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 |
Combinatorial properties of infinite sets. This is a corner-point of set theory and combinatorics.
2
votes
Accepted
"Arithmetically diverse" infinite binary string
Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$:
$$0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$
Observe that all but finitely many bloc …
5
votes
Accepted
Partitioning an infinite cardinal $\kappa$ into pairwise neighboring subsets
Yes. List the pairs $(\alpha,\beta)$ with $\alpha<\beta<\kappa$ as $(\alpha_\lambda,\beta_\lambda), \lambda<\kappa$.
Then construct the sets $B_\alpha\in\mathcal B, \alpha<\kappa$ as follows:
At stage …
2
votes
Accepted
Non-summable subsets of $[\omega]^{<\omega}$
Theorem: There is no such $E$.
Claim: for each $a\in E$ there exists $b\in E$ with $|b\setminus a|\ge 2$.
Proof of Claim: Let $a$ be a counterexample. Then all $b\ne a$ contain exactly one element eac …
9
votes
Accepted
Choice sets from above and below
Let $\cal S=\{\{1,2\},\{2,3\},\{3,1\}\}$.
Then $\cal S$ has no choice set, whatsoever.
So there is no asymmetry -- not every shy set is contained in a choice set, and not every gregarious set contai …
1
vote
Choice sets in "thick" sets of sets
Let $\kappa$ be any infinite cardinal.
Let $\cal S=\{\kappa\setminus\{x\}:0<x<\kappa\}$ and $C=\{0\}$.
Then all the given conditions are satisfied.