Consider the following statement due to F.Behrend (1946)


**Theorem:** Let $N$ be a large integer. Then there exists a subset $A\subset [1,N]$ with $\frac{|A|}{N}\geq \exp(-4\sqrt{\ln N})$ which does not contain any arithmetic progression of length three.


By $[1,N]$ I mean the set of integers from this interval.


Informally this theorem states that we can construct a *large set* of integers which lacks three-term arithmetic progression.


Let me ask the following question: Suppose we have that $\frac{|A|}{N}\geq \exp(-4\sqrt{\ln N})$ then why we say that the set $A$ is *large*? On one hand, $\exp(-4\sqrt{\ln N})\to 0$ as $N\to +\infty$. On the other hand we call it *large*?


I would be very thankful for explanation since I am novice in this field of math.



Behrend's Construction
The term "large" is subjective, of course, but arguably a subset of $\{1,2,\dots,N\}$ of size at least $N^{1-o(1)}$ can be regarded as "large". The set $A$ in the theorem is of that kind, since $\sqrt{\ln N}=o(\ln N)$. See also [small-o notation](https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation).


