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 $R$. 


*

*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. 


*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?

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