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.
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.