Is there a continuous surjection from $\mathbb R$ to $C_p([0,1])$? Let $C_p([0,1])$ denote the set of continuous functions $[0,1]\to\mathbb R$ with the subspace topology coming from the product $\mathbb R^{[0,1]}.$ The "p" stands for pointwise. Is this space a continuous image of $\mathbb R$?

Some observations:


*

*$C_p([0,1])$ is not locally compact

*To be an image of $\mathbb R$ a space must be $\sigma$-compact

*$C_p([0,1])$ is not Baire

*It would be equivalent to consider maps $[0,1]\to [0,1].$ (In one direction we can "clip" a function to $[0,1],$ and in the other direction every real-valued continuous function is $N$ times a $[0,1]$ valued function, and we can use interleaving to encode all functions hitting each multiple of $N$ into a single surjection from $\mathbb R.$)

*There is a nice diagonalization argument showing that $C_p(\mathbb R)$ is not a continuous image of $\mathbb R.$ See this answer and use the obvious continuous $C_p(\mathbb R)\to C_p(\mathbb N)$ coming from restriction to $\mathbb N$.

 A: A classical theorem in $C_p(X)$ theory, proved in Problem/Theorem 186 of "A $C_p$-Theory Problem Book (part1, Topological and Function Spaces)" by Tkachuk (which comes highly recommended if you're interested in spaces of the form $C_p(X)$):

Suppose $X$ is Tychonoff. Then the following are equivalent:
  
  
*
  
*$C_p(X)$ is $\sigma$-compact.
  
*$C_p(X)$ is $\sigma$-countably compact.
  
*$C_p(X)$ is locally compact.
  
*$C_p(X)$ is locally pseudocompact.
  
*$X$ is finite and discrete (so that $C_p(X) \simeq \mathbb{R}^n$ for some finite $n$).
  

A continuous image of $\mathbb{R}$ is $\sigma$-compact and $[0,1]$ is not finite, so we cannot have such a surjection.
The argument you referred to for the specific case $C_p(\mathbb{R})$ is conceptually simpler, but does not go straight through as $[0,1]$ does not have a nice infinite discrete $C$-embedded subspace like $\mathbb{N}$).
A: Henno Brandsma's answer give a reference for a more general question. Here I will just record a mostly self-contained proof for this specific case.
Given a function $\phi:\mathbb R\to C_p([0,1]),$ let $K_m$ be the 
image of $[-m,m]$ for each integer $m\geq 1.$ Inductively define a sequence of closed sets by $X_0=C_p([0,1])$ and for each $n\geq 1:$


*

*$m_n=\min\{m\geq 1\mid K_m\cap X_{n-1}\neq \emptyset\},$ or $m_n=\infty$ if there is no such $m$

*$X_n=\{f\in X_{n-1}\mid f(1/n)=(-1)^n/m_n\},$ where $1/\infty=0.$


So $m_n$ is a weakly increasing sequence, and $X_n$ is a weakly decreasing sequence of sets.
If $\max_n(m_n)=m<\infty,$ then $K_m\cap X_{n-1}\neq\emptyset$ never becomes empty for any $n.$ If $K_m$ were compact, then $(K_m\cap X_{n-1})_{n\geq 1}$ would be a non-increasing sequence of non-empty closed sets, so there would be some $f\in\bigcap_{n\geq 1}(K_m\cap X_{n-1}).$ But that gives $\liminf_{n\to \infty} f(1/n)\leq -1/m$ and $\limsup_{n\to\infty} f(1/n)\geq 1/m,$ which is not possible because $f$ must be continuous at $0.$ So in this case $K_m$ is not compact.
If $m_n\to\infty$ then $\bigcap_{n\geq 1} X_n$ contains the function $f:[0,1]\to\mathbb R$ that linearly interpolates between the points $(1/n,(-1)^n/m_n),$ setting $f(0)=0$ (note $1/m_n\to 0$). But for each $m$ we have $m_n>m$ for some $n,$ which gives $K_m\cap X_{n-1}=\emptyset,$ which implies $f\not\in K_m.$ So in this case $f\not\in\bigcup_{m\geq 1} K_m.$
In either case we have shown that $\{K_m\}$ is not a cover of $C_p([0,1])$ by compact sets. This means the original function $\phi$ is either not continuous or not surjective.
