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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.
7
votes
Is this version of Zorn's lemma provable in ZF?
Suppose that there is a Dedekind-finite set of reals, $A$. Enumerate the rational numbers as $\{q_n\mid n<\omega\}$, and for each $a\in A$ let $D_a\subseteq\omega$ be a recursively chosen subset such …
12
votes
Accepted
Partitioning a set of cardinality $\kappa$ into more than $\kappa$ disjoint subsets
Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.
The Div …
11
votes
Accepted
NBG, ZFC+I, and Global Choice
Care is needed here.
It is true that $\sf ZFC+I$ implies the consistency of $\sf NBG$. And it is true that $\sf ZFC+I$ does not prove Global Choice.
However, when we talk about consistency statements, …
6
votes
What is Gödel's pairing function on ordinals?
According to this .pdf file the definition is this:
Consider the canonical ordering on $\mathsf{Ord\times Ord}$:
$$(\alpha,\beta)\prec(\gamma,\delta)\iff\begin{cases}
\max\lbrace\alpha,\beta\rbrace\lt …
13
votes
When can we add choice to a model of ZF
There's no known condition, and this isn't very well researched in the literature.
Monro, in his paper on Dedekind finite sets, constructs a model with a proper class of Dedekind finite sets, so AC a …
18
votes
What choice principles does "every set is in bijection with a transitive set" imply?
You can say a bit more than just every infinite set surjects onto $\omega$. Every infinite set is Dedekind infinite.
The reason is simple. If $T$ is an infinite transitive set, then $T\cap V_\omega$ i …
50
votes
0
answers
2k
views
How many algebraic closures can a field have?
Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is …
12
votes
Can there be a proper class of Dedekind-finite cardinals?
Not only there can be a proper class of them, every set can be the image of a Dedekind-finite set.
This is embedded in the proof of the Morris model, and in https://arxiv.org/abs/1911.09285 it is made …
8
votes
1
answer
269
views
What is the consistency strength of "Singular worldly that is inaccessible in an inner model"?
In short, what can we say about the consistency strength of "$\kappa$ is a singular worldly and inaccessible in an inner model"?
Clearly, $0^\#$ exists since we have a singular cardinal which is regul …
5
votes
Accepted
Is reflection on Grothendieck universes equivalent to TG set theory?
The theory you suggest is significantly stronger than Tarski–Grothendieck. The latter theory is, essentially, $\sf ZFC$ augmented by "there is a proper class of inaccessible cardinals".
The theory you …
11
votes
Cardinal arithmetic under determinacy
Starting from $L(\Bbb R)$, we can take a symmetric extension which preserves $\sf DC$ and adds an $\omega_1$-amorphous set, just somewhere far above $\Theta$.
So we cannot prove that (1) or (2) hold f …
20
votes
Accepted
Proof/Reference to a claim about AC and definable real numbers
The argument, given to me by Hugh Woodin over a drink in Barcelona in September 2016, is a nice retort to "AC is obviously false since there cannot be a well-ordering of the real numbers". It is argua …
20
votes
2
answers
2k
views
What is $\omega_1^{CK}(\mathsf{Ord})$?
We know that if $\alpha<\omega_1^{CK}$ then there is some recursive $R$ such that $(\omega,R)$ has order type $\alpha$.
Let's consider now the ordinals, $\mathsf{Ord}$ with their natural order. This …
3
votes
Accepted
Turning linear ordering into well-ordering
You can't do this.
Working over the Solovay model, take a symmetric extension as follows. First, force by adding $\Bbb Q\times\omega_1$ many Cohen reals, with order automorphisms (lexicographic order, …
10
votes
Accepted
The equivalence of Dedekind-infinite and dually Dedekind-infinite as a weak form of (AC)
Tarski proved that if there is an infinite Dedekind-finite set, then there is one which is dually Dedekind-infinite as well.
Therefore, postulating that all dually Dedekind-infinite sets are Dedekind- …