Why does Frank Adams demand a finite CW-complex? On page 145 of J.F. Adams' "Stable Homotopy and Generalised Homology", there is a proposition:
Let $E$ be the suspension spectrum of a finite CW-complex $K$, and $F$ and spectrum (of topological spaces).  Then $[E,F]=\mathrm{colim_n}[\Sigma^nK,F_n]$.  The proof is not difficult, and it proceeds as follows:
Given two spectra maps which agree in the colimit, say $f$ and $g$, they agree on some finite level, $[\Sigma^pK,F]$.  Thus, the homotopy at that level can be suspended to create a homotopy of cofinite subspectra, which is exactly what we need to have a homotopy class of maps in $[E,F]$.  However, I cannot see why he requires that $K$ be a finite CW-complex.  It seems that we can find an upper bound of $f$ and $g$ regardless of that constraint, based on the fact that we are using a filtered colimit.  Does this make sense, or is there an obvious reason why one might choose a finite $K$?
Thanks!
 A: You stated that you figured it out, but for the public record let's say why anyway.
A map of spectra $X \to Y$, in Adams' terms, is defined in the "cells now, maps later" formalism.  Meaning, for every individual cell there is some suspension of it where the map is defined, but it may not be defined at first.  Note in particular that for each cell there is some finite stage at which the map is defined.
If $X$ is a suspension spectrum of a finite complex, then there are finitely many cells and so this is the same as, eventually, having a map defined $X_n \to Y_n$.  If $X$ is a suspension spectrum of an infinite complex, you can have a map which is only eventually defined on every cell, and never actually defined on all of $X_n$.
Note that this is necessary in order for $K \mapsto [\Sigma^\infty K, Y]$ to be a cohomology theory satisfying the wedge axiom.
This, by the way, was the subject of an error in an old edition of Hatcher's text (using a more modern viewpoint on spectra than Adams); see here for details.
