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I have a conceptual question: I know that the word "closed", depending on the context, can mean either the oposite of open or compact without boundary, but could the word "compact" mean compact without boudary?

This question is because I saw in a paper that $\int_M \Delta f dV=0$ where $f\in C^\infty(M)$ and $M$ is a compact minimal hypersurfaces of the Euclidean sphere.

No orientability assumption is assumed as well as any hypothesis about the boundary.

One more question: Does minimality imply orientability?

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    As far as I know, "closed" doesn't mean "not open" -- something can be not open and not closed, or both open and closed. But I am admittedly not aware of every context of the term.2012-12-13
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    Both depend on what is considered an open set. If the space is $X=[0,\infty)$ as a subset of the usual topology on $\mathbb{R}$, then $[0,1)$ is open (in the subspace topology). It is then usually the case that closed means the complement is open and compact means an open cover has a finite sub-cover. I cannot help on the second question.2012-12-13
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    Sorry, Ed Gorcenski, you are completely right, it was a misunderstanding(what I write at the beginning doesn't make any sense), what I wanted to say is "a set $A$ is closed when $A^c$ is open...".2012-12-13

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Regarding your latter question, usually hypersurface would mean co-dimension $1$ submanifold of Euclidean space. So in $\mathbb R^n$, $M$ would have to be $(n-1)$-dimensional. The Jordan-Brouwer separation theorem implies that if $M$ is compact, it has to be orientable.

That the integral of the Laplacian is zero, you get that from Stokes Theorem.

So minimality is not used for either, just compact and boundaryless, with co-dimension one.