Can a sufficiently large black hole be singularity-free?

Question

This came to me after reading that a black-hole that has the mass of the observable universe will also have an event horizon that covers the observable universe.

Since the definition of a black hole is that nothing can escape from it, does it actually require for it to have a single singularity of infinite density?

Or could you arrange a theoretical black hole from a super-galaxy, or a dense cluster of galaxies, where they have so much mass that you can't escape outside, but at the same time you can "live" inside without being ripped to pieces?

Answer 1

By the term black hole we normally mean one of four spacetime geometries, the Schwarzschild, Reissner–Nordström, Kerr or Kerr-Newman metrics. The universe is (we believe) approximately described by the Friedmann–Lemaître–Robertson–Walker metric, and it is not a black hole. The Big Bang is not the same as the singularity at the centre of a black hole.

For the simplest black hole, the Schwarzschild metric, once you are inside the event horizon every timelike path leads to the singularity. So not only is there no escape, but there is no way to remain permanently inside the black hole without hitting the singularity. For the charged and rotating black holes things are more complicated, because there are timelike paths that take you through the event horizon, miss the singularity and back out again. However it remains the case that (a) you can never return to your starting point and (b) there are no stable orbits inside the event horizon - you either hit the singularity or are ejected.

So the simple answer to your question is that you cannot arrange a black hole that allows you to live permanently inside the event horizon.

As a side issue, it is not the case that you cannot escape from the observable universe. Rather the reverse actually. Assuming the expansion of the universe carries on accelerating it will approach a de Sitter geometry. In this case there is a cosmological horizon that prevents anything outside the observable universe from entering it. However anything inside the observable universe can escape through the horizon (though from our perspective it would take an infinite time to do this).

Answer 2

Singularities exist in theoretical 'perfect' solutions to General Relativity, but when you look at actual natural Kerr-like objects spinning in a noise filled background of GR waves and other incoming radiation and matter, its likely that no physical real singularities exist.

Brandon Carter, referring to spinning black holes (all real black holes spin):

Thus we reach the conclusion that a timeline or null geodesic or orbit cannot reach the singularity under any circumstances except in the case where it is confined to the equator, cos() = 0 .... Thus as symmetry is progressively reduced, starting from the Schwarchild solution, the extent of the class of geodesics reaching the singularity is steadily reduced likewise, … which suggests that after further reduction in symmetry, incomplete geodesics may cease to exist altogether
Kerr Fields, Brandon Carter 1968. (NB: PDF)

No hair theorems don't make singularities any more likely as they speak to the ring - down time of a black hole in a perfectly quiet background. Its the natural stochastic infall of stuff that keeps real singularities from forming.