Does a finite suspension spectrum make a space finite? Suppose that $X$ is a space whose suspension spectrum $\Sigma_+^\infty(X)$ is dualizable in the stable homotopy category.  I believe this is equivalent to saying that $\Sigma_+^\infty(X)$ is (weakly) homotopy equivalent to a finite cell spectrum.  What does this imply about $X$?  In particular, does it imply that $X$ is weakly equivalent to a finite cell complex, as a space?
 A: Take an acyclic group, i.e. a group $G$ with trivial homology. Choose $G$ which is not finitely presented, so that its classifying space $X=BG=K(G,1)$ cannot be finite. Since $\Sigma X\simeq \star$ is contractible, $\Sigma^{\infty}_{+}(X)=S$ is the sphere spectrum (dualizable).
A: No. In the stable homotopy category a retract of a finite cell spectrum is again a finite cell spectrum, but in the weak homotopy category of spaces a retract of a finite cell complex is not necessarily a finite cell complex; there is an obstruction in the kernel of $K_0\mathbb Z[\pi_1(X)]\to K_0\mathbb Z$.
EDIT: 
For a simply connected space, finite generation of the direct sum of its integral homology groups implies that it is equivalent to a finite complex. Thus a connected space must become finite after one suspension if its suspension spectrum is finite. The same then follows without assuming connected.
Therefore finiteness of $\Sigma^\infty X$ is equivalent to finiteness of $\Sigma X$, and (as shown by Fernando's answer) this is strictly weaker than finiteness of $X$.
