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Results tagged with axiom-of-choice
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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.
4
votes
How much Dependent Choice is provable in $Z_2$? And what about Projective Determinacy?
As explained below, the "amount" of Dependent choice provable in Second Order Arithmetic is precisely $\Sigma^1_{2}$.
$\mathrm{Z}_2$ (second order arithmetic) proves $\Sigma^1_{2}$-DC (where DC = De …
- 17.7k
10
votes
Can a countable union of two-element sets be uncountable?
The point of this answer is to draw attention to the easy proposition below and the historical remark that follows it. In this answer "countable" means countably infinite (the finite case is trivial …
- 17.7k
18
votes
Accepted
SPOT as a conservative extension of Zermelo–Fraenkel
In plain terms, the conservativity of SPOT over ZF means that if a particular statement S in the language of ZF is provable in SPOT, then ZF can already prove S (with a possibly different proof). Note …
- 17.7k
16
votes
Is Global Choice conservative over Zermelo with Choice?
UPDATE: Elliot Glazer outlined a proof of nonconservativity of Z + GC over Z in his answer (on this page) to this MO question back in November of 2021; he has now (Dec 20, 2023) posted a paper on arX …
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10
votes
Accepted
Are there known ways to posit definable global choice in ZF without positing V=L?
Two comments/answers:
(1) By an old theorem of Roguski, for any $\Sigma_2^{\text{ZFC}}$ sentence $\phi$, the theories $\text{ZFC} + \phi$ and $\text{ZFC + V=HOD} + \phi$ are equiconsistent.
Rog …
- 17.7k
5
votes
Are all models of ZF + DC + "All set of reals are lebesgue measurable" also models of CH?
The question seems to be an open research problem; it was posed in 2011 on MO (and has remained unanswered), see:
Lebesgue Measurability and Weak CH
- 17.7k
15
votes
Accepted
What axioms are stronger than the Axiom of choice?
The axiom "every set is constructible" (denoted V = L), and the axiom "every set is definable from an ordinal parameter" (denoted often as V= HOD, and sometimes as V= OD) each implies AC, and each is …
- 17.7k
8
votes
Does ZFC prove the universe is linearly orderable?
Joel nicely answers Asaf's question.
Here I just want to add some footnotes
to his answer.
1. Joel' argument shows that $ZFC$ cannot even prove the definable class form of axiom of choice for …
- 17.7k
10
votes
Accepted
Is Lebesgue/Borel non-measurability actually caused by non-uniqueness?
The answer below has been edited in light of other answers and comments.
There are all sorts of models of $ZFC$ in which every set is definable without parameters, including nonmeasurable sets; indeed …
- 17.7k
20
votes
Half Cantor-Bernstein without choice
Your teacher's intuition is the correct one.
First of all, recall that a set $S$ is said to be Dedekind finite if there is no bijection between $S$ and any proper subset of $S$. In the early days of …
- 17.7k