On Thursday, March 20, 2025 at 11:38:17 AM UTC-6 Brent Meeker wrote: "No" isn't an argument. It's just a claim. My argument is based on set theory and topology. If an infinite set can be contained in a countable set of finite sets,
But that's not the case. The number for finite sets is, hypothetically, infinite. Space is a continuum, an order alpha1 infinity. We should get back to what is actually shown by the FLRW model. It assumes the universe isotropic and so can be characterized by a scale factor, a. So the only variables are a and time t. Parameters are pressure and mass/energy density which depend on a. Our present state is taken to be the boundary condition at a=1. The the solution can be propagated into the future and into the past. In the past a goes to zero. In the future it can expand toward and asymptotic limit, expand without limit, or contract to zero. All this is calculus, so it's assuming a continuum of spacetime. The set theory measure of every piece of spacetime is the same alpha1 infinity. Brent Concerning your last sentence above, we consider the observable region as finite, because it has a closed boundary, even though the number of events within are an alpha1 infinity. AG and if they represent spacetime, and each shrinks to zero, then so will the original infinite set. But maybe the infinite set of spacetime points cannot be contained in a countable set, in which case we'd have to use the Axiom of Choice. But I'm not sure if the infinite set of spacetime points can be covered or contained in an uncountable set created by applying the Axiom of Choice. In any event, you need an argument to establish your claim. AG -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/de2eefc3-29f6-4674-8b47-31bd57bb5b12n%40googlegroups.com.
