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 |
8
votes
Minimal subset of axioms for ZFC
As you indicate, Infinity is indispensable, since $V_\omega$ is a model of all the other axioms, and even if the background theory is full ZFC (i.e., if we pretend to live in a universe satisfying ZFC) … Partial conclusion so far: You need infinitely many axioms by Joel's answer, you don't need AC or Regularity, you don't need Separation if your version of Replacement is sufficiently strong, but you do …
20
votes
Accepted
Is it possible to show that an infinite set has a countable (infinite) subset, without using...
Short answer: No.
By countably infinite subset you mean, I guess, that there is a 1-1 map from the natural numbers into the set.
If ZF is consistent, then it is consistent to have an amorphous set …
4
votes
Result that follows from ZFC and not ZF but are strictly weaker than choice
An interesting dividing line, different from the ones you see in the cardinal choice axioms, is connected to the Tychonoff theorem saying that products of compact spaces are compact. …