• Lojcs@lemm.ee
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    6 months ago

    It still can be, just not on infinite precision as nothing can with fp.

    • holomorphic@lemmy.world
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      6 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 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.