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 6085
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.
3
votes
Accepted
Global or Relativised Dependent Choices
Given any $x$, let $\alpha_x$ be 0 if $\phi(x)$ fails. If it holds, for each $x'$ with $\phi(x')$ and $x'$ of set-theoretic rank less than or equal to $x$, let $\alpha^{x'}$ be the least ordinal $\alp …
- 32.5k
9
votes
Completeness of the club filter without AC
This is not much of an answer, but it may be useful:
a. It is an open problem (asked by Woodin) whether $\mathsf{ZF} + \mathsf{DC}$ suffices to prove that there is a regular cardinal larger than $\o …
- 32.5k
7
votes
Can we have an infinite sequence of decreasing cardinality all terms of which have equal siz...
This is consistent, at least under a rather tame large cardinal assumption. (One can also produce examples by manipulating Dedekind finite sets, but Asaf's answer addresses this. The answer here works …
- 32.5k
11
votes
Accepted
Distinct well-orderings of the same set
Clinton Conley has found a nice argument that solves the problem. I daresay that even in the presence of choice, it is nicer than the standard approach, and avoids having to separate the argument by c …
- 32.5k
12
votes
Are there known examples of sets whose power set is equal in size to power set of larger set...
Let me add some comments to Asaf's answer.
We are looking for situations with $|X|<|Y|$ and $|\mathcal P(X)|=|\mathcal P(Y)|$ (and choice fails).
Asaf rightly identifies that a very natural way of …
- 32.5k
54
votes
Accepted
Are there any non-linear solutions of Cauchy's equation $f(x+y)=f(x)+f(y)$ without assuming ...
It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurab …
- 32.5k
8
votes
Accepted
Comparability implies well-orderability?
It is open whether the continuum hypothesis for an infinite set $E$ implies the well-orderability of $E$. Of course, if $CH(E)$ holds, then the assumption in your (first) statement holds.
($CH(E)$ i …
- 32.5k
25
votes
Accepted
Is the statement that every field has an algebraic closure known to be equivalent to the ult...
Qiaochu, using the link I provided in my answer to this question, you find that this question is still open (or was, as of the mid 2000s, and I haven't heard of any recent results in this direction). …
- 32.5k
6
votes
Accepted
Does $\mathsf{SVC}^\ast$ exist?
For what is worth, I find no hits on Math Reviews for $\mathsf{SVC}^*$ and all hits for $\mathsf{SVC}$ are for the statement that says (in your notation) that there is an $S$ such that $\mathsf{SVC}(S …
- 32.5k
16
votes
Accepted
Hartogs number and the three power sets
Hi Asaf,
I thought about this a while ago. Of course, the question had been asked and solved before. Digging through the FOM archives for Spring 2009, I found (April 28, 2009; I fixed a typo in what …
- 32.5k
24
votes
Does constructing non-measurable sets require the axiom of choice?
As other answers point out, yes, one needs choice. The popular/natural examples of models of ZF+DC where all sets of reals are measurable are models of determinacy, and Solovay's model. They are relat …
- 32.5k
11
votes
Accepted
Does "$X \not\to (\omega)^\omega_2$ for every infinite $X$" imply ${\sf AC}$?
The answer is no, the statement that for every set $X$ we have $$X\not\to(\omega)^\omega_2$$ does not imply the axiom of choice.
This was shown by Kleinberg and Seiferas in 1973, see
MR0340025 (49 #4 …
- 32.5k
22
votes
Accepted
Hahn-Banach without Choice
The ultrafilter theorem is the statement that any filter on a set can be extended to an ultrafilter. It is perhaps more common to see it sated as the (Boolean) Prime ideal theorem: Every Boolean algeb …
- 32.5k
45
votes
Accepted
Does Arzelà-Ascoli require choice?
There is a canonical way of checking the literature for most questions of this kind. Since they come up with some frequency, I think having the reference here may be useful.
First, look at "Conseq …
- 32.5k
15
votes
Accepted
Axiom of choice: ultrafilter vs. Vitali set
Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is that the existence of a nonprincipal ultrafilter d …
- 32.5k