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 4213

forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.

27 votes
Accepted

Cardinality of the permutations of an infinite set

$k^k$. Easy that it's an upper bound. For lower bound split $X$ into two equinumerous subsets; there are $\ge k^k$ permutations swapping the two subsets.
Robin Chapman's user avatar
9 votes

Axiom of Infinity needed in Cantor-Bernstein?

One of the standard proofs avoids the Axiom of Infinity. It's based on the Tarski fixed-point theorem, see for instance www.cs.ucla.edu/~palsberg/course/cs232/papers/bernstein.pdf . But it does use th …
Robin Chapman's user avatar
5 votes
Accepted

Equality of two circular sets

In ZF minus the axiom of foundation there is no way of proving that all "circular" sets are equal. You could take a model of set theory allowing ur-elements and replace some or all of these by circula …
Robin Chapman's user avatar