# compact subsurfaces of bordered surfaces of infinite genus

• Feb 17th 2013, 01:17 AM
FrankD
compact subsurfaces of bordered surfaces of infinite genus
Hello,

I am a new user in this helpful forum and I start by posting a question that has been troubling me for quite some time now:

I want to find a proof for the following proposition:
"A bordered surface of infinite genus contains compact subsurfaces of arbitrarily large genus"

A few explanations to the definitions used:
A bordered surface $S$ is called of "infinite genus" if there is no bounded subset $A$ of $S$ such that $S-A$ is of genus zero.
"Bounded subset" means that the closure of $A$ is compact in $S$.
A "subsurface" is defined to be a closed region in $S$ whose boundary in $S$ consists of a finite number of nonintersecting simple closed curves

Now to the proof:
It seems to be trivial from a geometrical point of view, but writing it down is really difficult for me.
My idea was to use the reslationship between the genus and the Euler characteristic of a compact surface:
$g=1-1/2*(\chi + q)$ where $g$ denotes the reduced genus (we need this definition to cover orientable and non-orientable surfaces. For the orientable case $g$ is the normal genus and in the nonorientable case $2g$ is the normal genus) and $q$ the number of boundary curves. It follows that if $A$ and $A'$ are compact bordered surfaces with are joined along $r$ common boundary curves then: $\chi(A \cup A')=\chi(A)+\chi(A')$ and $g(A \cup A')=g(A)+g(A')+(r-1)$
I think that this last formula will help with the proof, but I can't really get it to work. Does anyone have an idea?

I greatly appreciate your help and am looking forward to great discussions on this board.
Frank