8

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 $\mathbb N$.
Dap
  • 25,701
  • 20
  • 52
  • If $f:\Bbb R\to C_p([0,1])$ is a continuous surjection, then $f([-n,n])$ is compact and $C_p([0,1])=\bigcup_n f([-n,n])$ is sigma-compact. I'm not sure about your specific topology, but very many infinite dimensional vector spaces are not sigma compact; they are too big. – s.harp Oct 20 '17 at 19:26
  • 1
    @s.harp in fact all infinite-dimensional topological vector spaces are not locally compact. So this provides the necessary proof. – Henno Brandsma Oct 20 '17 at 21:35
  • 1
    @HennoBrandsma: But the relevant property here is $\sigma$-compactness, not local compactness. And an infinite-dimensional topological vector space can be $\sigma$-compact. – Eric Wofsey Oct 21 '17 at 07:01
  • @EricWofsey not if its Baire like $C_p([0,1])$. So in this case it’s not $\sigma$-compact. – Henno Brandsma Oct 21 '17 at 07:10
  • Why is $C_p([0,1])$ Baire? – Eric Wofsey Oct 21 '17 at 07:13
  • @HennoBrandsma: I don't think it is Baire. The set of functions with at least $k$ upcrossings of $[-1,1]$ (i.e. there exist $x_1<\dots<x_{2k}$ with $f(x_i)<-1$ for $i$ odd and $f(x_i)>1$ for $i$ even) is open and dense, but the intersection is empty. – Dap Oct 21 '17 at 09:11
  • @EricWofsey I’ll check my books when I’m back home. There are some theorems on this that I cannot recall right now – Henno Brandsma Oct 21 '17 at 09:31
  • $C_p([0,1])$ is not Baire, for $C_p(X)$ to be Baire, $X$ has to be very specific (and $[0,1]$ is too nice). So that does not work, OK. How does @dap know there is some nice diagonalisation argument ? – Henno Brandsma Oct 21 '17 at 10:26
  • @HennoBrandsma: I've found the argument I was thinking of and added a link in the question (https://math.stackexchange.com/q/74779) – Dap Oct 21 '17 at 11:29

2 Answers2

4

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:

  1. $C_p(X)$ is $\sigma$-compact.
  2. $C_p(X)$ is $\sigma$-countably compact.
  3. $C_p(X)$ is locally compact.
  4. $C_p(X)$ is locally pseudocompact.
  5. $X$ 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 $\mathbb{N}$).

Henno Brandsma
  • 250,824
  • Just a side question. Is the $C_p(X)$ theory actually a viable area for a graduate student? It does not seem to be quite well known. If I hadn't wanted to look for a topology problem book, I would not have come across the theory at all. The author doesn't give any kind of motivation for the theory if I recall correctly. – Project Book Oct 21 '17 at 11:46
  • @ProjectBook It's a very lively and active subject. See the 4 problem books (which I have as well) by Tkachuk and Arhangelskii's book on it. Lots of nice open problems and interrelations with other parts of general topology. – Henno Brandsma Oct 21 '17 at 11:48
  • Okay, I guess my school just does not have a strong topology program. Thanks. – Project Book Oct 21 '17 at 11:52
  • @ProjectBook what researchers work at your school? – Henno Brandsma Oct 21 '17 at 12:14
  • I'm not sure if I can/should answer this question. – Project Book Oct 21 '17 at 12:19
4

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.

Dap
  • 25,701
  • 20
  • 52