The set of closed maps is closed, but not open in $C_S^\infty(\mathbb{R}, \mathbb{R})$.

Question

The set of closed maps is closed, but not open in $C_S^\infty(\mathbb{R}, \mathbb{R})$. (A map $f$ is closed if it takes closed sets onto closed sets.) $C_S^r(M, N)$ is the set of $C^r$ maps who's topology is defined by the strong basic neighborhood defined by equation (2) page 35 of Differential Topology by Hirsch.

I was thinking that to show that it's closed maybe we can show that the complement is open, but that doesn't' seem very feasible. To show that it is not open we can find a function $g$ such that any open ball around $g$ intersects the set in the the problem.

Answer 1

Claim: a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ is closed iff for $\alpha=\pm \infty$, as $x \rightarrow \alpha$, $f(x)$ is eventually constant or goes to infinity.

Proof: for $\Rightarrow$, we focus on the case $\alpha=\infty$. Assume $f$ is closed. Let $x_n$ be a sequence of real numbers going to infinity, then $\{f(x_n)|n \geq 1\}$ is closed. In particular, if $\ell \in \mathbb{R}$ is a limit point of $(f(x_n))_n$, then we must have $f(x_n)=\ell$ for infinitely many $n$.

Suppose $f \rightarrow \ell \in \mathbb{R}$, then the above shows that $f$ is eventually constant.

The set of limit points of $f$ at $\infty$ is an interval, so if $f$ is not eventually constant or does not go to infinity, there are $\ell \in \mathbb{R}$, $\epsilon >0$ such that every point of $[\ell-\epsilon,\ell+\epsilon]$ is a limit point of $f$. In particular, we can find a sequence $x_n \rightarrow \infty$ such that $f(x_n) \in [\ell+\epsilon/3^{n+1},\ell+2\epsilon/3^{n+1}]$, so that $f(x_n) \neq \ell$ for every $n$ but $f(x_n) \rightarrow \ell$. QED.

For $\Leftarrow$, we can check that a continuous $f$ is closed if $f$ and $f(-\cdot)$ are closed as functions $[0,\infty) \rightarrow \mathbb{R}$. So we’re reduced to showing that a continuous function $f:[0,\infty) \rightarrow \mathbb{R}$ is closed if it is eventually constant or it goes to $\infty$. Both cases are easy to see (the second, because $f$ is proper – the pre-image of a compact is compact – which is a stronger condition in eg metrizable spaces).

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So the goal is to show that

1. any neighborhood of $0$ contains non-closed functions.
2. any non-closed function has a neighborhood consisting only of non-closed functions.

For 2), by definition of the topology, we can work in $C^0_S$. We’ll focus only on behavior at $\infty$, and the types of singularity, leaving how to recompose them to the careful reader.

We can split into two cases, for $g$ a non-closed continuous function:

2a. $g$ converges to some finite $\ell$ but is not eventually constant.

2b. $g$ has more than one limit point at $\infty$.

In case 2b, $g$ has two distinct finite limit points $a$ and $b$. Let $\frac{|a-b|}{2} > \epsilon > 0$, then every function in $\{h| |h-g| <\epsilon\}$ has more than one finite limit point at $\infty$ so is non-closed.

In case 2a, there is a sequence $x_n$ such that $x_{n+1} > x_n+1$ such that $\delta_n=|f(x_n)-\ell| > 0$. Then $\{h| \forall n\forall y \in [x_n-1/2,x_n+1/2],\, |h-g|< \delta_n\}$ is open, any function $h$ in this set is such that $h(x_n) \neq \ell$, $h(x_n) \rightarrow \ell$, so is non-closed.

Regarding 1): it’s enough to show that for any $r \geq 0$, and any positive sequence $\epsilon_n$ going to zero, $U_{r,\epsilon}=\bigcap_{n,0 \leq s \leq r}{\{h| \forall y,\, n \leq |y| \leq n+1 \Rightarrow |h^{(s)}(y)| \leq \epsilon_n\}}$ contains a smooth function that is not stationary at zero.

Let $\chi$ be a smooth even non-negative function with support in $(-1/4,1/4)$ such that $\chi(0)=1$ and $s_n$ be a positive sequence with $(1+\|\chi\|_{C^r})s_n < \min_{1 \leq t \leq n+2}\,\epsilon_t$.

Define $\psi(x)$ by: if $x \in [n-1/2,n+1/2]$, $n \neq 0$, $\psi(x)=s_{|n|}\chi(x-n)$ (and $\psi(x)=0$ for $|x| \leq 1/2$). Then $\psi \in U_{r,\epsilon}$ and $\psi$ converges to zero, but $\psi$ is not stationary so isn’t closed.