The Schwarzchild interior solution is a static solution with no singularity which can be coupled to the Schwarzchild exterior to obtain a static, singularity-free universe. How does this universe dodge the Hawking-Penrose singularity theorems? There is no exotic matter or anything like that, that I can see.
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It is matched to the normal, exterior, solution at a radius greater than the Schwarzschild radius: thus there is no event horizon. So it is a (considerably idealised) solution that might describe something like a star, where the vacuum part of the field is the traditional Schwarzschild solution, but the field in the interior is not: consider the Sun, say.
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$\begingroup$ And if it is matched to the exterior solution inside the Schwarzchild radius then there is an event horizon and then a singularity theorem does apply? Which means that the solution can longer be static and collapse is inevitable? $\endgroup$ Mar 1, 2017 at 13:25
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$\begingroup$ @DzamoNorton It can't be matched inside the horizon, it has terms like $\sqrt{1-R_S/R_m}$ in it, where $R_m$ is the surface radius and $R_S$ the Schwarzschild radus. $\endgroup$– user107153Mar 1, 2017 at 13:34
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$\begingroup$ Thank you @tfb. Last question! So when a real ball of dust collapses through R=2M it is forced to go from having (approximately) a Schwarzchild interior solution to having some other interior solution? $\endgroup$ Mar 1, 2017 at 13:39
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$\begingroup$ Ah, never mind @tfb. A collapsing ball of dust is not static so it never had the Schwarzchild interior solution to start with. I think the penny has finally dropped! $\endgroup$ Mar 1, 2017 at 13:42
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2$\begingroup$ @DzamoNorton: a collapsing homogeneous ball of dust is described by the Oppenheimer-Snyder metric. $\endgroup$ Mar 1, 2017 at 16:25