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Results tagged with axiom-of-choice
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user 7206
An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty sets. The axiom of choice is related to the first of Hilbert's problems.
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 …
- 39.8k
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 …
- 39.8k
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 …
- 39.8k
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 …
- 39.8k
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, …
- 39.8k
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- …
- 39.8k
6
votes
Accepted
Long chains of Dedekind finite sets
The answer to all of these is yes. Monro constructed a model where for every ordinal $\alpha$ there is a Dedekind-finite set mapped onto $\alpha$.
This alone gives us arbitrarily long chains. Take suc …
- 39.8k
5
votes
Accepted
Weak Power Hypothesis and Dependent Choice
I've reduced the problem to "every infinite set is Dedekind infinite", which is a consequence of WPH if my memory serves me right (this was answered before on the site).
We'll show that well-ordered c …
- 39.8k
4
votes
Does $\mathsf{ZF}$ prove $\operatorname{Col}(\lambda,\kappa)$ preserves cardinals below $\la...
First of all, note that if $\lambda$ is a singular limit cardinal, then the forcing is terribly behaved already in $\sf ZFC$. Even if $\lambda$ is a successor, you'd still find that the same argument …
- 39.8k
6
votes
Exponentiation of Dedekind cardinals
Note that if $\frak m\leq^* n$, then $2^\mathfrak m\leq 2^\mathfrak n$, since given a surjection from $N$ onto $M$, the preimages define an injection between the power sets. So, if $\frak m\leq^* n\le …
- 39.8k
5
votes
Accepted
What are some "easy" violations of $\mathsf{SVC}$?
There are two type of models of $\lnot\sf SVC$.
Take a symmetric system and vary its "base cardinal" and take a product thereof. This was done in many papers:
Asaf Karagila, Embedding orders into th …
- 39.8k
5
votes
Accepted
References for the axiom of surjective comparability
Not too much that I'm aware of. It comes up in the case of $\omega$, since that would imply that even if Dedekind finite sets exist, they can at least be mapped onto $\omega$ (equivalently, "the power …
- 39.8k
10
votes
Accepted
Small Violations of Choice: Can we force AC without collapsing the cardinalities of ordinals?
The answer is no.
Firstly, let's get the obvious out of the way. The Feferman–Levy model is a symmetric extension in which $\omega_1$ is singular. If we force the axiom of choice, we must collapse the …
- 39.8k
6
votes
Class-theoretic division paradox
Let me just add one small example to Joel's very good answer, which deals with the a case where $\sf AC$ fails (for sets).
In your typical model of the form $L(\Bbb R)$ we will usually have the failur …
- 39.8k
8
votes
Automorphisms of algebraically closed fields without the Axiom of Choice
Läuchli constructed an algebraic closure of the rationals which does not have any non-identity automorphisms.
Läuchli, H., Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, 1-18 (1962). ZBL0108.0 …
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