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Let's denote the set of all computable functions from $\mathbb{N} \to \mathbb{N}$ as $F$. Now, by any Gödel numbering, $F \simeq \mathbb{N}$.

However, $\mathbb{N}^\mathbb{N} \simeq \mathbb{R}$.

It's obvious that not all functions over the natural numbers are computable, however, how would it be obvious to prove that there are as many uncomputable functions over the natural numbers as real numbers without using the statements above?

Andrés E. Caicedo
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univalence
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  • Not sure but maybe this could help https://math.stackexchange.com/questions/3053345/exactly-when-and-why-can-a-turing-machine-not-solve-the-halting-problem/3108034#3108034 – Marco Bellocchi Apr 18 '19 at 18:10
  • @MarcoBellocchi that's helpful and provides further information, thanks! :D – univalence Apr 18 '19 at 18:12
  • In ZFC there is the axiom of power set and if you assume there exists an enumeration of the subsets of $\Bbb{N}$ (same as functions $\Bbb{N} \to {0,1}$), you get a contradiction by Cantor diagonal argument. In term of computability the functions $f : \Bbb{N \to N}$ such that $n \mapsto f(n)$ is defined by a formula in ZFC and there exists a proof in ZFC that the formula defines uniquely a function, can be computably listed. The probem of definability is when we lack such a proof. – reuns Apr 18 '19 at 20:54
  • In a 1st order language there is only a countable number of formulas that can be used to compute a function. – William Elliot Apr 18 '19 at 22:13
  • @reuns this makes sense, it's interesting that it's really required to "make up a language" to list all computable functions. I'm wondering how this works in type theory, but this goes beyond this question's scopr – univalence Apr 19 '19 at 08:14

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Fix some noncomputable function $f:\mathbb{N}\rightarrow\mathbb{N}$. For each real $r\in (0,1)$, define a new function $$g_r: x\mapsto\begin{cases} f({x\over 2}) & \mbox{ if $x$ is even,}\\ \mbox{${x-1\over2}$th digit of $r$} & \mbox{ if $x$ is odd}\\ \end{cases}$$ (where if $r$ has two decimal expansions, we pick the eventually-$0$ one).

Since $f\le_Tg_r$, the function $g_r$ is noncomputable; meanwhile, the map $r\mapsto g_r$ is clearly injective. So we get an injection from $(0,1)$ to the set of noncomputable functions.

Noah Schweber
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