My question is the following:

I have an $\in$-chain of elementary submodels $\langle M_\xi\rangle_{\xi<\lambda^+}$ of $H_\theta$ for $\theta$ sufficiently large. I know that for any $M\prec H_\theta$, and for every $A\in M$ such that $A$ is countable in $M$, then $A\subseteq M.$ However, it would be great for my work if for every $A\in [\lambda^+]^\omega$, there is $\xi\in\lambda^+$ such that $A\in M_\xi.$ Do you know if it is true? and some easy arguments why yes, why not or why it depends?

Any good reference like "Everything you wanted to know about elementary submodels for set theory and you never dared to ask" would be also really appreciated.

  • $\begingroup$ This might be of use: E. C. Milner: The use of elementary substructures in combinatorics, Trends in discrete mathematics. Discrete Math., 136(1994), 243–252. $\endgroup$ Aug 1 '11 at 6:10
  • $\begingroup$ Thanks a lot, this is very close to what I was looking for! $\endgroup$ Aug 3 '11 at 20:26

Well, it could happen that the size of $\bigcup_{\xi\lneq\lambda^+}M_\xi$ is actually strictly below $(\lambda^+)^{\aleph_0}$, and in this case you cannot have what you want. This happens for example if the continuum hypothesis fails, $\lambda=\aleph_0$, and all the $M_\xi$ are countable, a not so uncommon situation.

On the other hand, if you allow the $M_{\xi}$ to be sufficiently large, i.e., if you allow $\bigcup_{\xi\lneq\lambda^+}M_\xi$ to be of size at least $(\lambda^+)^{\aleph_0}$, then you can at least arrange sequences $(M_\xi)_{\xi\lneq\lambda^+}$ satisfying what you want.

I once wrote something with the title "Applications of elementary submodels in general topology" which you can find at http://www.hausdorff-center.uni-bonn.de/people/geschke/publications and there is Alan Dow's [An introduction to applications of elementary submodels in topology, Topology Proceedings 13 (1988), pp. 17-72]. Also, the second volume of Discovering Modern Set Theory by Just und Weese (AMS Graduate Studies in Mathematics 18) has something about elementary submodels in set theory.

  • $\begingroup$ Thank you very much for your answer! It is very helpful. Actually, in your paper, I found exactly the hypothesis that I needed! Now I can continue after solving this sticky point. $\endgroup$ Aug 3 '11 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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