Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with axiom-of-choice
Search options answers only
not deleted
user 8133
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
An equivalent of the axiom of choice?
With such shared/common knowledge problems, even when AC isn't involved, I think a crucial first step is to get away from the "story" version. For example, since this is only interesting once the set …
- 21.1k
1
vote
A question about the Axiom of Choice and straight lines in the Euclidean plane.
(Too long to be a comment, but:) It is impossible to, say, make the union of $C$ either the complement of an open concave region (trivial) or the complement of an open circle (there's only one possibl …
- 21.1k
1
vote
Axiom of Choice and Vitali's theorem
(I believe this material is in Jech's giant set theory book, but I don't have it in front of me right now.)
I am almost entirely certain that the existence of a non-measurable set of reals is not equ …
- 21.1k
6
votes
Accepted
Why doesn't choice imply global choice (in NBG)?
Your first sentence is true (modulo the word "inductive"), but not in the way you mean: $ZFC$ proves the existence of many set well-orderings of each $V_\alpha$. Now, under some further assumption - s …
- 21.1k
1
vote
Accepted
Matching power series to infinity
I believe the answer to your simpler question is "no:" Fix an infinite sequence of sets $A_i$ ($i\in\omega$) for which no choice function exists. Now let $A$ be the free commutative ring generated by …
- 21.1k
2
votes
What axioms are stronger than the Axiom of choice?
The axiom of global choice. Technically this isn't really an axiom: global choice (GC) states that there is a formula $\phi(x, y)$ such that the relation $$ A\le_\phi B:= V\models \phi(A, B)$$ is a we …
- 21.1k
4
votes
Set theory question
I'm assuming you're asking about the theory $ZF+\neg AC+CH$. The answer is yes - but you have to be clear about what you mean by "$CH$."
First, a trivial example: we can start with a model of $ZFC+CH …
- 21.1k
1
vote
Accepted
Is Proper Class Choice equivalent to Global Choice?
Yes, this is still equivalent: given any relation $R$, consider the new relation $$R^{bigrows}=\{\langle x,y\rangle: \exists a,b(y=\langle a,b\rangle\wedge \langle x,a\rangle\in R)\}.$$ Basically, $R^ …
- 21.1k
5
votes
Accepted
The patterns of possibility for nontrivial automorphisms and nontrivial elementary embedding...
I think there's some confusion over automorphisms/embeddings which are "internal" vs. those which are "external." Some comments which I hope clear things up:
No model of ZF has nontrivial definable …
- 21.1k
14
votes
Independence of the countable axiom of choice
I'm going to play fast and loose with details here, but the outline is correct. To answer your new questions: no, there is no short proof. And this only shows one direction of the indpenedence of AC, …
- 21.1k
18
votes
Accepted
Cantor-Bernstein with "weakly injective" functions
No, it is not provable in $\mathsf{ZF}$.
It is consistent with $\mathsf{ZF}$ that there is a sequence of disjoint two-element sets whose union is not countable, i.e. $\vert A_i\vert=2$ but there is no …
- 21.1k
8
votes
Accepted
Amorphous proper classes in MK
Unless I'm missing something, the answer is no: we have a surjection $s$ from a given proper class to the class of ordinals - sending each element to its rank and then "collapsing" appropriately - and …
- 21.1k
4
votes
Relation between AC and the axiom of foundation
The following is an easy recipe for building a model of ZFA+AC with one atom, from a model $M$ of ZFC (it's trivial to modify this construction to include arbitrarily large sets of atoms, and even a p …
- 21.1k
8
votes
Accepted
Notation arb(x)
It depends how the remaining axioms of ZF are altered (or not). For example, do we broaden the scheme of replacement to apply to formulas of set theory which contain the symbol "$arb$"? If so, then as …
- 21.1k
4
votes
Accepted
Does choice always hold in a model of ZF with point-wise parameter-free definable sets?
The following fleshes out the comments above by Asaf and Andreas.
First, note that the idea you outline at the end will not work: it implicitly assumes that the relation "$\varphi$ defines $a$" is de …
- 21.1k