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 set-theory
Search options not deleted
user 517040
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.
11
votes
Extracting a common convergent indexing from an uncountable family of sequences
Using notation by the answer by Hamkins, I'll prove that $\mathbb c_{\mathbb R}\ge\min\{\mathfrak s,\mathfrak b\}$.
Thanks to Theorem 8.11 of Halbeisen's book Combinatorial Set Theory, we know that $\ …
- 300
9
votes
Extracting a common convergent indexing from an uncountable family of sequences
We prove that $\mathbb c=\mathbb c_{\mathbb R}$: clearly $\mathbb c\le\mathbb c_{\mathbb R}$ holds, hence it is enough to prove $\mathbb c\ge\mathbb c_{\mathbb R}$. This means: for all reflexive separ …
- 300