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 is called of "infinite genus" if there is no bounded subset of such that is of genus zero.

"Bounded subset" means that the closure of is compact in .

A "subsurface" is defined to be a closed region in whose boundary in 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:

where 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 is the normal genus) and the number of boundary curves. It follows that if and are compact bordered surfaces with are joined along common boundary curves then: and

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