Comment on Explain yourselves, comp sci.
Lojcs@lemm.ee 5 months agoIt still can be, just not on infinite precision as nothing can.
Comment on Explain yourselves, comp sci.
Lojcs@lemm.ee 5 months agoIt still can be, just not on infinite precision as nothing can.
holomorphic@lemmy.world 5 months ago
But the vector space of (all) real functions is a completely different beast from the space of computable functions on finite-precision numbers. If you restrict the equality of these functions to their extension ,defined as f = g iff forall x\in R: f(x)=g(x), then that vector space appears to be not only finite dimensional, but in fact finite. Otherwise you probably get a countable infinite dimensional vector space indexed by lambda terms (or whatever formalism you prefer.) But nothing like the space which contains vectors like F_{x_0}(x) := (1 if x = x_0; 0 otherwise) where x_0 is uncomputable.