Adding sets not containing arithmetic progressions of length three by forcing

Question

Consider the following forcing notion: conditions in $\mathbb{P}$ are pairs $(s, N),$ where:

1) $s\in 2^{<\omega}$,

2) $N\in \mathbb{N}$,

3) (by identifying $s$ with a subset of $lh(s)$) $s$ contains no arithmetic progressions of length $3$, and $\Sigma_{n\in s}1/n \geq N$.

The ordering is defined in the natural way: $(t, M)\leq (s,N)$ iff $t$ extends $s$ and $M\geq N.$

Now let $G$ be $\mathbb{P}$-generic over $V$, and let $R=\bigcup\{s: \exists N, (s,N)\in G \}.$ We can imagine $R$ as a subset of $\mathbb{N}$. The following is clear:

Claim 1. $R$ contains no arithmetic progressions of length three.

Given any finite $M$, consider the set $D_M=\{(s, N)\in \mathbb{P}: N\geq M \}.$ Then

Claim 2. If for each $N$, the set $D_N$ is dense in $\mathbb{P},$ then $R$ is a large set of natural numbers, i.e., $\Sigma_{n\in R}1/n=\infty$ (and hence $R$ witnesses a counterexample to the famous Erdos-Turan conjecture).

Question. Suppose $(s, N)\in \mathbb{P}.$ is there $t\in 2^{<\omega}$ extending $s$ such that $(t, N+1)\in \mathbb{P}?$

Of course a positive answer to the above question implies that all $D_M$'s are dense.

Remark. If such an $R$ exists in an extension, then it already exists in the ground model.

Answer 1

A "yes" answer to your question is equivalent to the statement "there exists a large set of natural numbers that admits no arithmetic progression of length three." I'm submitting the proof of this equivalence as an answer since I don't expect to see an actual answer unless it shows up in Annals too :)

So, to the proof. You've already noted the forward direction.

For the backward direction, suppose $A$ is a large set of natural numbers with no arithmetic progression of length three, $k\in\mathbb{N}$, and $s\subseteq k$ has no arithmetic progressions of length three. We'll show that there's a large set $B$ such that $B\cap k = s$, and $B$ has no arithmetic progressions of length three. This is clearly sufficient.

The "naiive" choice for $B$ is the set $s\cup (A\setminus k)$, which is large and has no length-3 AP's which are entirely below $n$ or entirely above $k$; but of course there may be an AP of length 3 which crosses $k$. There are only finitely-many APs of length 3 with two points in $s$, so we may remove the corresponding points from $A$ (if they exist) and still have a large set. So we'll assume that we've already done this, i.e. $A\cap k = \emptyset$, and there are no APs of length 3 with two points in $s$.

If the AP has two points in $A$, then it's more complicated. Suppose $i\in s$. Let $B_i$ be the set you get from $A$ by removing the possible third points, i.e. $$ B_i = A\setminus\{2n - i \;|\; n\in A\}$$ We'll show that $B_i$ is still large. Let $N$ be a large natural number. Note that $$ \sum_{n\in B_i\cap N} \frac{1}{n} \ge \sum_{n\in A\cap N} \frac{1}{n} - \frac{1}{2n-i}$$ If $n$ is large enough, say $n > 3i$, then $\frac{1}{2n - i} < \frac{2}{3n}$, so if $A\cap 3k = \emptyset$ then the above sum is at least $$ \sum_{n\in A\cap N} \frac{1}{3n} = \frac{1}{3} \sum_{n\in A\cap N} \frac{1}{n} $$ Hence, as the sums on the right go to $\infty$ as $N\to\infty$, it follows that $B_i$ is large.

Applying this process multiple times, once for each member of $s$, we eventually get a large set $B$ such that $s\cup B$ has no AP's of length 3.