The largest black hole The Schwarzschild radius involves an expression in terms of Newton's constant $G$, the mass $M$ inside a radius $r$, and the speed of light squared $c^2$. Current estimates of the universe's matter density are about six protons per cubic meter.  But, the $M$ inside a sphere goes up as $r^3$, while the "time curvature" coefficient is $1 - \frac{2GM}{c^2\, r}$.  So this coefficient is bound to hit zero for $r$ large enough.  The $M$ outruns the denominator as a function of $r$.
I calculated that this coefficient hits zero when $r$ equals 13.54 billion light years.  Question:  Is this any evidence for our universe being one very large black hole?   
 A: Aruns weak equivalence is an argument which goes like this: To make the temperature of a black hole go down, you need to add matter to the system. Using the following approximation we have
$m \rightarrow \infty$
Then the temperature goes to zero
$T \rightarrow 0$
And for a black hole with infinite mass, the curvature tends to zero as well!
$K \rightarrow 0$
As I have stated before though, you cannot really have a system like a vacuum reach absolute zero, when the vacuum is not perfectly Newtonian. To add to his extended weak equivalence, assume the following ~
The radius of a black hole is found directly proportional to its mass $R \approx M$. The density of a black hole is given by its mass divided by its volume $\rho = \frac{M}{V}$ and since the volume is proportional to the radius of the black hole to the power of three $V \approx R^3$ then the density of a black hole is inversely proportional to its mass radius by the second power $\rho \approx M^2$.
What does all this mean? It means that if a black hole has a large enough mass then it does not appear to be very dense, which is more or less the description of our own vacuum: it has a lot of matter, around $3 \times 10^{80}$ atoms in spacetime alone - this is certainly not an infinite amount of matter, but it is arguably a lot yet, our universe does not appear very dense at all. 
