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This question already has an answer here:

By meaningful I mean experimentally falsifiable.

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marked as duplicate by ACuriousMind, Kyle Kanos, user36790, AccidentalFourierTransform, Qmechanic Apr 4 '16 at 11:26

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Yes it's falsifiable and indeed has been falsified.

The geometry of spacetime is described by an equation called the metric that we get by solving Einstein's equation. We can get some information about the metric that describes the universe by studying the motion of objects like galaxies, galaxy clusters etc. When we do this we find the observations are broadly in line with a geometry called the FLRW metric. This is the metric that describes an isotopic, homogeneous expanding universe. By contrast the various metrics that describe black holes predict behaviour that is completely different to anything we observe.

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    $\begingroup$ What metric you use to describe inside the blackhole? $\endgroup$ – aQuestion Apr 3 '16 at 18:14
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    $\begingroup$ @aQuestion: If we are inside an already formed black hole the metric would be the Schwarzschild, Reissner-Nordstrom, Kerr or Kerr-Newman geometries. If we are in a collapsing region on its way to becoming a black hole the metric would be the Oppenheimer-Snyder geometry. $\endgroup$ – John Rennie Apr 3 '16 at 19:10
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    $\begingroup$ Hi @JohnRennie- would it be fair to say that we can only put an upper bound on the size of a black hole, and distance from the singularity, required to be consistent with observation? Since, as I understand, it, the larger a black hole is the less deviation from the background geometry one experiences inside the event horizon. Presumably this bound would be much larger than the observable universe. $\endgroup$ – Rococo Apr 4 '16 at 0:29
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    $\begingroup$ @Rococo: every observation we've made, from Planck to the the deep sky survey is consistent with an (approximately) FLRW geometry. It would take an extraordinarily contrived geometry to reproduce all this with us inside a black hole. $\endgroup$ – John Rennie Apr 4 '16 at 5:14
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    $\begingroup$ @PyRulez: even for an observer inside a Schwarzschild black hole the singularity is not naked. There is no timelike trajectory from the singularity to the observer's eye. An observer falling into a black hole, however massive, observes an apparent horizon that retreats before them but always wraps the singularity. For a rotating black hole things get more complicated and I'm not sure whether the singularity would be visible but I would guess not. $\endgroup$ – John Rennie Apr 4 '16 at 5:18

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