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8 votes
0 answers
241 views

Topological applications of $\mathfrak{p}=\mathfrak{t}$

I'm working on the Malliaris-Shelah's recent result of $\mathfrak{p}=\mathfrak{t}$, but I'm more interested in what possible topological applications can derivate from this equality. … So, my question is: are there any interesting consequence of $\mathfrak{p}=\mathfrak{t}$ in Topology? …
Alexei0709's user avatar
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 $\mathfrak p< \mathfrak b < \mathfrak q$? … $\min\{\mathfrak b, \mathfrak q \}\in \{\mathfrak p,\mathfrak q \}$ ? …
Alexander Osipov's user avatar
4 votes
3 answers
395 views

Problem understanding a passage of the proof of $\mathfrak{p}=\mathfrak{t}$ involving forcing

generic over $\mathfrak{M}$ and $\mathfrak{M}[G]$ the generic model extending $\mathfrak{M}$ and containing $G$ obtained using forcing. … Then $\mathfrak{M}[G]$ is also a model of ZFC and $G\in\mathfrak{M}[G]$. Given $A\subset B$ and $x\in B$, use the notation $x\leq A$ to denote $\forall a\in A[x\leq a]$. …
Cla's user avatar
  • 775
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? …
Taras Banakh's user avatar
  • 41.8k
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 $\mathfrak{d} < \mathfrak{c}$. …
Sergio Garcia's user avatar
5 votes
1 answer
154 views

Can the Boolean group $C_2^\omega$ be covered by less than $\mathfrak b$ nowhere dense subgr...

by closed subsets of Haar measure zero in $C_2^\omega$ (the upper bound $\mathrm{cov}_H(C_2^\omega)\le\mathfrak r$ is proved, for example, here). … 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? …
Taras Banakh's user avatar
  • 41.8k
8 votes
2 answers
479 views

Relations between two tower numbers

It is clear that $$\mathfrak t\le\hat{\mathfrak t}\le\mathfrak c.$$ Martin's Axiom implies $\mathfrak t=\hat{\mathfrak t}=\mathfrak c$. … Is there any non-trivial upper or lower bound for the cardinal $\hat{\mathfrak t}$? In particular, is $\mathfrak b\le\hat{\mathfrak t}$? Problem 2. …
Taras Banakh's user avatar
  • 41.8k
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 $\mathfrak d_{\omega, \omega}=\mathfrak d$). …
user avatar
4 votes
1 answer
178 views

What is the smallest cardinality of a maximal ultrafamily of infinite subsets of $\omega$?

It can be shown that $$\max\{\mathfrak s,\mathfrak a\}\le\mathfrak {uf}\le\mathfrak c,$$ where $\mathfrak a$ is the smallest cardinality of a maximal infinite almost disjoint family of infinite sets in … Is $\mathfrak{uf}<\mathfrak c$ consistent? Added in Edit. …
Taras Banakh's user avatar
  • 41.8k
4 votes
1 answer
220 views

Indecomposable ordinals and pseudointersection

Then for any collection $\{B_i\in U: i<\mathfrak{m}\}$ it is true that there exists $B\subset \alpha$ of order type $\alpha$ such that $B-B_i$ is finite for all $i<\mathfrak{m}$. … Here $\mathfrak{m}$ is the least cardinal such that Martin's Axiom holds below $\mathfrak{m}$ (i.e. meeting any $\beta$ many dense sets for any $\beta<\mathfrak{m}$). …
Otto's user avatar
  • 1,006
13 votes
2 answers
1k views

On Hamkins' answer to a problem by Michael Hardy

Thus it's clear that $\mathfrak{r}\mathfrak{r}\geq\mathfrak{i}\mathfrak{i}$. … More precisely is it possible to have $\mathfrak{i}\mathfrak{i}<\mathfrak{r}\mathfrak{r}<\mathfrak{c}$? Is it possible also to have $\mathfrak{i}\mathfrak{i}=\aleph_0$? …
Rahman. M's user avatar
  • 2,381
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 $< \mathfrak{d}$? …
Matteo Casarosa's user avatar
7 votes
1 answer
431 views

Existence of a specific mad family

It's possible to show that $\omega<\mathfrak h\leq \mathfrak c$, that $\mathfrak h$ is regular and it's consistent that $\mathfrak h<\mathfrak c$. … The height of $\mathcal T$ is $\mathfrak h$. Every node of $\mathcal T$ has exactly $\mathfrak c$ successors. …
user avatar
7 votes
0 answers
443 views

The minimum cardinality of an almost disjoint reaping family

Note that $\mathfrak{r}\leq\mathfrak{u}$ by an easy argument. Claim 2: $\mathfrak{r}\leq\kappa\leq\mathfrak{u}$. Proof. … 3) What is the value of $\kappa$ in the Goldstern-Shelah model of $\mathfrak{r}<\mathfrak{u}$? …
Todd Eisworth's user avatar
5 votes
1 answer
356 views

Calculate the $\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$ cofinalities of th...

It is clear that $$\max\{{\uparrow\downarrow}(\mathfrak P),{\downarrow\uparrow}(\mathfrak P)\}\le \min\{{\downarrow}(\mathfrak P),{\uparrow}(\mathfrak P)\}.$$ I would like to know the values of the … Calculate the $\downarrow$-cofinality ${\downarrow}(\mathfrak P)$ of the poset $\mathfrak P$. In particular, is ${\downarrow}(\mathfrak P)=\mathfrak c$? Or ${\downarrow}(\mathfrak P)=\mathfrak d$? …
Taras Banakh's user avatar
  • 41.8k

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