(\mathcal{M})$ vs. $\mathfrak{b}$ vs. $\mathfrak{s}$
For example, the classical results state that the following inequalities are relatively consistent:
$\omega_1=\mathfrak{s}<\mathfrak{b}=\omega_2$ (Balcar-Pelant-Simon '80);
$\omega_1=\mathfrak{b}<\ … mathfrak{s}=\omega_2$ (Shelah '84);
$\omega_1=\text{cov}(\mathcal{M})<\mathfrak{b}=\omega_2$ (Bartoszyński '96);
$\omega_1=\mathfrak{b}<\text{cov}(\mathcal{M})=\omega_2$ (Miller '81). …
- 1,151
5
votes
1
answer
204
views
$\mathfrak{p}=\mathfrak{b}=\mathfrak{a}=\aleph_1$ and $\mathfrak{d}=\mathfrak{c}=\kappa$
Question:
Then we can say in $M[K]$ that:
$(i)$ $\mathfrak{p}=\mathfrak{b}=\mathfrak{a}=\aleph_1$ and $\mathfrak{d}=\mathfrak{c}=\kappa$ ? … $\mathfrak{b}$, is the least size of an un$\mathfrak{b}$ounded family. …
- 181
10
votes
0
answers
495
views
Is there a model of set theory in which $\mathfrak p< \mathfrak b < \mathfrak q$?
Is there a model of set theory in which $\mathfrak p< \mathfrak b < \mathfrak q$? … $\min\{\mathfrak b, \mathfrak q \}\in \{\mathfrak p,\mathfrak q \}$ ? …
- 1,101
5
votes
1
answer
154
views
Can the Boolean group $C_2^\omega$ be covered by less than $\mathfrak b$ nowhere dense subgr...
It is known that $\max\{\mathrm{cov}(\mathcal N),\mathrm{cov}(\mathcal M)\}$ can be strictly smaller than $\mathfrak b$ (this happens, for example, in the Laver and Mathias models).
Problem. … Is $\mathrm{cov}_H(C_2^\omega)<\mathfrak b$ consistent? What is the value of $\mathrm{cov}_H(C_2^\omega)$ in the Laver (or Mathias) model? …
- 41.8k
5
votes
1
answer
246
views
Generalizing The Cardinal Characteristics of the Continuum
|=2^{|\alpha|}$
For cardinals $\kappa,\mathfrak{b}_\kappa>\kappa$ (this is shown by generalized diagonalization)
The last two points combined make $\mathfrak{b}_\kappa=\kappa^+$ if $\kappa^+=2^\kappa$ … (or if GCH is assumed)
$\mathfrak{b}_\omega$ is clearly, in this case, $\mathfrak{b}$, and the same is true with $\mathfrak{d}_\omega$. …
- 675
8
votes
Accepted
Can the forcing-absolute fragment of SOL have a strong Lowenheim-Skolem property?
Claim 1: Given a structure $\mathfrak{B}$ and a set $B$ of ordinals coding $\mathfrak{B}$, and given an element $x\in\mathfrak{B}$ and a second order formula $\varphi$, if "$\mathfrak{B}\models\varphi( … Claim 2: $\pi\upharpoonright\bar{\mathfrak{B}}:\bar{\mathfrak{B}}\to\mathfrak{B}$ is elementary with respect to forcing absolute second order formulas. …
- 9,902
8
votes
1
answer
391
views
Does "agreement on cardinalities" imply second-order elementary substructurehood?
{Card}}\mathfrak{B}$.") … $\mathfrak{A}\preccurlyeq_\mathcal{L}\mathfrak{B}$. …
- 21.1k
15
votes
2
answers
513
views
Nontrivial signed measure on Lebesgue measurable sets being trivial on Borel sets
It is hard (if ever possible) to show $(\mu|_{\mathfrak{B}(\mathbb{R})})^+ = \mu^+|_{\mathfrak{B}(\mathbb{R})}$ and $(\mu|_{\mathfrak{B}(\mathbb{R})})^- = \mu^-|_{\mathfrak{B}(\mathbb{R})}$. … Background: I am trying to prove (or disprove) that if $\mu$ and $\lambda$ are signed measures on $\mathfrak{L}(\mathbb{R})$, then $\mu|_{\mathfrak{B}(\mathbb{R})} = \lambda|_{\mathfrak{B}(\mathbb{R})} …
- 807
3
votes
Accepted
$\text{cov}(\mathcal{M})$ vs. $\mathfrak{b}$ vs. $\mathfrak{s}$
=\lambda<\mathfrak{c}=\mu$,
2) $\mathfrak{s}=\mathfrak{b}=\kappa<\mathrm{cov}(\mathcal{M})=\lambda<\mathfrak{c}=\mu$. … http://projecteuclid.org/euclid.jsl/1294170995
4) $\mathfrak{s}=\mathrm{cov}(\mathcal{M})=\aleph_1<\mathfrak{b}=\kappa<\mathfrak{c}=\lambda$. …
- 791
5
votes
Accepted
Does "agreement on cardinalities" imply second-order elementary substructurehood?
"), whereas $\mathfrak{A}$ does not satisfy this, so $\mathfrak{A}\not\equiv_{\mathrm{SOL}}\mathfrak{B}$, and hence $\mathfrak{A}\not\preccurlyeq_{\mathrm{SOL}}\mathfrak{B}$. … {B}}$ and $\varphi^{\mathfrak{B}}\cap\mathfrak{A}$ both have cardinality continuum. …
- 9,902
5
votes
Accepted
$\kappa$-scales and the continuum
It is not difficult to show that a scale exists if and only if $\mathfrak{b} = \mathfrak{d}$, and then the minimal length of a scale is the common value of these two cardinal characteristics (which is … In fact, the only inequalities that must hold are
$$\aleph_1 \leq \mathfrak{b} = cf(\mathfrak{b}) \leq cf(\mathfrak{d}) \leq \mathfrak{d} \leq \mathfrak{c}.$$
Using Hechler's Theorem (see this answer of …
- 44.4k
6
votes
1
answer
297
views
What is the height (or depth) of $[\mathbb{N}]^\infty$?
It is easy to see that
"$\mathfrak{t}=\mathfrak{b}$ or $\mathfrak{b}<\mathfrak{d}$"
implies
"$\mathfrak{b}<\mathfrak{ht}$".
Question 1. … Is it consistent that
"$\aleph_1=\mathfrak{t}<\mathfrak{b}=\mathfrak{c}=\aleph_2<\mathfrak{ht}$"? …
- 3,104
16
votes
1
answer
738
views
Is there a pair of non-isomorphic structures each of which is isomorphic to an ultrapower of...
{A}^I/\mathcal{U}\cong\mathfrak{B}$ and $\mathfrak{B}^J/\mathcal{F}\cong\mathfrak{A}$? … {U}\cong \mathfrak{B}^I/\mathcal{U}$). …
- 12.4k