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$\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\CR}{\mathcal{C}\left(\RR\right)}$

There are as many continuous functions than real numbers: $\left|\left\{f, f:\CR \rightarrow \RR\right\}\right|=|\RR|$.

It is easy, for a machine, to approximately scroll over every real number from $\text{-}\infty$ to $\text{+}\infty$, provided we choose arbitrary, finite bounds and an arbitrary, finite precision. (btw cheers, Turing ;)

Since there are bijections between $\CR$ and $\RR$, is there not, similarly, a natural way to scroll over $\CR$? How would these —necessary— choices of "bounds and precision" translate?

My guess is that there may be many possible approaches (scroll over polynomials coefficients and degrees, over coefficients of periodic functions and the length of their sum, etc.), but then..

How come this problem seems more difficult than just $\textit{spanning}\ \RR$ since $\RR$ and $\CR$ are equipotent?
How come it seems to need more variables and more arbitrary choices?

[EDIT:] as suggested in the comments, here would be the requirements for such a "natural scrolling":

Let $s:\left\{\begin{array}{ll} \RR \to \CR \\ r \mapsto s_r \end{array}\right.$ be called a "scrolling" of $\CR$.

$\forall x \in \RR$, we can define a trace function, tracking the evolution of the image of $x$ by $s_r$ while scrolling: $t_x:\left\{\begin{array}{ll} \RR \to \RR \\ r \mapsto s_r(x) \end{array}\right.$

Can we have both:

  • $s\ \text{bijective}$
  • $t_x \in \CR \forall x \in \RR$ ?
iago-lito
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    If you can implement any bijection $\Bbb R\leftrightarrow \mathcal C(\Bbb R)$ (and this might be the hard part), then you can use it to "scroll" over $\mathcal C(\Bbb R)$. But the result will probably look very strange. Are you looking for an especially nice way? What do you mean with "spanning $\Bbb R$"? – M. Winter Oct 20 '17 at 12:13
  • @M.Winter I can think of no implementation involving only one scrolled real variable and then approximating every function in $\mathcal{C}(\mathbb{R})$.. For instance, if I were to use polynomials, I'd have to pick one degree $n \in \mathbb{N}$ then scroll over $n$ coefficients in $\mathbb{R}^n$.. which, bizarrely, seems more complicated, why? – iago-lito Oct 20 '17 at 12:18
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    You seem to be looking for a function on $\mathbb{R}$ that selects from the space of continuous functions in a "nice way" - suggesting that it be surjective, and continuous in some topology on the function space. Rather like a space filling curve. I don't think there is one that will satisfy you. – Ethan Bolker Oct 20 '17 at 12:23
  • @EthanBolker Interesting. And why? Are these requirements somehow contradictory? – iago-lito Oct 20 '17 at 12:24
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    I think they are contradictory. Perhaps edit your question to include formal requirements like the ones I've suggested and see if a functional analyst can come up with an example or a proof of contradiction. – Ethan Bolker Oct 20 '17 at 12:29
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    @iago-lito Try to formalize "scrolling" through a set $X$. I would do it with a continuous map $\Bbb R\to X$. This definition already includes $\Bbb R$, so no wonder it works so good in $\Bbb R$, e.g. you can just choose the identity $\mathrm{id}:\Bbb R\to \Bbb R$ to "scroll" through $\Bbb R$. So the reason why it works so good for the reals is that the term "scrolling" internally uses this set (and its topology). Or do you have another formal definition in mind? – M. Winter Oct 20 '17 at 13:12
  • @EthanBolker I have added formal requirements to the OP, do you think they can be fulfilled? ;) – iago-lito Oct 20 '17 at 14:07

3 Answers3

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There is no surjective function $s:\mathbb R\to \mathcal C(\mathbb R)$ such that $t_n:r\mapsto s_r(n)$ is continuous for each integer $n.$

There are no continuous surjective maps $[-n,n]\to\mathbb R$. So if $t_n$ is continuous we can pick some $y_n > t_n([-n,n])$ for each positive integer $n.$ Let $f$ be any real function with $f(n)=y_n$ for each $n.$ Suppose $s_r=f$ for some $r.$ There is some integer $n>|r|,$ but $y_n>t_n(r)=s_r(n)=f(n)=y_n,$ a contradiction.

(I remember reading this argument on this site or mathoverflow, possibly by Brian M. Scott, but can't find it now.)

More abstractly, the property this argument uses is that $\mathbb R$ is $\sigma$-compact, and $\mathbb R^{\mathbb N}\subset \mathcal C(\mathbb R)$ is not $\sigma$-compact, and a continuous image of a $\sigma$-compact space must be $\sigma$-compact. I am curious whether the result holds for functions on a compact interval.

Dap
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Just an observation.

Even tough all trace maps are continuous, the diagonalized trace map $x\mapsto t_x(x)$ cannot be continuous. If it were continuous, then so would $f(x)=t_x(x)+1$. So there would be some $r\in\Bbb R$ with $s(r)=f$. But then

$$f(r)=t_r(r)+1=\phi(r)(r)+1=f(r)+1,$$

Contradiction.

M. Winter
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  • Interesting.. maybe it's a good direction towards finding a contradiction in the very requirements. I feel like they are somehow impossible to fulfill. – iago-lito Oct 20 '17 at 15:14
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The point is that $\Bbb R$ is linearly ordered in a continuous manner. You can scroll through it and at every step the new element is close to the old one in the usual metric. You need to define a metric on the space of continuous functions to use this. An obvious one to think of is the integral of the absolute difference between the functions. That isn't a metric because it doesn't converge for all pairs of functions but we can just insist that functions that are the image of close reals must have an integral that converges. This is a continuity restriction. Now you can use the proof that a bijection between $\Bbb R$ and $\Bbb R^2$ cannot be continuous. You are looking for a bijection between $\Bbb R$ and $\Bbb {R}^\omega$ which cannot be continuous for the same reason.

Ross Millikan
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