6
$\begingroup$

In contrast to classic results for arithmetic progressions of arbitrary length in one set at least of any finite partition of $\mathbb N$, it is easy to construct a partition in two sets of integers $A$ and $B$ such that neither $A$ nor $B$ contains an infinite arithmetic sequence. Actually (with choice, of course), one has a much stronger result : if $(A_i)_{i\in I}$ is a family of subsets of $E$ with $\#A_i=\#I=\#E$ (infinite), there exists a partition of $E$, $E=F\cup G$, such that for all $i\in I$, $A_i$ is not included in $F$ nor in $G$ (the not very hard proof using a diagonal argument must be well-known). Is this kind of anti-Ramsey result part of some more general theory?

$\endgroup$

1 Answer 1

5
$\begingroup$

According to P. Erdõs, A. Hajnal: On a property of families of sets, Acta Math. Acad. Sci. Hungar. 12 (1961), 87--123 (see page 90) the stronger result you formulated is a theorem of Bernstein from F. Bernstein, "Zur Theorie der trigonometrischen Reihen", Leipz. Bet. (Berichte über die Verhandlungen der Königl. Sächsischen Gesellschaft der Wissenschaften zu Leipzig. Math.phys. Klasse) 60 (1908), 325--338" MR, eudml. If you are looking for further results, "property B" is the keyword you need.

I presented a minicourse on some related properties of families of sets on the 7th Young Set Theory Workshop in 2014, so my slides may contain some results you are interested in.

$\endgroup$
3
  • 2
    $\begingroup$ I conjecture that the part in parentheses (beginning "Beriehte") should be "Berichte über die Varhandlungen der Königl. Sächsischen Gesellschaft der Wissenschaften zu Leipzig. Math.phys. Klasse". $\endgroup$ Nov 1, 2015 at 17:43
  • $\begingroup$ Thank you, Andreas. The copy-paste from a pdf file changed some letter. $\endgroup$ Nov 1, 2015 at 18:52
  • $\begingroup$ I have made the change suggested by Andreas. $\endgroup$
    – LSpice
    Nov 1, 2015 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.