Sudip Kumar Acharyya, Kshitish Chandra Chattopadhyay, Pritam Rooj
Suppose $F$ is a totally ordered field equipped with its order topology and $X$ a completely $F$-regular topological space. Suppose $\mathcal{P}$ is an ideal of closed sets in $X$ and $X$ is locally-$\mathcal{P}$. Let $C_\mathcal{P}(X, F)=\{f\colon X\rightarrow F~|~ f$ is continuous on $X$ and its support belongs to $\mathcal{P}\}$ and $C_\infty^\mathcal{P}(X, F)=\{f\in C_\mathcal{P}(X, F)~|~ \forall \varepsilon>0$ in $F$, $cl_X\{x\in X: |f(x)|>\varepsilon\}\in \mathcal{P}\}$. Then $C_\mathcal{P}(X, F)$ is a Noetherian ring if and only if $C_\infty^\mathcal{P}(X,F)$ is a Noetherian ring if and only if $X$ is a finite set. The fact that a locally compact Hausdorff space $X$ is finite if and only if the ring $C_K(X)$ is Noetherian if and only if the ring $C_\infty(X)$ is Noetherian, follows as a particular case on choosing $F=\mathbb{R}$ and $\mathcal{P}=$ the ideal of all compact sets in $X$. On the other hand if one takes $F=\mathbb{R}$ and $\mathcal{P}=$ the ideal of closed relatively pseudocompact subsets of $X$, then it follows that a locally pseudocompact space $X$ is finite if and only if the ring $C_\psi(X)$ of all real valued continuous functions on $X$ with pseudocompact support is Noetherian if and only if the ring $C_\infty^\psi(X)=\{f\in C(X)~|~ \forall\varepsilon>0, cl_X\{x\in X: |f(x)|>\varepsilon\}$ is pseudocompact $\}$ is Noetherian. Finally on choosing $F=\mathbb{R}$ and $\mathcal{P}=$ the ideal of all closed sets in $X$, it follows that: $X$ is finite if and only if the ring $C(X)$ is Noetherian if and only if the ring $C^*(X)$ is Noetherian.
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