A question concerning the isomorphic type of continuous functions

Question

let $S$ be the ring of all functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ which are continuous outside a bounded open interval containing zero (depended on $f$). Is it possible to consider $S$ as (ring isomorphic to) $C(X)$ for some completely regular Hausdorff space $X$?

Answer 1

This is not possible.

First, let me argue (similarly to the first paragraph of Pietro Majer's answer) that the sup-norm can be defined purely in terms of the ring structure. Indeed $$\|f\|=\sqrt{\,\inf\big\{{\textstyle\frac b a}\,\big|\,\,a,b\in\mathbb N,\,\,b-af^2 \text{ is a square}\big\}}\in\mathbb R_{\ge 0}\cup\{\infty\}.$$

If a ring is isomorphic to $C(X)$, then it is complete with respect the the above defined sup-norm.

I'll now show that $$S:=\{f:\mathbb R\to \mathbb R\,\,|\,\,f \text{ continuous outside } [-R,R] \text{ for } R\gg 1\}$$ is not complete with respect to the sup-norm, and can therefore not be isomorphic to a ring of the form $C(X)$.

For that, I'll exhibit a Cauchy sequence $(f_n)$ of elements of $S$ such that the limit $f$ is not in $S$.

Let $f$ be the function given by $f(x)=\lfloor\, 1+|x|\,\,\rfloor^{-1}$ and let $f_n:=f\cdot \chi_{[-n,n]}$, where $\chi_{[-n,n]}$ is the characteristic function of the interval $[-n,n]$. Then $f_n\to f$, $f_n\in S$, but $f\not \in S$.

Answer 2

This is an extended comment. I am not sure how much it helps.

Claim. Suppose that $S\cong C(X)$ as $\mathbb R$-algebras for some completely regular space $X$. Then the set $Y$ of isolated points of $X$ is dense.

Pf. In $S$, the primitive idempotents are exactly the elements of the form $\chi_{\{t\}}$. These elements have the following properties:

1. $f\chi_{\{t\}} = f(t)\chi_{\{t\}}\in \mathbb R\chi_{\{t\}}$ for all $f\in S$.
2. If $f=0$ iff $f\chi_{\{t\}}$ for all $t\in \mathbb R$.

So if $S\cong C(X)$ as $\mathbb R$-algebras, then the set $E$ of primitive idempotents of $C(X)$ satisfy the analogous properties. Now idempotents of $C(X)$ are characteristic functions of clopen sets. We claim if $\chi_U$ is primitive, then $U$ is a singleton.

Indeed, if $x\neq y\in U$, we can find $f\in C(X)$ with $f(x)=1$ and $f(y)=0$. But then $f\chi_U\neq c\chi_U$ for any real number $c$, contradicting that $C(X)$ satisfies the analog of 1.

Next we claim that the set $Y$ of isolated points of $X$ is dense. Let $V$ be any open set of $X$. Let $x\in V$. Then there is $f\in C(X)$ such that $f(X\setminus V)=0$ and $f(x)=1$. By the analogue of 2, there is an isolated point $u$ with $f\chi_{\{u\}}\neq 0$. But then $u\in V$.

Answer 3

I think it is not possible. Such a ring isomorphism $\Phi$ should also preserve the order structure, because in both rings non-negative elements are exactly the squares; as a consequence, it must also preserve the constant functions, since it preserves the constant $1$. In other words, $\Phi$ is an ordered $\mathbb{R}$-algebras isomorphism.

In both rings, characteristic functions of singletons can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ such that any positive element smaller than $u$ is a scalar multiple of $u$ ( that is "$0\le v\le u$ implies $v=\lambda u$ ").

Note that the ring $S$ contains all characteristic functions of singletons of $\mathbb{R}$.

Since $\Phi$ preserves the ordered $\mathbb{R}$-algebras structure, if $u:=\chi_{\{t\}}$ is a characteristic function of a singleton of $\mathbb{R}$, then $\Phi(u)$ is also a characteristic function of a singleton $\chi_{\{x\}}$ of $X$.

This way we have defined an injective map $\phi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\Phi(\chi_{\{t\}})= \chi_{\{\phi(t)\}}\, .$ Note that by the order properties of $\Phi$, if $f\in S$ vanishes at $t_0\in\mathbb{R}$, then $\Phi(f)$ vanishes at $\phi(t_0)\in X$ (reason: if $f(t_0)=0$ then $f ^2\ge\lambda\chi_{\{t_0\}}$ for no $\lambda>0$, so $\Phi(f )^2\ge\Phi(\lambda\chi_{\{t_0\}})=\lambda\chi_{\{\phi(t_0)\}}$ for no $\lambda>0$, hence $\Phi(f)(\phi(t_0))=0$). Since $\Phi(c)=c$ for any constant function, we also have $\Phi(f(\phi(r)))=f(r)$ for any $f\in S$ and $r\in\mathbb{R}$ (reason: if $c:=f(r)$, the function $f-c$ vanishes in the point $r$, so that $\Phi( f -c) = \Phi(f) - c$ vanishes in $\phi(r)$, that is $\Phi( f)(\phi(r)) = f(r)$ for all $f\in S$ and $r\in \mathbb{R}$). So $\Phi^{-1}(u)=u\circ\phi$ for any $u\in C(X)$ (as $\Phi$ is bijective). However this yields a contradiction.

Let $\{q_n\}_{n\in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$. Then the (normally convergent) series $\sum_{n\in\mathbb{N}} 2^{-n} \chi_{\psi(q_n)}$ represents an element $u$ of $C(X)$ that for any $t\in\mathbb{R}$ vanishes at $\phi(t)$ if and only in $t$ is irrational; hence $u\circ \phi\in S$ vanishes exactly on the irrationals, a contradiction.