Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying Shelah's proof (actually written by Uri Abraham) that adding one generic real implies the existence of a Suslin tree (available in this link, I think that freely for everyone.)

The notion of forcing is the set finite functions from $\omega$ to $\omega$, with stronger being "extending", then we construct a tree on $\omega_1$ by defining some functions by using the functions, and ensuring that the result gives us a Suslin tree.

At one point, the claim is that if $X$ is an uncountable anti-chain in the tree (in the generic extension, of course) then there exists $p$ such that $p\Vdash X$ is an uncountable anti-chain.

Then, it says, we can find $q$ stronger than $p$ for which $Y=\{\alpha | q\Vdash\alpha\in X\}$ is uncountable.

That last statement is unclear to me. I'm sensing that this is something relatively simple like a pigeonhole argument, but I'm uncertain how to deduce it.

share|cite|improve this question
up vote 6 down vote accepted

Consider a fixed generic extension obtained by a generic where $p$ belongs. We have that $X$ is uncountable. For each $\alpha\in X$ there is $q\in G$ that in $V$ forces $\alpha\in X$. We may assume $q\le p$ by further extending if necessary. This shows that $A_\alpha=\{q\mid q\le p\land q$ forces in $V$ that $\alpha\in X\}$ is nonempty for each $\alpha\in X$. Since Cohen forcing is countable, the same $q$ must be in uncountably many of these $A_\alpha$, and we are done.

There is another way of presenting the argument that avoids talking about forcing extensions: For each $q\le p$ let $A_q$ be the set of $\alpha$ that $q$ forces to be in $X$. If each $A_q$ is countable, then (again, because Cohen forcing is countable) there is an $\alpha$ such that any ordinal forced to be in $X$ by a condition extending $p$ must be strictly below $\alpha$. But then $p$ itself must force $X$ to be contained in $\alpha$, contradicting that p forces $X$ to be uncountable.

share|cite|improve this answer
@Asaf, independently of Shelah's argument, I highly recommend that you also study the very elegant proof of the Suslin tree result by Stevo Todorcevic. – Andrés E. Caicedo Jan 23 '11 at 19:31
Is it the one appearing in Jech? Where the forcing notion is normal trees? I have read it as well, I just need to study this proof as well. Also, I had a hunch that posting this question here will get you to answer faster than catching Uri on the phone at this hour :-) – Asaf Karagila Jan 23 '11 at 19:36
@Asaf: Stevo shows a nice general construction that to each real assigns a tree. If the real is Cohen, the tree is Suslin. The argument is in section 6 of Stevo Todorcevic, "Partitioning pairs of countable ordinals", Acta Math. 159 (1987), no. 3-4, 261-294, and should also be in his book on Coherent sequences. – Andrés E. Caicedo Jan 23 '11 at 19:59
@Asaf: It takes some getting used to. There are some really beautiful ideas in his papers, so the effort pays off. I'm sure there are accounts of the proof in other places. Jean Larson has written some notes on it, for example. If you email her, she may send them to you. – Andrés E. Caicedo Jan 23 '11 at 20:04
I know that Uri Abraham (who happens to be my advisor) has done some work on walks and such, and we did discuss those ideas more then once in a biweekly seminar we have during most of the academic year. I will refer the question to him if I meet him tomorrow. – Asaf Karagila Jan 23 '11 at 20:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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