Could we live in a black hole?

Is the Universe a Black Hole?

My question: do we live in a black hole? That sounds bizarre at first, but the following rough estimate does not make it seem impossible: The manageable universe has around 100 billion galaxies with an average of 100 billion stars (solar masses). Since the diameter of a black hole (SL) is proportional to its mass, the diameter of a corresponding SL would be around 1023 Kilometers or ten billion light years (possibly even more with dark matter and energy). But that is roughly the size of the known universe! So: are we living in a black hole?
(Herbert Haupt)

In the book "Modern Physics" by Gehrke and Koberle on p. 27 there is the equation for the temporal development of the scale factor ("world radius"). If one assumes a flat universe and one puts the necessary density in the equation, then one easily arrives at the fact that the expansion speed of the universe is equal to the escape speed from the expanding universe. Does that have any meaningful physical meaning?
(Günter nephew)

Over the years, more than half a dozen thoughtful readers have noticed that the mass of a (spatially) flat universe that is within its Hubble radius is equal to the mass of a black hole with that radius as the Schwarzschild radius. The Hubble radius is the distance which the expansion rate ν according to the famous formula ν = H0D. formally reached the speed of light, whereby H0 is the Hubble constant measured today. If the Hubble radius is viewed as the radius of the universe, then this equality suggests that the universe could be viewed as a black hole.

First the short calculation that leads to the above-mentioned finding. The mass density ϱcr of a spatially flat universe results from the Hubble constant according to the general theory of relativity H0 and the Newtonian gravitational constant G to
$$ {\ varrho} _ {cr} = \ frac {3 \ space {{H} _ {0}} ^ {2}} {8 \ space \ pi \ space G}. $$
A sphere with the Hubble radius R.Hubble = c/H0 therefore closes the crowd
$$ M = \ frac {4 \ space \ pi} {3} \ left (\ frac {c} {{H} _ {0}} \ right) ^ {3} {\ varrho} _ {cr} = \ frac {c ^ {3}} {2 \ space G \ space {H} _ {0}} $$
a. Their Schwarzschild radius rS. is
$$ {r} _ {S} = \ frac {2 \ space G \ space M} {c ^ {2}} = \ frac {c} {{H} _ {0}} = {R} _ {\ text {Hubble}}, (1) $$
so just equal to the Hubble radius. Can it be a coincidence? Is the Universe a Black Hole? The answer to both questions is a resounding no, for the following reasons:

This article is featured in Stars and Space September 2020

& squ; A black hole has a center and is enclosed in a horizon, which is defined by the fact that no light can penetrate it to the outside. This does not apply to the Hubble Radius because light can easily cross it in either direction.

& squ; It depends on the observer of a black hole where its horizon lies. Only for a static observer at an infinite distance does the radius of the horizon coincide with the Schwarzschild radius. An observer approaching the black hole always has the horizon between him and the black hole, even if he has already crossed the Schwarzschild radius inward.

& squ; It is crucial for the creation of a black hole that the existing mass is compressed into a narrow space. In the universe, on the other hand, the existing mass is distributed extremely evenly without us being able to recognize a border or a center.

& squ; However, the result (equation 1) is no coincidence. This is because the only length scale that can be formed from a mass, the gravitational constant and the speed of light is straight G M / c2 is and because at the same time the only length scale that can characterize a homogeneous, isotropic, expanding universe, the Hubble radius R.Hubble = c/H0 is. Equation 1 shows that these two length scales agree up to a factor of two. First of all, this means that the dynamics of the universe are determined by gravity. The factor two comes from the fact that the calculation assumed that the universe is spatially flat.