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.