Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results for "mathfrak p" tagged with set-theory
Search options questions only
not deleted
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.
12
votes
1
answer
533
views
Building the real from Dedekind finite sets
It is also known that the reals may contain a Dedekind finite set, so they are Dedekind finite union of disjoint sets (if $\mathfrak p<\mathfrak c$ is Dedekind finite then $\frak p\times c=c$ so we may … partition $\Bbb R$ into $\frak p$ many parts)
Is it possible to get a "better" construction of the reals using Dedekind finite sets? …
- 1,676
5
votes
1
answer
369
views
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. …
- 1,101
3
votes
0
answers
78
views
Is $\mathfrak q_0$ equal to the smallest cardinality of a second-countable $T_1$-space which...
It is known that $\mathfrak p\le\mathfrak q_0\le\mathfrak b$. … The standard proof of the inequality $\mathfrak p\le\mathfrak q_0$ gives a bit more:
Any second-countable $T_1$-space of cardinality $<\mathfrak p$ is a $Q$-space. …
- 41.8k
14
votes
2
answers
666
views
Are there interesting examples of theorems proved using ‘height’ extensions?
$\mathfrak{p}=\mathfrak{t}$, remarkable cardinals, Todorčević and Farah's book "Some applications of the method of forcing"). …
- 613
6
votes
0
answers
404
views
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. …
- 1,782
2
votes
0
answers
77
views
When are classes with prescribed reducts "pseudo"-elementary?
Another necessary condition is that $\{0\},\{1\} \in \mathfrak L(H) \subseteq \{\{0\},\{1\}\} \cup \mathfrak P(\mathbb N_{\geq 2})$ . … set $\mathfrak P(\mathbb N)$, while for an element $L \in \mathfrak P(\mathbb N)$, we have $\mathfrak L(L)$ if and only if $L \in \mathfrak L(H)$. …
- 365
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 ${\downarrow}(\mathfrak P)=\mathfrak d$? …
- 41.8k
1
vote
1
answer
140
views
Thinning directed sets ${\frak P}$ of partitions of $\omega$ with no ${\frak P}$-discrete su...
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 … Let $\mathfrak P$ is a family of partitions of $\omega$. …
- 48.8k
3
votes
0
answers
122
views
The existence of $T$-ultrafilters in ZFC
A partition $\mathcal P$ is called finitary if $\sup_{P\in\mathcal P}|P|$ is finite.
Let $\mathfrak P$ is a family of partitions of $\omega$. … An infinite subset $T\subset\omega$ is called $\mathfrak P$-thin if for any partition $\mathcal P\in\mathfrak P$ there exists a finite set $F\subset T$ such that for any $P\in\mathcal P$ the intersection …
- 41.8k
11
votes
1
answer
627
views
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 …
- 41.8k
3
votes
1
answer
137
views
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
2
votes
1
answer
135
views
Compactifications with remainder $[0,\omega_1]$ and convergent sequences
The statement $(\star)$ does not hold under $\omega_1<\mathfrak p$. That is why I am asking only about the consistency of $(\star)$.
Remark 2. …
- 41.8k
2
votes
0
answers
116
views
Cardinal characteristics of the ideal of $\sigma$-continuity of the Pawlikowski function
On the other hand, it can be shown that $\mathrm{non}(\mathcal I_P)\ge\mathfrak p$.
It is well-known that the strict inequality $\mathfrak p<\mathrm{non}(\mathcal M)$ is consistent. … Which of the inequalities $\mathfrak p<\mathrm{non}(\mathcal I_P)$ and $\mathrm{non}(\mathcal I_P)<\mathrm{non}(\mathcal M)$ is consistent?
Problem 2. …
- 41.8k
2
votes
0
answers
120
views
Two small uncountable cardinals related to Q-sets
It is clear that $\mathfrak q_2\le \mathfrak q_1\le \mathfrak q_0$. By analogy with the proof of Theorem 2 in this paper, it can be shown that $\mathfrak p\le\mathfrak q_2$. … So, we have the inequalities
$$\mathfrak p\le\mathfrak q_2\le\mathfrak q_1\le\mathfrak q_0.$$
The cardinal $\mathfrak q_0$ has been studied in the literature. …
- 41.8k
5
votes
0
answers
102
views
Universal and strong $Q$-sets
\mathbb R$ form a Polish space $P$ and every subset $A\subset f^{-1}(X)$ there exists a $\sigma$-compact set $C\subset P$ such that $C\cap f^{-1}(X)=A$. … It is known that $$\mathfrak p\le\mathfrak q_0\le \min\{\mathfrak b,\mathrm{non}(\mathcal N)\}.$$
Using the inequality $\mathfrak q_0\le\mathfrak b$, it can be shown that every subset $X\subset\mathbb …
- 41.8k