Hacker News new | ask | show | jobs
by baddox 3423 days ago
Indeed, if you consider a real number to just be a decimal representation formed by concatenating all the natural numbers in a set (not at all a rigourous construction of the reals, but hopefully sufficient for this argument), the equivalence is clear.
2 comments

I've always found dedekinds construction (where sqrt 2 is represented by the set of rationals {x in Q : x^2 < 2}) to be a nice natural place where powersets of countable infinite sets come up (in addition to being a great way to construct R)
I tend to prefer considering the sets to encode bitstrings (encode a set as sum(2 ^ -x for x in X)), but yes the equivalence between computable sets and computable numbers is straightforward.