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Suppose $K$ is some nice space, for example $\mathbb R$ or $\mathbb C$. Let $X$ be a set and $C$ a ring of functions $X \to K$. Is there any way to determine, from the algebraic structure of $C$, whether there exists a topology $\tau$ on $X$ under which $C$ is exactly the ring of continuous $K$-valued functions on $(X,\tau)$?

In the simplest case I can think of, we might restrict our question to when we want $(X,\tau)$ to be a compactum. Then we should be able to recover the points of $X$ from the maximal ideals of $C$, which would have to be indexed by the points of $X$. The ideal $I_x = \{f \in C \colon f(x)=0\}$. For that idea to exist at all I think it's enough that some function in $C$ vanishes at each $x \in X$.

But what if the topology is not required to be compact? Is anything known about this question? Is there any more easily-checkable answer beyond, "Equip $X$ with the weak topology generated by $C$ and see if there are any other continuous functions"?

Edit: Of course some spaces share a ring of continuous functions. Any non-realcompact space has the same function ring as its realcompactification, and as every intermediate space. So it is possible that the topology $\tau$ may not be unique. My question is not about finding the topology itself, but just about showing one exists at all.

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  • $\begingroup$ It seems to me that $C([0,\omega_1))$ and $C([0,\omega_1])$ are indistinguishable algebraically. $\endgroup$ – Tomek Kania Nov 28 '15 at 16:01
  • $\begingroup$ Even so, if I gave you either of $C[0,\omega_1)$ or $C[0,\omega_1]$ (they are the same), could you look at the ring and say whether it is the ring of continuous functions for some (maybe not unique) space? $\endgroup$ – Daron Nov 28 '15 at 16:16
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    $\begingroup$ That depends on what structure you want to put on either the topological or the algebraic side. Supposing your ring is a commutative $C^*$ algebra you can apply Gelfand duality to construct a compact hausdorff space. In this case the ring of functions are the continuous one. If you have a real f.g. algebra, you can construct an affine scheme over $\mathbb R$. The ring of functions would be polynomial functions. That is all part of the bigger picture of the duality between spaces and algebras ncatlab.org/nlab/show/Isbell+duality $\endgroup$ – Georg Lehner Nov 28 '15 at 16:42
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This characterization problem was studied by F.W. Anderson, Trans. Amer. Math. Soc. 103 (1962), 249--271, and A.W. Hager, Proc. London Math. Soc. (3) 19 (1969), 233--257, among others. So there are answers, but I wouldn't call them "easily checkable".

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