this post was submitted on 28 Apr 2024
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It still can be, just not on infinite precision as nothing can with fp.
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 countably 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.