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
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.