3D surfaces of infinite genus How might one show that the set of connected 3D surfaces with infinite genus (up to homeomorphism) is countably infinite? 
We could either use proof by contradiction or come up with a way to count them? What would be a relativelt simple way to show this?
Thanks.
 A: $\newcommand{\RR}{\mathbb R}$
$\newcommand{\HH}{\mathbb H}$
$\newcommand{\Ends}{\mathop{\rm Ends}}$
Let $F$ be an infinite genus surface, properly embedded in $\RR^3$.  Then up to homeomorphism of surfaces (not ambient homeomorphisms) there are uncountably many such surfaces.
Proof. Note that $\RR^3$ is homeomorphic to $\HH^3$.  So we may consider surfaces properly embedded in $\HH^3$ where there is more "room".  Let $\HH^2$ be a totally geodesic plane in $\HH^3$.  Let $T = T_3$ be a proper embedding of the regular $3$--valent tree into $\HH^2$.  Take a small neighborhood of $T$ in $\HH^3$ to get a strange embedding of a three-ball $B$ into $\HH^3$. 
Now, at every vertex of $T$ drill out a small tube from $B$ and take the boundary of the resulting infinite genus handlebody to get an infinite genus surface $F = F(T)$, properly embedded in $\HH^3$.  Note that the topological space $\Ends(F)$, the ends of $F$, are an invariant of the homeomorphism type of $F$.  In this example, $\Ends(F)$ is homeomorphic to the Cantor set. 
By taking a subtree $T' \subset T$ we can obtain another surface $F' = F(T')$.  As before, the $\Ends(F')$ is homeomorphic to the Gromov boundary of $T'$.  Finally, since there are uncountably many pairwise non-homeomorphic compact subsets of the Cantor set, we find the required uncountable set of pairwise non-homeomorphic embedded surfaces. QED
This answer was partly prompted by Agol's answer to a previous question.
