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Results for "mathfrak d" 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.
5
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1
answer
204
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$\mathfrak{p}=\mathfrak{b}=\mathfrak{a}=\aleph_1$ and $\mathfrak{d}=\mathfrak{c}=\kappa$
Where
$\mathfrak{c}=2^{\aleph_0}$ the size of the continuum.
$\mathfrak{d}$, is the least size of a $\mathfrak{d}$ominating family. … $\mathfrak{a}$, is the least size of an infinite m$\mathfrak{a}$d family. …
- 181
16
votes
1
answer
605
views
The dominating number $\mathfrak{d}$ and convergent sequences
My question is thus about relations between $\mathfrak{z}$ and $\mathfrak{d}$, especially I am interested in the following:
Question: Is it consistent that $\mathfrak{d}<\mathfrak{z}$ ($<\mathfrak{c}$ … Recall that the cofinality of measure $\text{cof}(\mathcal{N})$ is not less than $\mathfrak{d}$. …
- 1,151
2
votes
1
answer
118
views
Is there a subset of irrationals of size $\mathfrak{d}$ whose image, under any bijection to ...
Given any bijection $\varphi$ between the irrationals and $\omega^\omega$, and a subset $A \subseteq \mathbb{R} \smallsetminus \mathbb{Q}$ of size $\mathfrak{d}$ , under which properties $\varphi(A)$ … I guess we should assume $\mathfrak{d} < \mathfrak{c}$. …
- 105
1
vote
1
answer
148
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Scales and concentration
Then $S=\{s_{\alpha} : \alpha<\mathfrak{d}\}$ is called a $\mathfrak{d}$-scale if for every $\alpha<\beta<\mathfrak{d}$, $s_{\beta}\nleq^{*} s_{\alpha}$. … It is obvious that every scale is $\mathfrak{d}$-concentrated on $[\mathbb{N}]^{<\infty}$, but every $\mathfrak{d}$-scale? …
- 21
6
votes
Accepted
Comparing bornologies for domination/escaping
Note that $\mathfrak{b}=\mathfrak{d}$ is equivalent to the existence of a $<^\ast$-increasing sequence $(f_\alpha)_{\alpha<\mathfrak{d}}$ which is cofinal in $(\mathcal{N},{<^\ast})$, where $f <^\ast g … \end{aligned}$$
Note that $D$ is dominatable iff $D \subseteq D_\alpha$ for some $\alpha < \mathfrak{d}$, and that $E$ is escapable iff $E \subseteq E_\alpha$ for some $\alpha < \mathfrak{d}$. …
- 44.4k
5
votes
Accepted
$\kappa$-scales and the continuum
$\mathfrak{d}$ the minimal cardinality of a cofinal family in $(\omega^\omega,{<^*})$. … In fact, the only inequalities that must hold are
$$\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
157
views
Cofinal trees in $({}^\omega \omega , \leq^\ast )$
So, I know that the existence of a scale (that is, a linear cofinal set in $({}^\omega \omega , \leq^\ast )$, where $\leq^\ast $ is eventual domination, is equivalent to $\mathfrak{b} = \mathfrak{d}$, … A scale is a tree of width $1$ and height $\mathfrak{d}$. If $\mathfrak{b} < \mathfrak{d}$, can I at least force the existence of a cofinal tree of width $< \mathfrak{d}$? …
- 505
3
votes
2
answers
256
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How to increase unbounding and dominating numbers for $(\kappa^\lambda,\leq^*)$
Set
$\mathfrak{b}_\kappa^\lambda:=\min\{|F|:F\subseteq \kappa^\lambda\text{ and }\neg\exists y\in \kappa^\lambda\forall x\in F(x\leq^* y)\}$,
$\mathfrak{d}_\kappa^\lambda:=\min\{|D|:D\subseteq \kappa … I'm studying the cardinals $\mathfrak{b}_\kappa^\lambda$ and $\mathfrak{d}_\kappa^\lambda$. …
- 31
3
votes
Accepted
Generalizing The Cardinal Characteristics of the Continuum
Don Monk has a paper describing generalized $\mathfrak b$ and $\mathfrak d$ as you describe. … I think in his notation the numbers you describe are $\mathfrak b_{\alpha, \alpha, \alpha}$ and $\mathfrak d_{\alpha, \alpha, \alpha}$
Generalized ${\frak b}$ and ${\frak d}$. Notre Dame J. …
- 884
15
votes
3
answers
615
views
Dominating families in bigger cardinals
The smallest cardinality of a dominating family is $\mathfrak d$, and it is well known that $\omega_1\leq \mathfrak d\leq \mathfrak c$ and that it is consistent that $\mathfrak d$ may be almost everything … (so $\mathfrak d_{\omega, \omega}=\mathfrak d$). …
user46747
8
votes
Accepted
Cardinality of cofinal set of normal functions $f \colon \omega_1 \to \omega_1$
For the second question: Let $\mathfrak d(\kappa)$ be the smallest number functions needed to dominate all functions from $\kappa$ to $\kappa$. … In particular, both $\mathfrak d(\aleph_1)=2^{\aleph_1}$ and $\mathfrak d(\aleph_1)< 2^{\aleph_1}$ are consistent.
(This specific result for $\aleph_1$ may be older, though.) …
- 14k
5
votes
1
answer
246
views
Generalizing The Cardinal Characteristics of the Continuum
The bounding number $\mathfrak{b}$ and dominating number $\mathfrak{d}$ could be easily generalized for ordinals $\alpha$, as follows:
$$\forall f,g\in\alpha^\alpha(f\leq_{\alpha}g\Leftrightarrow|\{\beta … (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
11
votes
Accepted
Decomposing $\mathbf{\Pi}^1_1$ sets into closed sets
Now the Baire space is the union of $\mathfrak d$ (the dominating number) compact sets. … It is well known that $\mathfrak d=\aleph_1$ is consistent with the failure of CH.
This happens for instance in the so called Sacks model. …
- 16.2k
12
votes
Accepted
Models of $\mathsf{ZFC}$ with neither $P$- nor $Q$-points
It is provable in ZFC that
$$
\aleph_1\leq\text{cov}(\mathcal B)\leq\mathfrak d\leq\mathfrak c.
$$
Theorem 9.25 says (among other things) that if $\mathfrak d=\mathfrak c$ then there exists a P-point. … So in order to have neither a P-point nor a Q-point, you'd need
$$
\aleph_1\leq\text{cov}(\mathcal B)<\mathfrak d<\mathfrak c,
$$
which implies $\mathfrak c\geq\aleph_3$. …
- 73.1k
7
votes
Accepted
${\frak b}$ and ${\frak d}$ defined with $\leq$ instead of $\leq^*$
Using "$\le$" instead of "$\le^*$," I claim $\mathfrak{b}'=\omega$ and $\mathfrak{d}'=\mathfrak{d}$. … To see $\mathfrak{d}'=\mathfrak{d}$, suppose $S$ is a set of functions with $$(*)\quad\forall f\in\omega^\omega\exists s\in S(f\le^*s);$$ I'll build an $S'$, with $\vert S'\vert=\vert S\vert$, such that …
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