классика на фесбуке

я к вам пишу чего же боле
я уронила в речку мяч

(Alex Gabriel posted)

fb libra bashing

 src


"What is laid bare for the world to see is an architecturally schizophrenic code artifact claiming to be a new reliable platform for global payment infrastructure. Yet the actual implementation diverges from this goal in bizarre ways when one actually dives into the codebase. I’m sure there is an interesting story about the internal corporate politics of this project and as such I thought it apt to do some diligence on what I see as a truly strange set of architectural choices that break the entire system and put consumers at risk."

peano, sets, foundations

We had a rather confusing discussion with gonchar lxe sassa_nf

I think I was wrong here and there.

So, the issue was, can we declare that a "subset" of natural numbers is either finite or countable?
First, the question has no sense. Peano arithmetic has no notion of "finite", "countable", "set of".
What we can prove though, is that, e.g., for any natural number n there is a prime natural number p where p = n + k for some natural number k. Which can be interpreted roughly as "prime numbers are not limited", or "there is no natural number that can enumerate all prime numbers".

Now, about the sizes.

We may dive into a set theory, e.g. ZFC. In this ZFC we can model Peano numbers, and we can, using Infinity Axiom and Separation Axiom Schema, produce "a set of all natural numbers". This set is infinite in a certain sense (there's more than one definition of finite and infinite). In any case, due to Choice Axiom (and others) and some theorems, a subset of ℕ is no bigger than ℕ. If it's bounded, it's finite; if it's not bounded, it can be counted anyway (thanks lxe for the hint).