Stage bounds imply completeness and finite multiplicative convergence

Use the supplied realization under [A1]. Regard $X$ as a scheme over the prime field of its standing field, so the stage-bound supplier in [F1] applies to this actual restricted tower. Choose its integer $B=B(X)\ge 0$. Its contract covers every qcqs scheme of finite valuative dimension, including the empty scheme. Use that bound uniformly: for any fixed integer $n$, set $U_n=\max (1,B+n+1)$ and $L_n=0$. If $r\ge U_n$ then $n<r-B$, so ${\pi }_n{F}^r=0$. If $r\le L_n$, the map to $K$ is an equivalence by [F1]. These are precisely the high/low hypotheses of [F2], including actual vanishing of stage groups rather than only their images. The resulting additive convergence and displayed associated-graded identification are therefore those of [F2].

Completeness also follows from the stage bound. The sequential limit of spectra is the fiber of $1-\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}:{\prod }_{r\ge 0}{F}^r\to {\prod }_{r\ge 0}{F}^r$. For any fixed degree $n$, the groups ${\pi }_n{F}^r$ and ${\pi }_{n+1}{F}^r$ are zero for sufficiently large $r$ by step 1.1. On their finite-support products $1-\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}$ is an isomorphism, with inverse the finite geometric sum of the nilpotent shift. The fiber long exact sequence thus gives ${\pi }_n{\lim }_r{F}^r=0$ for every $n$, hence the limit spectrum is zero. Restricting to stages $r\ge 0$ is cofinal for this tower limit.

The image filtration on each $K_n$ is full for $r\le 0$ and zero for $r\ge U_n$, so it is finite. The product identification in [F2] makes the convergence in step 1.1 multiplicative with exactly this finite filtered abutment. This proves all claims.