Irrational rotation is ergodic but not weakly mixing

By [F3] and [F4], $e^{2\pi i}=(e^{\pi i}{)}^2=1$, and the same holds for every integer multiple of $2\pi i$ by multiplication and inversion. Thus $f(x)=e^{2\pi ix}$ is well defined across the circle cut. By the Cartesian formula [F4], it equals $\cos (2\pi x)+i\sin (2\pi x)$ and is continuous by [F5] (the limit from below at 1 equals its value at 0), has $\mathrm{\mid }f\mathrm{\mid }=1$ by [F4], and belongs to both $L^1(\lambda )$ and $L^2(\lambda )$ because the circle has mass one. The addition law gives $f\circ R_{1\mathrm{/}2}=-f$. By [F6] and [F7], if $m=\int fd\lambda$ then $m=\int f\circ R_{1\mathrm{/}2}d\lambda =-m$, so $m=0$.

For every $n\ge 0$, the addition law and integer-period identity give $f(R_{\alpha }^{n}x)=e^{2\pi in\alpha }f(x)$. Hence $\int (f\circ R_{\alpha }^n)\bar{f}d\lambda =e^{2\pi in\alpha }\int \mathrm{\mid }f{\mathrm{\mid }}^{2}d\lambda =e^{2\pi in\alpha }$. The centered correlation of [F2] is the same because the mean is zero. Its modulus is one by [F4], so for every $N\ge 1$, $N^{-1}{\sum }_{n=0}^{N-1}\mathrm{\mid }{C}_n(f,f)\mathrm{\mid }=1$. It cannot tend to zero; the necessary implication in [F2] disproves weak mixing. Irrationality supplies ergodicity by [F1]. Countable choice is inherited from [F1] and [F6]; only the implication from weak mixing to vanishing absolute $L^2$ correlation averages in [F2] is needed. No Fourier-series completeness or spectral theorem is used.