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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.
2
votes
Global or Relativised Dependent Choices
Another economical way to think about it is to use the
Reflection theorem (although unwinding this amounts just to
Andres' argument). Namely, by the Reflection Theorem there
is an ordinal $\theta$ suc …
- 236k
6
votes
Maximal chains and antichains of statements weaker than AC
The partial order of statements weaker than the axiom of choice is simply the upper-cone above AC in the Lindenbaum algebra of ZF statements. That is, it is a quotient of the Lindenbaum algebra by the …
- 236k
5
votes
Cantor's diagonal argument and ZF
Let me point out that in the case $X=\mathbb{N}$, the assertion seems to be simply equivalent to the existence of an injection $\omega_1\to \mathbb{R}$. Ricky has cleverly proved the forward implicati …
- 236k
2
votes
Can iterating countable unions give every set? (ZF)
If we assume ZF plus the assertion that $\omega_1$ is regular (which is provable from countable choice), or that indeed there is any uncountable regular cardinal $\delta$,
then such a set $S$ exists. …
- 236k
7
votes
Accepted
For models of ZF, if for some $A$ we have $L[A] = L$, what can we deduce about $A$?
Regarding your last question, it is easy to see that if
$A\cap L=\varnothing$, then $L[A]=L$, because at every stage, if we
have agreement $L_\alpha[A]=L_\alpha$ so far, then having
$A$ as a predicate …
- 236k
12
votes
Accepted
Relations of axioms of choice
Yes, the axiom of choice is equivalent to the assertion that $\text{AC}_X$ holds for every definable $X$.
One usually has to take a little care with foundational matters when definability is involve …
- 236k
5
votes
Why doesn't choice imply global choice (in NBG)?
The other answers are great. Let me point out, however, that one doesn't need the inaccessible cardinal in Asaf's argument, because one can force over any ZFC model, and there is a pure-forcing argume …
- 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
13
votes
Accepted
Non-constructive existence proofs without AC?
If ZF is consistent, then the answer is no.
ZF proves that there is an $x$ such that $P(x)$: either $x$ is an $L$-generic Cohen real or there is no $L$-generic such real. (Here, by $L$ I mean the co …
- 236k
41
votes
Accepted
Are all sets totally ordered ?
In the paper Dense orderings,
partitions and weak forms of choice, by Carlos G. Gonzalez FUNDAMENTA MATHEMATICAE 147 (1995), the author states the following theorem, where AC is the Axiom of Choice, …
- 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
3
votes
Linear space with (Hamel) basis and the axiom of choice
Yes, it is true that AC is equivalent to the assertion that every vector space has a basis, and this is discussed in all the usual treatments of equivalents to the axiom of choice. For example, the re …
- 236k
4
votes
The role of the rigid relation principle ($RR$) in the Kunen inconsistency
Question 2 is an immediate consequence of the theorems in my paper with Justin.
J. D. Hamkins and J. Palumbo, The rigid relation principle, a new weak choice principle, Mathematical Logic Quarterly …
- 236k
3
votes
Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory?
Here is a partial answer in the case of complete Boolean algebras, which I claim do all arise as Lindenbaum algebras. Let $\mathbb{B}$ be a complete Boolean algebra, and suppose that $M$ is any $\math …
- 236k
15
votes
Accepted
On surjections, idempotence and axiom of choice
The answer is no.
First, I argue that it is consistent with ZF that a Dedekind
finite set $A$ can map onto $A\times 2$, and much more.
To see this, begin with any infinite Dedekind finite set $B\su …
- 236k