Skip to main content

New answers tagged

2 votes

Is there an uncountable extension of the Ramsey set $[\omega]^2$?

If $N$ is an infinite proper subset of $\omega$ then the family $\mathcal A=[\omega]^2\cup\{A\subseteq\omega:A\not\subseteq N\}$ is a Ramsey family and $|\mathcal A|=2^{\aleph_0}$. On the other hand, ...
bof's user avatar
  • 13.4k
4 votes
Accepted

Is there an uncountable extension of the Ramsey set $[\omega]^2$?

Yes. Just take $\mathcal{A}$ to be $[\omega]^2$ together with the powerset of the even integers.
Tony Huynh's user avatar
  • 32.1k
9 votes
Accepted

Maximal Ramsey families

No. Consider the family $\mathcal{R}$ of all sets of size 1. It is certainly a Ramsey family. Consider a Ramsey extension $\mathcal{R}_1\supset \mathcal{R}$. Take a map $f$ which maps all sets from $\...
Fedor Petrov's user avatar
4 votes
Accepted

Non-Ramsey function $f:[\omega]^{<\omega}\to\{0,1\}$

For $P \subset \omega$ finite, let $f(P) = \#P \bmod{2}$, i.e., $f(P) = 0$ if $P$ has even cardinal, and $1$ otherwise.
Najib Idrissi's user avatar

Top 50 recent answers are included