Consider the following properties for a subset $A$ of $\mathbb{N}$:

(1) $A$ is large: $\sum_{n \in A}$$ 1\over n$$=\infty,$

(2) $A^\infty=\limsup \frac{|A \cap \{ 1, \dots, n\}|}{n} >0$,

(3) $A_\infty=\liminf \frac{|A \cap \{ 1, \dots, n\}|}{n} >0.$

By a conjecture of Erdős-Turán, if $A$ is large, then it contains arithmetic progressions of any given (finite) length.

By a result of Szemerédi, if $A^\infty >0,$ then $A$ contains infinite arithmetic progressions of length $k$ for all positive integers $k$.

Let's consider these properties for generic subsets of $\omega$ added by forcing (I assume they do not contain $0$) and let me say a few examples:

(a) If $r$ is Cohen, then $r$ is large, $r^\infty=1, r_\infty=0$ and for any $K,L$, we can find $M$ such that $M, M+L, M+2L, \dots M+KL$ are in $r$.

(b) If $r$ is Random, then $r$ is large, and $r^\infty=r_\infty=$$1\over 2.$ So by Szemerédi's result, it contains arbitrary large arithmetic progressions.

Question 1. Is there a direct proof, without using of Szemerédi's result, that $r$ contains arbitrary large arithmetic progressions?

(c) If $r$ is Mathias, related to an ultrafilter $U$, then the properties of $r$ depends on $U$.

It is possible to say the same results for some other generic reals, but as there are many generic reals that I do not know, I would like to ask a more general question:

Question 2. Suppose that $r$ is a generic real added by forcing (we assume it does not contain $0$). Disuss if $r$ is large, and what are $r^\infty$ and $r_\infty$. Also say if $r$ contains arbitrary large arithmetic progressions or not (I would rather direct proofs instead of referring to some known results (say for example, as $r^\infty>0,$ it contains arbitrary large arithmetic progressions)).

  • $\begingroup$ Is there a reason you phrase the Erdos-Turan and Szemerédi conclusions differently? Aren't they exactly the same? (A sequence that contains APs of arbitrary length must contain infinitely many APs of any given length, and the converse result is trivial.) $\endgroup$ Feb 18, 2016 at 5:43
  • $\begingroup$ There is no reason, I just stated them based on Wikipedia. $\endgroup$ Feb 18, 2016 at 7:07
  • $\begingroup$ @MohammadGolshani (+1) Very interesting question! As far as I can remember you have asked a few other related questions so far. Maybe it is a good idea to add some links to them as well. $\endgroup$ Feb 19, 2016 at 15:53

1 Answer 1


Nice question! I happened to be thinking about some similar things a few weeks ago. Here is what I found:

Cohen and random reals:

Cohen and random reals have just about any Ramsey-theoretic property you could want. In fact, they will satisfy the conclusion of the central sets theorem, which implies automatically that they will also satisfy the conclusions of van der Waerden's Theorem, Schur's Theorem, and Hindman's Theorem (and more). Let me briefly explain why.

If $r$ is random, then $r$ must be thick, which means that it contains arbitrarily long intervals (already, this is much stronger than the conclusion of van der Waerden's Theorem). This is not hard to prove, and you certainly don't need anything as powerful as Szemeredi's Theorem to do it. Informally: if you flip a coin over and over and over again, you expect that (for any fixed $n$) you will eventually get $n$ heads in a row. Formally: you may check that, for each $n$, the set $$D_n = \{r : r \text{ contains a sequence of }n \text{ consecutive } 1s\}$$ has measure $1$ in $2^\omega$. $D_n$ is also a dense $G_\delta$, which shows that Cohen reals are also thick.

If $r$ is thick, it automatically satisfies the conclusion of van der Waerden's Theorem. It is known that if a set $A \subseteq \mathbb{N}$ is thick, then $\overline{A}$ contains a minimal idempotent in the semigroup $(\beta \mathbb{N},+)$, which is equivalent to $A$ being a central set. If you're not familiar with central sets, then you still might be familiar with the ultrafilter-based proof of Hindman's Theorem, and you'll know that any set in a (not necessarily minimal) idempotent ultrafilter will satisfy the conclusion of Hindman's Theorem (and Schur's Theorem is implied by Hindman's).

Other kinds of reals:

For Mathias reals arising from an ultrafilter $U$, you're right to say that the Ramsey-theoretic properties of the real will depend on $U$. More specifically, it seems that the generic real will satisfy the conclusion of van der Waerden's Theorem if and only if every member of $U$ does. This is the case, for example, when $U$ is a minimal ultrafilter. However, Grigorieff reals (when $U$ is a $P$-point) will not satisfy the conclusion of van der Waerden's Theorem, because it is known that every $P$-point contains reasonably sparse sets not containing arbitrarily long progressions.

It seems impossible to say anything positive about Sacks reals. Given a set $A \subseteq \mathbb{N}$, it's not hard to find a perfect tree $T$ that forces your Sacks real to be contained in $A$. Thus a Sack's real can be contained in any very sparse set you like, and it will not have nice Ramsey-theoretic properties, it will not be large, and it will have $A^\infty$ and $A_\infty$ both zero. It is also not too hard to find a perfect tree $T$ that forces your Sacks real to be thick. So Sacks reals can also have very nice Ramsey-theoretic properties -- it just depends on the particular Sacks real you are looking at. A similar analysis applies to Prikry-Silver reals.

  • $\begingroup$ Thank you for your nice answer. I knew that for most tree like forcing notions, the largeness depends on the condition. $\endgroup$ Feb 27, 2016 at 8:29

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