 Classical Roth bound for three-term progressions

Write $r_3(N)$ for the maximum cardinality of a subset of $[N]$ containing no nonconstant three-term arithmetic progression. There is an absolute constant $C$ such that every three-term-progression-free subset $A\subseteq [N]$ satisfies $|A|\le CN/\log \log N$ for $N\ge 3$. A quantitative Roth density-increment step gives, for subset density $\delta$ and $N\ge 4\cdot 10^6{\delta }^{-4}$, a arithmetic progression of length at least ${\delta }^2\sqrt{N}/2000$ and increased subset density at least $\delta +{\delta }^2/64$. The increment decreases reciprocal subset density by at least $1/65$, so fewer than $\lceil 65/\delta \rceil$ iterations are possible. If $N_j$ is the successive length, then $\log N_{j+1}\ge \frac{1}{2}\log N_j-\log (2000/{\delta }^2)$, using the initial subset density to bound all later ones. Hence $\log N_j\ge 2^{-j}\log N-2\log (2000/{\delta }^2)$. If $\delta \log \log N$ exceeds a sufficiently large absolute constant, all these intervals remain above the density increment threshold, a contradiction. This proves the displayed bound. The result is a classical quantitative version of Roth theorem on three-term arithmetic progressions, without asserting an optimal bound.