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8
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0
answers
241
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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? …
10
votes
0
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495
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Is there a model of set theory in which $\mathfrak p< \mathfrak b < \mathfrak q$?
Is there a model of set theory in which $\mathfrak p< \mathfrak b < \mathfrak q$? … $\min\{\mathfrak b, \mathfrak q \}\in \{\mathfrak p,\mathfrak q \}$ ? …
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 $\forall a\in A[x\leq a]$. …
2
votes
0
answers
81
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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? …
2
votes
1
answer
118
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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 $\mathfrak{d} < \mathfrak{c}$. …
5
votes
1
answer
154
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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? …
8
votes
2
answers
479
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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. …
15
votes
3
answers
615
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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 $\mathfrak d_{\omega, \omega}=\mathfrak d$). …
4
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1
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178
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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. …
4
votes
1
answer
220
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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 $\beta<\mathfrak{m}$). …
13
votes
2
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1k
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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 $\mathfrak{i}\mathfrak{i}=\aleph_0$? …
6
votes
1
answer
157
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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 $< \mathfrak{d}$? …
7
votes
1
answer
431
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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. …
7
votes
0
answers
443
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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 $\mathfrak{r}<\mathfrak{u}$? …
5
votes
1
answer
356
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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 ${\downarrow}(\mathfrak P)=\mathfrak d$? …