Difference between rings of continuous functions on $[0,1]$ and $(0,1)$.

Consider $X=[0,1],Y=(0,1)$. Denote $C(W)=$set of real-valued continuous functions on $W$. Here $X$ and $Y$ are inherited the standard topology from $R$.

1. What is the difference between two rings $C(X),C(Y)$? Clearly $C(Y)$ has functions which are not considered continuous on $X$ as $\frac{1}{(x-1)^r},\frac{1}{x^r}$ for $r\in R_{>0}$. But I can always restrict the functions on $X$ to $Y$ to obtain a continuous map by considering $Y$ as a subspace of $X$. So I have $C(X)\xrightarrow{Res^X_Y}C(Y)$ and this should be mono.

The compactness of $[0,1]$ requires the function bounded whereas $C(Y)$ does not have this property.

1. How does the ring structure recognize topological difference or do we pass the topological difference through the ring structure like what we have done in sheaf of regular functions through some sort of localization technique?

2. Is there a generic recipe to construct the structure of these rings from $X$ or $Y$?

• When you ask in question 1 "What is the difference", can you elaborate what exactly you are looking for? – Kyle Aug 7 '17 at 5:19
• How does the ring structure recognize topological difference or do we pass the topological difference This is a very ambitious question that probably requires an entire textbook to answer. You could take a look at Rings of continuous functions by Gillman and Jerrison and then refine your question. – rschwieb Aug 7 '17 at 13:06
• @KyleGannon Sorry for the late response. So I am aware of ring structural difference. However, If I look at $X=R$, with $Y=X-pt$ for zariski topology. I would expect localization from open set. In general setting this is completely different. In open set $(0,1)$ the rings of continuous functions has more information than that of $[0,1]$. So the difference is whether topologically different space must have different ring structure? Should I expect 1-1 correspondence for classes upto homeo? – user45765 Aug 7 '17 at 16:58
• @rschwieb I was not aware this was so ambitious as I thought a bit more about a problem on Falko's Field theory's chpt 4 exercises when I realize there is something more to be asked. – user45765 Aug 7 '17 at 16:59

I will only answer question 1 and only from the model theorist's perspective: $C(X)$ is not elementary equivalent to $C(Y)$ (i.e. we can find a sentence in the language of rings with is true in one but false in the other). This implies that they are also not isomorphic (your remark above demonstrates that they are not "canonically" isomorphic).

Let $\rho(x) =(\exists y)(x\cdot y=1)$; $\varphi(x)=(\exists y)(y^{2}=x)$; and $\psi(x)= \varphi(x) \wedge (\forall y \neq \mathbf{0})(\varphi(y)\to\neg\varphi(x-y))$. $\rho(x)$ is the collection of elements which have an inverse, $\varphi(x)$ is the collection of elements which are squares, and $\psi(x)$ is the collection of elements which are both squares and for any other (non-zero) square, their difference is not a square (we think of $\psi(x)$ as elements which are positive and close to $0$. For instance, the identity function from $X \to X$ satisfies this condition).

Claim 1: Let $f \in C(X)$ or $f \in C(Y)$. Then, $f$ is a square iff $f \geq 0$.

Claim 2: If $f \in C(X)$ and $\psi(f)$ is true, then $\exists x_0$ such that $f(x_0) = 0$. This follows from the fact that $X$ is compact. However, this is not true for $g \in C(Y)$. Consider $g = id_Y$.

Claim 3: If $f \in C(X)$ and $\psi(f)$ is true, then $f$ does not have a multiplicative inverse in $C(X)$ (since $f(x_0) = 0$ for some $x_0$). However, this is not true in $C(Y)$ as remarked above.

Claim 4: $C(X)\models\forall x(\psi(x)\to\neg\rho(x))$ and $C(Y) \models \exists x(\psi(x) \wedge \rho(x))$.

Therefore, the two structures are not elementary equivalent in the language of rings.

Furthermore, we can define a partial order on these structures (in the ring language) by saying that $f \leq g \iff \varphi(g -f)$. Now, $C(X)$ is archimedean with respect to this partial order (and the unique image of $\mathbb{Z}$ mapped into this structure) while $C(Y)$ is not.