The special case $k=3$: Roth's theorem

Step 1 (Fourier count of progressions). Suppose $A \subseteq \{1,\dots,N\}$ has density $\delta$ and no nontrivial 3-term progression. Writing $\hat{1_A}(\theta) = \sum_{n \in A} e(\theta n)$ for the Fourier transform of the indicator of $A$, the number of triples $(x, x+r, x+2r) \in A^3$ can be written as an integral of $\hat{1_A}(\theta)^2 \, \overline{\hat{1_A}(2\theta)}$ over $\theta$.

Step 2 (the pseudorandom case). If every nonzero Fourier coefficient of $1_A$ were small compared to $\delta^2$, the integral would be dominated by the $\theta = 0$ term alone, which already forces roughly $\delta^3 N^2$ triples — far more than the trivial ones — contradicting the assumption that $A$ has none.

Step 3 (density increment). So some nonzero coefficient must be large; this means $1_A$ correlates with the linear phase $e(\theta n)$, i.e. $A$ is noticeably biased on some arithmetic progression (or Bohr set). Restricting to that sub-progression produces a shorter interval on which the density of $A$ has increased by a fixed multiplicative factor.

Step 4 (iteration and conclusion). Density cannot increase past 1, so after finitely many rounds of this density-increment step the process must terminate — meaning the pseudorandom case of Step 2 is eventually forced, which produces the missing 3-term progression and contradicts the no-progression assumption. Roth's original bookkeeping gives $N(3,\delta) \lesssim \exp(\exp(1/\delta))$-type thresholds; this was improved over decades, and Kelley and Meka's 2023 argument pushes the bound down to nearly Behrend's classical construction, $N(3,\delta) \lesssim \exp\!\big(-c(\log N)^{1/12}\big) N$-type density, essentially closing the gap for this case.