I am reading this note on super-Riemann surfaces. In the second paragraph of section 7.4.1 (page 87), there is a statement that I am trying to understand:

The compactified moduli space of closed oriented Riemann surfaces has no boundary.

However, as it is mentioned, the compactified moduli space of surfaces with boundary is an orbifold with boundary.

Can someone please elaborate these points and that why the compactified moduli space of closed oriented Riemann surfaces does not have boundaries and is just compact? A good pedagogical reference on the Deligne-Mumford compactification is highly appreciated.

why the compactified moduli space of closed oriented Riemann surfaces does not have boundaries and is just compactAs written, this is a tautology --- it is compact because it is the compactification of something. Maybe a better way to ask the question is why the Deligne--Mumford compactification ismodular, that is, why the points added in actually correspond to something geometric. A reference that I like is Harris--Morrison,Moduli of Curves. $\endgroup$ – Bort Nov 14 '18 at 10:42An analytic construction of the Deligne-Mumford compactification of the moduli space of curvesprojecteuclid.org/download/pdf_1/euclid.jdg/1406552251 $\endgroup$ – Maxime Scott Nov 14 '18 at 15:31