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Results tagged with axiom-of-choice
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user 1946
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.
17
votes
2
answers
2k
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Do the surreal numbers enjoy the transfer principle in ZFC?
The surreal field $\newcommand\No{№}\No$ is definable in ZFC, and it is easy to see that the surreal order is $\kappa$-saturated for every cardinal $\kappa$, precisely because we fill any specified ga …
- 13.1k
10
votes
What is the relationship between non-existence of those kinds of singular sets and AC?
Yes, it is equivalent to axiom of choice to say that there is no supersingular set.
It is easy to see with AC that there is no supersingular set.
Conversely, suppose that there is no supersingular set …
- 236k
11
votes
Accepted
If existence of a pre-isomorphism implies existence of an isomorphism, would AC follow?
You say
Now using axiom of choice one can easily prove that for any distinct sets if there exists a pre-isomorphism between them then there exists an isomorphism between them.
But this is not true. …
- 236k
21
votes
What choice principles does "every set is in bijection with a transitive set" imply?
Here is one instance, although not with a "classical" choice principle.
Namely, your principle TC implies the rigid relation principle RR, a weak choice principle introduced by Justin Palumbo and myse …
- 236k
7
votes
Hereditarily countable sets in Antifounded ZF
Update. This answer does not answer the question that was asked, since Jech is using what had seemed to me as an idiosyncratic definition of hereditary countable. But upon reflection, I find his defin …
- 236k
14
votes
Proof/Reference to a claim about AC and definable real numbers
Unfortunately, the claim you have stated is not true. Regardless of the axiom of choice, every real is definable from a countable sequence of ordinal parameters, since the real is definable from the s …
- 236k
11
votes
Accepted
Complexity of definable global choice functions
There is no such phenomenom for $n\geq 2$.
The reason is that if a model of ZFC has a definable choice function, of any complexity, then it actually has one of complexity $\Delta_2$. This is because t …
- 236k
8
votes
Accepted
Would strengthening Foundation and Choice in NBG, make it equi-consistent with MK?
The answer is no.
This follows from a modification of Kameryn's answer at your other question. Namely, KM implies the existence of a transitive model of NBG, and transitive models will always satisfy …
- 236k
16
votes
Accepted
Long chains of amorphous cardinalities
If $A$ is amorphous, then every subset of $A$ is either finite or cofinite in $A$. Since every cardinal below $A$ is determined by a subset of $A$, it follows from this that the cardinals below $A$ ar …
- 236k
13
votes
Accepted
Class-theoretic division paradox
The answer is yes.
First, let us observe the following lemma. Let us work in GBc, that is, Gödel-Bernays set theory with the axiom of choice, but only choice for sets, and not global choice.
Lemma. If …
- 236k
39
votes
3
answers
3k
views
Can one show that the real field is not interpretable in the complex field without the axiom...
We all know that the complex field structure $\langle\mathbb{C},+,\cdot,0,1\rangle$ is interpretable in the real field $\langle\mathbb{R},+,\cdot,0,1\rangle$, by encoding $a+bi$ with the real-number p …
- 20.5k
10
votes
Accepted
Is there a strict limit on choice principles in $\sf ZFC$?
The answer is no. Every theorem of ZFC is equivalent to a choice principle. I will prove this is true, but nevertheless also I will be first to agree that the manner in which this is true is trivial a …
- 236k
9
votes
Accepted
Is the Ordering Principle equivalent to a selection principle?
Here is one way to view the so-called ordering principle as a selection principle.
Theorem. The following are equivalent over ZF set theory:
Every set admits a linear order.
For every set $X$, there …
- 236k
13
votes
Accepted
Logical strength of a statement about vector spaces
This can be proved with just countable choice, so it is not equivalent to DC, since countable choice is known to be weaker than DC.
For each $n$, let $B_n$ consist of all $n$-tuples of independent vec …
- 236k
7
votes
Is there a class choice principle over MK that is equivalent to class well ordering over MK?
Usually the notation WO refers to the well-order principle for sets only, and this is usually taken as part of KM. So let me refer to your global-well-order principle as the class-well-order principle …
- 11.3k