Epsilon–delta limits work just fine on metric spaces. Do you even math?The ironing. You need complete spaces otherwise the limit can be outside the space and you cannot take the distance between the limit and elements of your series because your distance function must have both arguments from the space.
fixed lower bound on distanceThat is a separate and entirely legitimate question. If your distance function is either at least LOW or exactly 0, so that there are no distances in the open interval (0,LOW), then your convergent series can only be of a very particular form. For a[i] to converge to LIM, for any delta there must be a starting N over which dist(a[i>N],LIM)<delta. Pick delta=LOW/2 and you get a starting N over which the distances are under LOW and therefore 0, so a[i>N]=LIM. So the only convergent series are those that after a finite number of terms switch to a constant and never deviate again. So you get craziness like picking some nonzero X and having a[i]=X/2**i not being convergent. Your space only consists of isolated points and its interior is empty. So you won't be taking any nonconstant limits, like those needed for derivatives, and you don't get any proper calculus.
Not sure why it's separate when that was the entirety of my point.It's separate from the completeness problem from >>9 that it was posted as replying to. Both problems -- lack of completeness and "fixed lower bound" -- lead, through different routes, to the same result: inability to take commonly needed limits.
Maybe you're are confusing infinite resolution with infinite bounds1. The distance between any two points in a metric space is finite, because the distance function is defined from the product SxS with values in the nonnegative reals, so a volume is either unbounded or its bounds are automatically finite.
since the universe itself is discrete.And as soon as you present your mathematically consistent proof of this to the larger scientific community, your Nobel prize is a foregone conclusion.
I mistakenly assumed that you had the capacity to infer that the Nobel prize would evidently be in physics and that mathematical consistency would be required between your proposed model and your experimental proof that it is not continuous. It is not enough to show empirically that Newtonian gravity is untenable, e.g. precession of Mercury, someone also has to propose a mathematically consistent alternative. Then the only way you don't get a Nobel prize for that is if you already have one and they don't give you two in the same discipline.mathematically consistent proof of this to the larger scientific communityThere is no Nobel prize in mathematics
doltThe ironing.