# A seemingly wrong definition of convergence of spectral sequences in Bott & Tu?

After introducing exact couples, Bott & Tu defines spectral sequences as follows:

A sequence of differential groups $\{E_r,d_r\}$ in which each $E_r$ is the homology of its predecessor $E_{r-1}$ is called a spectral sequence. If $E_r$ eventually becomes stationary, we denote the stationary value by $E_\infty$, and if $E_\infty$ is equal to the associated graded group of some filtered group $H$, then we say that the spectral sequence converges to $H$.

It seems to me that the concept of convergence above isn't compatible with the proof of the following theorem

Theorem 14.6. Let $K=\oplus_{n\in\mathbb Z}K^n$ be a graded filtered complex with filtration $\{K_p\}$ and let $H_D^*(K)$ be the cohomology of $K$ with filtration given by $H(K)=F_0\supseteq F_1\supseteq F_2\supseteq\dotsb$, where $K\supseteq K_p$ induces a map from $H(K_p)$ to $H(K)$, whose image is denoted as $F_p$. Suppose for each dimension $n$ the filtration $\{K_p^n\}$ has finite length. Then the short exact sequence $0\to\oplus K_{p+1}\to\oplus K_p\to\oplus(K_p/K_{p+1})\to0$ induces a spectral sequence which converges to $H_D^*(K)$.

In the sequel, the proof checks that the spectral sequence converges for every dimension, but without uniformity. It seems that the concept of convergence should be modified. I don't know what kind of modification is okay for the rest of book? I don't want a definition with full generality.

Any help? Thanks!

You're right. One does not want that $E_r$ as a whole becomes stationary but usually $E_r$ comes with a grading $E_r = \bigoplus_{p,q\in \mathbb Z} E^{pq}_r$ and then you want that $E^{pq}_r$ becomes stationary. $E_\infty$ is then the graded abelian group $\bigoplus_{p,q\in \mathbb Z} E^{pq}_\infty$. If this graded abelian group is the same as the associated graded group of a filtered group $H$, we say that $E_r$ converges to $H$.