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Results tagged with set-theory
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user 4600
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.
8
votes
Is it consistent with ZFC (or ZF) that every definable family of sets has at least one defin...
Yes,
fix a definable relation $\le_L$ that well-orders all of $L$.
If $V=L$ then every definable nonempty set $A$ has a definable member $a$, namely:
$a :=$ the $\le_L$-least element of $A$.
- 24.8k
1
vote
non(Meager) in Random times Random extension
I'm curious to hear the set-theoretic answer. In the meantime it's interesting to note that in the computability-theoretic setting the answer seems to be "no".
There the analogue would be:
Suppose $A …
- 24.8k
3
votes
Accepted
What are the sufficient conditions for a set P to be the set of all majority subsets of a se...
I think this is equivalent (for an odd number of elements of $S$) to:
$Q$ is closed under supersets;
For any set $A\subseteq S$, exactly one of $A\in Q$, $A^c\in Q$ holds;
For any permutation $\pi$ …
- 24.8k
4
votes
Intuition for the infinite cardinals p and t (now that p = t)?
A couple of simple examples; they don't have cardinality $\mathfrak p$ or $\mathfrak t$ but depending on your knowledge they may or may not help.
They are families of sets of natural numbers that I'l …
- 24.8k
1
vote
A stronger notion of injectivity
If $A$ and $B$ are countable and equipped with probability measures $\mu_A$, $\mu_B$ that give each element positive probability then any measure-preserving map is a bijection. (I guess per @MonroeEsk …
- 24.8k
10
votes
4
answers
2k
views
Axiom of Infinity needed in Cantor-Bernstein?
Can one prove the Cantor-Bernstein (or Schröder-Bernstein) theorem without using the Axiom of Infinity?
- 24.8k
1
vote
Accepted
Is there a name for this cardinal?
It appears in the Hitting Set Problem, and so it would make some sense to call $\tau(\omega)$ the hitting set cardinality.
- 24.8k
6
votes
Accepted
Translates of meager sets
No, there is no such set.
The situation for meager sets is dual to that described by Pietro Majer in a comment on Translates of null sets,
"I was vaguely thinking to Hausdorff measures w.r.to gaug …
- 24.8k
5
votes
${\frak b}$ and ${\frak d}$ defined with $\leq$ instead of $\leq^*$
${\frak b}'=\aleph_0$ since you can take $S$ to be the collection of constant functions.
${\frak d}' = {\frak d} + \aleph_0$ since you can take any family $S$ realizing $\frak d$ and close it under f …
- 24.8k
5
votes
Accepted
Partitioning an infinite cardinal $\kappa$ into pairwise neighboring subsets
Yes. List the pairs $(\alpha,\beta)$ with $\alpha<\beta<\kappa$ as $(\alpha_\lambda,\beta_\lambda), \lambda<\kappa$.
Then construct the sets $B_\alpha\in\mathcal B, \alpha<\kappa$ as follows:
At stage …
- 24.8k
1
vote
Accepted
Models of ZFC and the Borel hierarchy
@FrançoisG.Dorais commented:
By reading the axioms (i.e. every axiom is satisfied, as a formalized statement) it's a $\Pi^0_{\omega+1}$ set (since satisfying a first-order sentence is $\Pi^0_n$ fo …
- 24.8k
8
votes
1
answer
479
views
VC dimension of standard topology on the reals
Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is an open set $U$ with $D=S\cap U$?
I'm asking merely out of curiosity, but I'll mention that thi …
- 24.8k
10
votes
2
answers
1k
views
$\aleph$ looks like $\mathbb N$?
We all know the notation $\aleph_\lambda$ for the $\lambda$th (or, I guess, $\lambda+1$st) infinite cardinal number; in particular $\aleph_0$ is the cardinality of the the set of natural numbers $\mat …
- 24.8k
9
votes
Accepted
Choice sets from above and below
Let $\cal S=\{\{1,2\},\{2,3\},\{3,1\}\}$.
Then $\cal S$ has no choice set, whatsoever.
So there is no asymmetry -- not every shy set is contained in a choice set, and not every gregarious set contai …
- 24.8k
9
votes