(no title)
denotational | 1 year ago
Direct:
Make one of the sets uncomputable, at which point the equality of the sets cannot be decided. This happens when the real defined by the Dedekind cut is itself uncomputable. BB(764) is an integer (!) that I know is uncomputable off the top of my head. The same idea (defining an object in terms of some halting property) is used in the next proof.
Via undecidability of Cauchy reals:
Equality of Cauchy reals is also undecidable. The proof is by negation: consider a procedure that decides whether a real is equal to zero; consider a sequence (a_n) with a_n = 1 if Turing machine A halts within n steps on all inputs, 0 otherwise; this is clearly Cauchy, but if we can decide whether it’s equal to 0, then we can decide HALT.
Cauchy reals and Dedekind reals are isomorphic, so equality of Dedekinds must also be undecidable.
Hopefully those two sketches show what I mean by decidable; caveat that I’m not infallible and haven’t been in academia for a while, so some/all of this may be wrong!
denotational|1 year ago
I meant BB(748) apparently.
To elaborate on this point a bit, I specifically mean uncomputable in ZFC. There may be other foundations in which it is computable, but we can just find another n for which BB(n) is uncomputable in that framework since BB is an uncomputable function.
thaumasiotes|1 year ago
But you're arguing that equality of Dedekind reals is undecidable based on a problem that occurs when you define a particular "Dedekind real" only by reference to some property that it has. If you had a representation of the values as Dedekind reals, it would be trivial to determine whether they were or weren't equal. You're holding them to a different standard than you're using for the integers and rationals. Why?
Let's decide a question about the integers. Is BB(800) equal to BB(801)?
It sure seems like it isn't. How sure are you?