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 questions only
not deleted
user 17064
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.
10
votes
3
answers
574
views
Terminology for a set that does not surject onto $\omega$ (in ZF)
Short question: Is there a standard term for a set $F$ such that there does not exist a surjection $F \twoheadrightarrow \omega$ (in the context of ZF)?
More detailed version: Consider the following f …
- 32.5k
12
votes
2
answers
424
views
Trading Choice for Comprehension (or Replacement)
This question is basically a request for clarification about a remark made by Sam Sanders in a comment to another question: IIUC what he's saying, there are statements that can be proved either with a …
- 32.5k
25
votes
1
answer
2k
views
In what ways is ZF (without Choice) "somewhat constructive"
Let me summarize what I think I understand about constructivism:
"Constructive mathematics" is generally understood to mean a variety of theories formulated in intuitionist logic (i.e., not assuming …
- 32.5k
5
votes
0
answers
290
views
Reference for countable and uncountable algebraic closures of $\mathbb{Q}$ in ZF
The following facts seem to be part of the folklore (where $\mathsf{ZF}$ means Zermelo-Fraenkel set theory with no axiom of choice):
it is consistent with $\mathsf{ZF}$ that there exists an uncountab …
- 32.5k
20
votes
1
answer
1k
views
Axiom of Choice versus V=L in opposition to large cardinals
Consider the following two observations:
The axiom $V=L$ is incompatible with large cardinal axioms that are somehow "too large", like measurable cardinals.
The axiom of Choice is incompatible with …
- 32.5k