TITLE: Do compact complex manifolds fall into countably many families?
QUESTION [39 upvotes]: Do there exist countably many proper holomorphic submersions of complex manifolds $\mathcal{X}_n \to B_n$ such that every compact complex manifold appears as a fiber in at least one of the families?  
(Conventions: Assume that each $B_n$ is Hausdorff, connected, and paracompact; these conditions imply that $B_n$ is also second countable.  For the question at hand, one could equivalently require each $B_n$ to be an open subset of $\mathbf{C}^m$ for some $m$ depending on $n$.)

REPLY [3 votes]: This is an attempt to express a variant of Jason's answer
in slightly more down-to-earth terms, so that maybe less has to be referred to his experts.  (This answer is too long for a comment, so I made it a community wiki.)
It seems to me that there is really only one difficult point in the argument beyond the existence of the Kuranishi deformation space, namely the "Theorem" below.
Step 1: Nash's article "Real algebraic manifolds" in Annals of Math. (1952) 
proved that every connected compact $C^\infty$ real manifold 
is diffeomorphic to a component of a real algebraic variety.
There are only countably many families of real algebraic varieties, 
and Ehresmann's fibration theorem (1950) shows that
the diffeomorphism type is locally constant within each family.
Thus there are only countably many diffeomorphism types.
Therefore we now restrict attention to one.
Step 2: Let $M$ be a compact $C^\infty$ real manifold,
let $T$ be its tangent bundle,
and let $E$ be the bundle with fibers $E_x := \operatorname{End} T_x$.
Fix $r$ not too small, 
and let $C^r(M,E)$ be $\{\text{$C^r$ functions $M \to E$}\}$
with the $C^r$ compact-open topology (Hirsch, Differential topology, 2.1),
which is second countable.
The space $\operatorname{Comp}$ 
of complex structures (i.e., integrable almost complex structures) on $M$
is a subset of $C^r(M,E)$; give it the subspace topology,
so it too is second countable.
Explicitly, complex structures $I$ and $I'$ are close
if uniformly on $M$ they are within $\epsilon$ (with respect to some metric)
and the same holds for their derivatives up to order $r$.
Step 3: On the other hand, 
let $\mathcal{M}$ be the set of isomorphism classes of 
compact complex manifolds whose underlying real manifold 
is diffeomorphic to $M$.
Kuranishi proved that each compact complex manifold $X$
admits a versal deformation over 
a pointed connected complex analytic space $(B,0)$.
Each such versal deformation gives rise to a subset of $\mathcal{M}$, 
namely the set of isomorphism classes of the fibers;
define a topology on $\mathcal{M}$ in which these subsets form 
a basis of neighborhoods of the point $[X] \in \mathcal{M}$ determined by $X$.
Step 4: Let $\Phi \colon \operatorname{Comp} \to \mathcal{M}$
be the map sending $(M,I)$ to its isomorphism class as a complex manifold.
Then $\Phi$ is open because for any open subset 
$U \subset \operatorname{Comp}$ and $[X] \in \Phi(U)$,
a choice of $C^\infty$ trivialization of a versal deformation over $B$
defines a continuous lifting $B \to \operatorname{Comp}$ 
of $B \to \mathcal{M}$.
The key point is now the following (which seems to be a consequence of what Jason's experts claimed):
Theorem: $\Phi$ is continuous.
(In other words, given a versal deformation of $X = (M,I)$,
any complex structure $I'$ sufficiently close to $I$ in the $C^r$ sense
defines a complex manifold $(M,I')$ isomorphic to a fiber of the deformation.)
Given this, $\mathcal{M}$ is second countable
because it is the image of a second countable space under an open continuous map.
This means that $\mathcal{M}$ is covered by countably many subsets
each arising from a family over a complex analytic subspace 
in $\mathbf{C}^n$ for some $n$.
Each such base can be stratified into complex manifolds, so we are done.