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 $\displaystyle $S$$ is called of "infinite genus" if there is no bounded subset $\displaystyle $A$$ of $\displaystyle $S$$ such that $\displaystyle $S-A$$ is of genus zero.

"Bounded subset" means that the closure of $\displaystyle $A$$ is compact in $\displaystyle $S$$.

A "subsurface" is defined to be a closed region in $\displaystyle $S$$ whose boundary in $\displaystyle $S$$ 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:

$\displaystyle $g=1-1/2*(\chi + q)$$ where $\displaystyle $g$$ 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 $\displaystyle $2g$$ is the normal genus) and $\displaystyle $q$$ the number of boundary curves. It follows that if $\displaystyle $A$$ and $\displaystyle $A'$$ are compact bordered surfaces with are joined along $\displaystyle $r$$ common boundary curves then: $\displaystyle $\chi(A \cup A')=\chi(A)+\chi(A')$$ and $\displaystyle $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