Skip to main content
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
Search options not deleted user 7206

This tag is for questions about proving that some statement is independent from a theory, meaning it is neither provable nor refutable from that theory. Common examples are the continuum hypothesis from the axioms of ZFC, and the axiom of choice from the axioms of ZF.

14 votes
Accepted

Relationship between AC, WO, and Zorn's lemma in ZF-Powerset

This is a classic theorem of Zarach, that it is consistent that ${\sf ZF}^-$ holds with the Axiom of Choice, but not every set can be well-ordered. Zarach, Andrzej, Unions of ${\sf ZF}^-$models wh …
Asaf Karagila's user avatar
  • 39.8k
8 votes
Accepted

Relationship between fragments of the axiom of choice and the dependent choice principles

The idea is to mimic the permutation models as given in Jech. One can then ask, "Well, in Jech he chooses some set of objects in the full universe, and shows it has a support. But in forcing we don't …
Asaf Karagila's user avatar
  • 39.8k
5 votes
Accepted

Implications of the existence of a pair of surjective functions, without Axiom of Choice

No, and here is a counterexample. Suppose that $|\Bbb R|<|[\Bbb R]^\omega|$, that is, there are more countable subsets of reals than reals. This is indeed possible, e.g. if all sets of Lebesgue measu …
Asaf Karagila's user avatar
  • 39.8k
5 votes
Accepted

Dedekind-"finiteness" for arbitrary limit cardinals

Start with your favourite model of $\sf ZFC$, your favourite regular cardinal $\mu$, and your favourite limit cardinal $\lambda>2^\mu$. Now consider the ${<}\mu$-support product $\prod_{\alpha<\lambda …
Asaf Karagila's user avatar
  • 39.8k
4 votes

Independence of the countable axiom of choice

First of all, the easy answer. We can prove that the axiom of choice implies the axiom of countable choice, quite easily. So by showing that the axiom of choice is consistent with the axioms of $\sf …
Asaf Karagila's user avatar
  • 39.8k