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Results for "mathfrak t" 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
votes
Short proof of $\frak p=t$
While not short, A Measure Theoretic Proof of $\mathfrak p=\mathfrak t$ gives a proof that does not rely on model theoretic or proof theoretic techniques. …
- 151
4
votes
Philosophy of forcing and ctm
Todorčević and Farah's book, Malliaris and Shelah's result that $\mathfrak{p}=\mathfrak{t}$). …
- 613
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
2
votes
Game versions of the tower number $\mathfrak t$
For problem 1: yes, $\mathfrak{t}_I \leq \mathfrak{t}_J \leq \mathfrak{c}$. … In fact, this argument shows that $\mathfrak{t}_I \leq \mathfrak{t}_J \leq \mathfrak{s}$, where $\mathfrak{s}$ denotes the splitting number. …
- 18.5k
2
votes
1
answer
152
views
Game versions of the tower number $\mathfrak t$
It is easy to see that $\mathfrak t\le \mathfrak t_I\le\mathfrak t_J\le\mathfrak c^+$.
Problem 1. Is $\mathfrak t_I\le\mathfrak c$? $\mathfrak t_J\le\mathfrak c$?
Problem 2. … Is the strict inequality $\mathfrak t<\mathfrak t_J$ (resp. $\mathfrak t<\mathfrak t_I$ or $\mathfrak t_I<\mathfrak t_J$) consistent? …
- 41.8k
4
votes
Accepted
A permutation group inducing a topologically transitive action without dense orbits on $\ome...
Under $\mathfrak t=\mathfrak c$, every topologically transitive continuous action of a group $G$ on $\omega^*$ has a dense orbit.
Proof. … By the definition of the tower number $\mathfrak t$ and the equality $\mathfrak t=\mathfrak c>\alpha$, there exists an infinite subset $V_\alpha\subseteq\omega$ such that $V_\alpha\subseteq^* U_\beta$ …
- 41.8k
3
votes
Accepted
Relations between two tower numbers
Thus, in his model $\mathfrak{t}=\hat{\mathfrak{t}}=\aleph_1<\aleph_2=\mathfrak{b}=\mathfrak{c}$, and so the last question in your first problem has a negative answer. … Thus, this gives a model where $\mathfrak{t}<\hat{\mathfrak{t}}$.
Edit 3/17/20
I tracked down another reference (due to Dordal [2]) with some more information on this question. …
- 7,125
5
votes
0
answers
216
views
Improving a Lindstrom-y fact about $\mathcal{L}_{\omega_1,\omega}$?
{T}\subseteq\mathcal{L}\cap\mathbb{A}$, if every $\mathbb{A}$-finite subset of $\mathfrak{T}$ has a model then $\mathfrak{T}$ has a model. … Barwise compactness then gives us a model of the whole theory $\mathfrak{T}$. …
- 21.1k
4
votes
0
answers
212
views
Discrete version of Arzela-Ascoli theorem
$\mathfrak{t\le scp}$, where $\mathfrak{t}$ is the tower number.
Proof. Assume that $A=\{a_\xi \mid \xi<\kappa\}$ for $\kappa<\mathfrak{t}$. … For example, can we prove $\mathfrak{scp=t}$ or $\mathfrak{scp\le h}$ (where $\mathfrak{h}$ is a shattering number)? I would appreciate your help! …
- 3,042
8
votes
2
answers
479
views
Relations between two tower numbers
Consider two small uncountable cardinals:
$\mathfrak t=\min\{|T|:T\subset[\omega]^\omega$ is a tower$\}$;
$\hat{\mathfrak t}=\sup\{|T|:T\subset[\omega]^\omega$ is a regular tower$\}$. … It is clear that $$\mathfrak t\le\hat{\mathfrak t}\le\mathfrak c.$$
Martin's Axiom implies $\mathfrak t=\hat{\mathfrak t}=\mathfrak c$. …
- 41.8k
8
votes
0
answers
241
views
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? …
- 181
6
votes
Accepted
Is each Parovichenko compact space homeomorphic to the remainder of a soft compactification ...
It is true, in ZFC, that $\omega_1+1$ is soft-Parovichenko but "all compactifications with remainder $\omega_1+1$ are soft" is equivalent to $\mathfrak{t}>\omega_1$. …
- 11.4k
12
votes
Accepted
Is the smallest $L_\alpha$ with undefinable ordinals always countable?
${\mathfrak t}$ is the least $\beta$ such that there is a $\gamma<\beta$ with $L_\gamma \prec L_\beta$. That ${\mathfrak t} \leq$ the least such $\beta$ is obvious. … On the other hand, if $X \subset L_{\mathfrak t}$ is $\subseteq$-least with $X \prec L_{\mathfrak t}$, then $X \not= L_{\mathfrak t}$; hence if $\sigma \colon L_\gamma \cong X$, then either $\sigma$ is …
- 1,808
2
votes
Problem understanding a passage of the proof of $\mathfrak{p}=\mathfrak{t}$ involving forcing
At the beginning of the proof we can suppose by passing to a Levy collapse that $\mathfrak{t} = 2^{\aleph_0}$. …
- 2,146
4
votes
3
answers
395
views
Problem understanding a passage of the proof of $\mathfrak{p}=\mathfrak{t}$ involving forcing
a passage of the proof of Claim 14.7 of the paper "Cofinality spectrum theorems in model theory, set theory, and general topolgy" by Malliaris and Shelah, or equivalently Proposition 4D of Fremlin "p=t, … generic over $\mathfrak{M}$ and $\mathfrak{M}[G]$ the generic model extending $\mathfrak{M}$ and containing $G$ obtained using forcing. …
- 775