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Results for "mathfrak p" 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.
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Does $\mathfrak{P}(A)\cong\mathfrak{P}(B)$ imply $A\cong B$? [duplicate]
Does $\mathfrak{P}(A)\cong\mathfrak{P}(B)\implies A\cong B$ hold for arbitrary sets $A, B$?
Notation:
We write $\mathfrak{P}(M)$ to denote the power set of a set $M$. …
8
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241
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Topological applications of $\mathfrak{p}=\mathfrak{t}$
{p}$ has some topological property, then $\mathfrak{t}$ also has it). … So, my question is: are there any interesting consequence of $\mathfrak{p}=\mathfrak{t}$ in Topology? …
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5
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1
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204
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$\mathfrak{p}=\mathfrak{b}=\mathfrak{a}=\aleph_1$ 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 $\mathfrak{d}=\mathfrak{c}=\kappa$ ? … $\mathfrak{p}$, is the least size of family $\mathcal{E}\subseteq [\omega]^\omega$ such that $\mathcal{E}$ has the SFIP and there does not exist any $\mathfrak{p}$seuod-intersection of $\mathcal{E}$. …
- 181
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1
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The Parovichenko cardinal, is it equal to $\max\{\aleph_2,\mathfrak p\}$?
These two results yield the inequality $\mathfrak P\ge\max\{\aleph_2,\mathfrak p\}$.
So, under CH we have $\mathfrak P=\aleph_2>\mathfrak c=\mathfrak p=\aleph_1$. … Is it consistent that $\mathfrak P>\max\{\aleph_2,\mathfrak p\}$? …
- 41.8k
12
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1
answer
1k
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Hausdorff gaps and $\mathfrak{p}=\mathfrak{t}$
Recently Malliaris and Shelah proved that $\mathfrak{p}=\mathfrak{t}$
(see: http://math.uchicago.edu/~mem/Malliaris-Shelah-CST.pdf). … of the form $\omega^\omega/G$
in the case $\mathfrak{p}<\mathfrak{t}$. …
user44578
10
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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 $\mathfrak p< \mathfrak b < \mathfrak q$? … $\min\{\mathfrak b, \mathfrak q \}\in \{\mathfrak p,\mathfrak q \}$ ? …
- 1,101
3
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2
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295
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Short proof of $\mathfrak{p}=\mathfrak{t}$ by Juris Steprans [duplicate]
I have just read this question Short proof of $\frak p=t$.
The link present in the answer about the proof given by Steprans doesn't work anymore. … Can you provide me the updated link to the proof by Steprans of $\mathfrak{p}=\mathfrak{t}$? …
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5
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1
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369
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Is it true that $\mathit{MA}(\omega_1)$ iff $\omega_1<\mathfrak{p}$?
Recall that
$\mathfrak{p}=\min\{|F|: F$ is a subfamily of $[\omega]^{\omega}$ with the sfip which has no infinite pseudo-intersection $\}$. … The cardinal $\mathfrak{q}_0$ defined as the smallest cardinality of a subset of $\mathbf{R}$ which is not a $Q$-space.
Q1. Is it true that $\mathit{MA}(\omega_1)$ iff $\omega_1<\mathfrak{p}$ ?
Q2. …
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4
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1
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Countable support iterations of proper forcings with fusion (axiom A), that make $\mathfrak{...
What are some proper forcing notions with fusion (satisfying axiom A but not countably closed), which when iterated $\aleph_2$ times with countable support, preserve cardinals and make $\mathfrak{p}=\aleph … For example, the prototypical such forcing, if in the above we replace $\mathfrak{p}$ with:
$\mathfrak{b}$, we get Laver forcing,
$\mathfrak{d}$, we get Miller forcing,
$\mathfrak{c}$, we get Sacks …
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3
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1
answer
137
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Is $\mathfrak p=\omega_1$ equivalent to the existence of a Hausdorff gap without infinite ps...
The definition of the small uncountable cardinal $\mathfrak p$ implies that under $\mathfrak p>\omega_1$ each Hausdorff gap has an infinite pseudointersection. … What does happen under $\mathfrak p=\omega_1$?
Problem. Is $\mathfrak p=\omega_1$ equivalent to the existence of a Hausdorff gap without infinite pseudointersection? …
- 41.8k
6
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404
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Is $2^\mathfrak{m}<2^\mathfrak{n}\Rightarrow\mathfrak{m}<\mathfrak{n}$ equivalent to the axi...
Lindenbaum and Tarski assert in ``Communication sur les recherches de la théorie des ensembles'' without proof that $\mathfrak{p}^\mathfrak{m}<\mathfrak{p}^\mathfrak{n}\wedge\mathfrak{p}\neq0\Rightarrow … This is not difficult to prove, but I do not know whether the statement for $\mathfrak{p}=2$ is also equivalent to the axiom of choice. …
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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 ${\downarrow}(\mathfrak P)=\mathfrak d$? …
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13
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1
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354
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Is the Martin's axiom number $\mathfrak m$ regular
This is because
$$\mathfrak m \leq \mathfrak p \leq \mathfrak c$$
where $\mathfrak p$ is the pseudo-intersection number. … Hence $\mathfrak m = \mathfrak c$ implies $\mathfrak m = \mathfrak p$ and $\mathfrak p$ can be shown to be regular. …
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3
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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 by forcing theorem, $\mathfrak{M}[G]\vDash A\subset G$ if and only if there exists a $p\in G$ such that $p\Vdash \check{A}\subset \check{G}$. …
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627
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A new cardinal characteristic (related to partitions)?
A family $\mathfrak P$ of partitions of $\omega$ is called directed if for any two partitions $\mathcal A,\mathcal B\in\mathfrak P$ there exists a partition $\mathcal C\in\mathfrak P$ such that each set … An infinite subset $D\subset\omega$ is called $\mathfrak P$-discrete if for any partition $\mathcal P\in\mathfrak P$ there exists a finite set $F\subset D$ such that for any $P\in\mathcal P$ the intersection …
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