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Results tagged with set-theory
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user 7206
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.
6
votes
How strong is limitation of size + generalized continuum hypothesis?
I might be missing something, but if $V$ is a model of $\sf ZFC$, then $V_{\omega+\omega}$ is a model of Zermelo's set theory (and then some, since it also has regularity).
If $V\models\sf GCH$ then …
- 39.8k
3
votes
Accepted
Is any axiom system for sets categorical?
Of course not, since you can always take a permutation of $M$ and redefine $\in$ as the transport of structure defined by the permutation.
But even if you mean "up to isomorphism", the answer is stil …
- 39.8k
3
votes
When do we have a bijection between a proper class A and its power set class P(A)?
It suffices to show that for $A=\sf Ord$. Since in that case global choice holds and the rest follows.
Fix a pairing function for ordinals, namely a bijection between $\sf Ord$ and $\sf Ord\times Ord …
- 39.8k
1
vote
Order in bijective-equivalent collections of proper classes in set-theory
In a previous question Joel Hamkins linked to a model obtained by adding Cohen subsets to every regular cardinal. This model had the property that the universe could not be linearly ordered.
In parti …
- 39.8k
7
votes
Accepted
How big is the proper class of all sets?
Both the statement are true, which is fine because we talk about relative consistency here.
If $\sf ZFC$ is consistent then we know how to generate a model of $\sf ZFC^-+\it G\neq V$.
On the other h …
- 39.8k
2
votes
Accepted
About the rank of sets
Note that every function $f$ is a subset of $A\times B$, and so the rank of $C$ is at most $\newcommand{\rank}{\operatorname{rank}}\rank(\mathcal P(A\times B))$.
It is possible that $\rank(A\times B) …
- 39.8k
3
votes
Question about HOD
There's not enough information in the question. If $M\models V=L$, then in $M$ we have that $\mathrm{HOD}^M=L^M=M$, and therefore $\mathrm{HOD}^{\mathrm{HOD}^M}=L^M=M$ as well.
In particular we have …
- 39.8k
13
votes
ZF + the reals are the countable union of countable sets consistent
T. Jech, The Axiom of Choice. This particular proof appears in Chapter 10.
Essentially, the forcing goes through collapsing all the $\aleph_n$ (for finite $n$) to be countable, so in the full generic …
- 39.8k
4
votes
Accepted
A question about "paradoxical" sentences in the language of ZF set theory.
Here are two examples:
Let $F(x)$ be the statement "There exists $y$ such that $x$ is the power set of $y$", or formally, $$F(x)=\exists y\forall u(u\subseteq y\leftrightarrow u\in x).$$
Since every …
- 39.8k
16
votes
Accepted
Cardinalities of which there exists partitions of a set containing elements of the same size
This is equivalent to the axiom of choice.
If the axiom of choice holds, then given $A$ and $B$ which are infinite, then $|A\times B|=\max\{|A|,|B|\}$. So let's say $|A|$ is the maximal one, then thi …
- 39.8k
2
votes
A simple proof for a case where: $\mathbf{L}_\mu \models ZF^-$?
Here is a possible route for a proof, but I'm not 100% sure if the idea holds up. I'm posting it here as CW so others can make adjustments if necessary.
Step 1: $\mu$ is an admissible ordinal.
Th …
11
votes
Accepted
Is there a "natural" bijection between models of $ZFC$ and $ZF\neg C$?
Well, the simple answer is that if there is a single model, there is a proper class of models, of each cardinality possible.
But let's instead restrict to transitive models. Then the answer is consid …
- 39.8k
6
votes
AC and Krull's theorem equivalence
Originally the proof was given by Hodges, from 1979 [1].
Banaschewski gave a new proof of the theorem in 1994 [2].
Bibliography:
Hodges, W., Krull implies Zorn. Journal of the London Mathematica …
- 39.8k
4
votes
Accepted
Existence of a branch of limit ordinal length
No. Aronszajn trees are the classical example here. The formal definition of an Aronszajn tree is simply a tree of height $\omega_1$ where every level is countable. This tells you nothing about your r …
- 39.8k
7
votes
1
answer
426
views
$\kappa$-scales and the continuum
In Jech's Set Theory he defines a $\kappa$-scale as a family of functions $\langle f_\alpha\colon\omega\to\omega | \alpha < \kappa \rangle$ for which:
$f_\alpha < f_\beta$ except maybe for a finite …
- 39.8k