Roth's Theorem Finitary and Infinitary forms are equal

Question

Theorem1 Let A be a subset of positive integers with positive upper density then A contains a 3 term arithmetic progression.

Theorem2 For any $\delta>0$ there exists $N_{0}$ such that for every $N\geq N_{0}$ and every $A\subseteq\{1,2,...,N\}$ with $|A|\geq \delta N$, A contains a 3 term arithmetic progression.

My aim is to show that these theorems are equal but I think my proof has a gap to showing theorem1 implies theorem2.

Idea:

Suppose theorem1 holds and assume that theorem2 does not hold. i.e. there exists a $\delta>0$ and for all $N$, there exists a subset $A_{N}$ of $\{1,2,...,N\}$ with $|A_{N}|\geq \delta N$ but $A_{N}$'s does not contain a 3 term arithmetic progression. Let $f_N$ be a characteristic function of $A_{N}$. Each characteristic function is an element of $\{0,1\}^\mathbb{N}$ and we know that $\{0,1\}^\mathbb{N}$ is compact w.r.t. product topology on $\{0,1\}^\mathbb{N}$. Then there exists a subsequence $f_{N_{k}}$ of $f_{N}$ which is convergent. Let $f_{N_{k}}$ converges to f where f is the characteristic function of a set $A\subseteq \{1,2,...,N\}$ and also A does not contain a 3 term arithmetic progression.

Then as $f_{N_{k}}\rightarrow f$ implies $|A_{N_{k}}\cap\{1,2,...,N\}|\rightarrow|A\cap\{1,2,...,N\}|$.

This gives us,

\begin{equation} $ \limsup_{N->\infty}\frac{|A\cap\{1,2,...,N\}|}{N}=\limsup_{N->\infty}\frac{|A_{N_{k}}\cap\{1,2,...,N\}|}{N} > \delta $ \end{equation} which contradicts to theorem1.

Question: Is my proof wrong? If it is wrong, where is the gap, can you explained to me?

Thanks for any advice.

Answer 1

You are on the right track with your proof, but here is an issue: you write

$$"\limsup_{N \to \infty} \frac{|A \cap \{1,2,...,N\}|}{N} = \limsup_{N \to \infty} \frac{|A_{N_{k}} \cap \{1,2,...,N\}|}{N}."$$

However, since $A_{N_k} \subseteq \{1,2,...,N_k\}$, the righthand side of this equality is 0 (only $N$, not $N_k$, is going to infinity in this expression). Thus

$$\limsup_{N \to \infty} \frac{|A_{N_{k}} \cap \{1,2,...,N\}|}{N} \not> \delta.$$

Using convergence in $\{0,1\}^{\mathbb{N}}$ to find the set $A \subseteq \mathbb{N}$ was a good idea, but you'll have to finish the proof differently.