# Why compact surfaces can be regarded as region without boundary?

I have been reading DoCarmo and feel quite confused by that he mentioned several times that compact surfaces can be regarded as regions without boundary, which is used in the proof of a corollary of Gauss-Bonnet and several other places.

But I can't figure out why this is the case.

Thanks!

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Here are some compact surfaces: the sphere $S^2$, the torus $S^1 \times S^1$, the Klein bottle... For contrast, any open subset of $\mathbb{R}^2$ is not a compact surface, and in fact, no compact subset of $\mathbb{R}^2$ is a surface. –  Zhen Lin Dec 19 '11 at 10:05

I don't have DoCarmo's book on my desk, but in Thorpe's "Elementary topics in differential geometry" it is explained on page 177 what an $n$-surface with boundary is. It is the object $M$ you get when you take the zero-set $S:=f^{-1}(0)$ of a function $f:\ {\mathbb R}^{n+1}\to{\mathbb R}$, cut $S$ along disjoint smooth boundary curves and throw away the unwanted part of $S$ (plus some technical conditions).
It seems that by a "region without boundary" DoCarmo means such an $M$ where no cutting has been done, such that in formulas where there appear integrals along the boundary curves these integrals are trivially zero. A priori this has nothing to do with compactness: The $(x,y)$-plane in $(x,y,z)$-space is such an $M$.