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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.
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
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
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
3
votes
2
answers
295
views
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 $\mathfrak{p}=\mathfrak{t}$? …
- 775
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
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
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
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
4
votes
3
answers
395
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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
8
votes
1
answer
818
views
Is the smallest $L_\alpha$ with undefinable ordinals always countable?
Let $\mathfrak{t}$ be the least ordinal such that $L_{\mathfrak{t}}$ has undefinable ordinals; i.e. there is an $\alpha<\mathfrak{t}$ such that $L_{\mathfrak{t}}$ cannot define $\alpha$. … {t}$ is eventually writable, then for every $\alpha<\mathfrak{t}$, $\alpha$ can be defined in $L_{\mathfrak{t}}$ as $\mathcal{O}^{++}(n)$ for some $n$
That's a contradiction, so there is some ordinal below …
- 1,252
6
votes
1
answer
297
views
What is the height (or depth) of $[\mathbb{N}]^\infty$?
Then $\mathfrak{t}<\mathfrak{ht}\le\mathfrak{c}^+$. … Is it consistent that
"$\aleph_1=\mathfrak{t}<\mathfrak{b}=\mathfrak{c}=\aleph_2<\mathfrak{ht}$"? …
- 3,104
4
votes
Accepted
When is there an unbounded tower in $[\mathbb{N}]^\infty$?
Since the random reals added constitute a non null set of size $\aleph_1$, standard arguments show that $V^P \models \mathfrak{p} = \mathfrak{t} = \aleph_1$. … So $V^p \models \aleph_1 = \mathfrak{t} < \mathfrak{b} = \kappa = 2^{\aleph_0} $ plus BT. …
- 9,631
3
votes
Accepted
What is the height (or depth) of $[\mathbb{N}]^\infty$?
In the Hechler model, $\aleph_1 = \mathfrak{t} < \mathfrak{b} = \mathfrak{c} = \aleph_2 < \mathfrak{ht}$. … That $\aleph_1 = \mathfrak{t} < \mathfrak{b} = \mathfrak{c} = \aleph_2$ is discussed in Section 11.6 of Blass's handbook article. …
- 18.5k
7
votes
0
answers
553
views
Recent application of model theory in set theory by Shelah-Malliaris
Recently one of the oldest open problems in set theory about the cardinal invariants of the continuum (i.e the question of whether $\mathfrak{p}=\mathfrak{t}$) was solved by Shelah and Malliaris (see " …
user39869
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 $V_\alpha\subseteq^* U_\beta$ …
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