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As shown here, the set of the continuous real functions $(ℝ,ℝ)$ has the same cardinality as the set of the real numbers $ℝ$.

Considering that it is easy to define a well-behaved distance in the set $(ℝ,ℝ)$, the immediate question would be: can you construct a continuous mapping $Φ: ℝ → (ℝ,ℝ)$ that is also surjective?

Since all the functions would be continuous, they'd be completely determined by their restriction to the rational numbers $Φ: ℚ → (ℚ,ℝ)$; by playing around with pairs of countable indices, it is easy to approximate a given continuous function - I wonder if there is a diagonal argument that can be used to prove or disprove the existence of a continuous mapping that can approximate any continuous function.

Bernard
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Gianluca Calcagni
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  • What metric do you put on $C(\mathbb R, \mathbb R)$? Also I think there might be a problem in your intuition because you get an injection $C(\mathbb R, \mathbb R) \rightarrow C(\mathbb Q , \mathbb R)$ which might not be an homeomorphism because there might be continuous functions $\mathbb Q \rightarrow \mathbb R$ which cannot be continously extended. – blamethelag Oct 26 '21 at 19:51
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    If you define distance in which $C(\mathbb{R} . \mathbb{R} )$ becomes a complete metric vector space then it is impossible. It follows from the fact that then $C(\mathbb{R} . \mathbb{R} )=\bigcup_n f([-n.n] ) $ becomes a first category which contradicts the Theorem of Baire. –  Oct 26 '21 at 19:58
  • @blamethelag thanks for your answer. I considered two different distances, but it seems to me the choice makes little difference.

    First choice: given two continuous functions $f_j:ℝ→ℝ$ for j=1,2, rescale the codomain of the functions to obtain $f'j:ℝ→(-1,1)$ (e.g. by applying $x→sign(x)\frac{x^2}{1+x^2}$), define the distance as $\limsup{x \in ℝ}|f'_1(x) - f'_2(x)|$

    Second choice: rescale the codomain of the functions, select a complete enumeration of the rational numbers $q:\mathbb{N}→\mathbb{Q}$, define the distance as $\sum_{n\in\mathbb{N}}\frac{|f'_1(q(n)) - f'_2(q(n))|}{2^n}$

     – Gianluca Calcagni Oct 27 '21 at 10:52
  • @MotylaNogaTomkaMazura thanks for your answer as well. Indeed, choosing a complete metric vector may be an issue - but that is not needed in my question. Are you implying that the existence of the mapping $Φ$ would imply that the distance must form a complete metric space? I don't see such relation (the mapping goes backwards in respect to the definition of "distance"), but maybe I am missing something obvious here. – Gianluca Calcagni Oct 27 '21 at 10:57

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