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We denote the ring of all continuous real-valued functions on $X$ by $C(X)$. The ring of all bounded continuous real valued functions on $X$ is denoted by $C_b(X)$. One of the goals of the study of $C(X)$ is to connect algebraic properties of the ring $C(X)$ with the topological properties of the space $X$.

For a simple and well-known connection there is the following theorem:

Theorem: The topological space $X$ is connected iff the ring $C(X)$ has no idempotent element other than $0$ and $1$.

My question comes from the relation between ring homomorphisms of the rings $C(X)$, $C(Y)$ and topological spaces $X$, $Y$.

Q1: Let $X$ and $Y$ be two $T_{3\frac{1}{2}}$topological spaces. If $X$ is a zero dimensional space and $C(X)$ is ring isomorphic to $C(Y)$ (i.e. $C(X)\cong C(Y)$) can we deduce that $Y$ is also a zero dimensional space?

(we recall that a topological space $X$ is zero dimensional if it has a base of clopen subsets.)

The same question can be asked by exchanging the ring $C(X)$ with the Banach algebra $C_b(X)$ as follows:

Q2: Let $X$ and $Y$ be topological spaces with the previous assumption. If $X$ is a zero dimensional space and $C_b(X)$ is ring isomorphic to $C_b(Y)$ (i.e. $C_b(X)\cong C_b(Y)$) can we deduce that Y is also a zero dimensional space?

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  • $\begingroup$ @Qiaochu: The theorem is trivial and holds for every ringed space. $\endgroup$
    – Martin Brandenburg
    Commented Jun 15, 2012 at 14:15
  • $\begingroup$ @Martin, could you expand on your comment? I don't think about T3.5 spaces that are not locally compact Hff very often, so the question is not so obvious to me. $\endgroup$
    – Yemon Choi
    Commented Jun 15, 2012 at 18:17
  • $\begingroup$ A locally ringed space $X$ is connected if and only if $\mathrm{\Spec}(\mathscr{O}_X(X))$ is connected, and the latter holds if and only if $\mathscr{O}_X(X)$ is connected, i.e., has no idempotents other than $0$ or $1$. @Martin Does this hold for every ringed space? $\endgroup$
    – Keenan Kidwell
    Commented Jun 15, 2012 at 19:34
  • $\begingroup$ @Martin: whoops. I got a little confused there. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 15, 2012 at 20:43
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    $\begingroup$ I was refering to what AliReza has named a "Theorem" in his question. Qiaochu has deleted his comment that this might not be true for geneal $X$. [just to clarify my comment above] @Keenan: Perhaps ringed spaces with nontrivial stalks. $\endgroup$
    – Martin Brandenburg
    Commented Jun 16, 2012 at 14:49

2 Answers 2

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If $X$ and $Y$ are realcompact spaces (for example, if they are compact, or $\sigma$-compact, or are separable metric spaces), then the isomorphism of $C(X)$ and $C(Y)$ implies the homeomorphism of $X$ and $Y$. See L. Gillman and M. Jerison, Rings of Continuous Functions, Van Nostrand, 1960 (Theorem 8.3). This implies the requested zero-dimensionality result. For compact sets, there is also the Banach-Stone theorem stating that $X$ and $Y$ are homeomorphic if $C(X)$ and $C(Y)$ are linearly isometric.

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  • $\begingroup$ Thank you dear Anatoly Kochubei. I think you Put an Extra condition in the Question. You suppose that $X$ and $Y$ are two real-compact spaces. But the space $X$ in my Question has only the zero-dimensionality condition.As you Know there is a non real-compact zero dimensional space.(fore example [$0, ω_1$))which is not real compact. For the second answer You suppose the compactness, But you Know that if $C_b(X)$ is isomorphic with $C_b(Y)$ we could not deduce that $C(X)$ is isomorphic to $C(Y)$. Please defend your answer in general case, not with the special case. $\endgroup$
    – Ali Reza
    Commented Jun 15, 2012 at 14:39
  • $\begingroup$ I just gave the conditions, under which your questions get positive answers. Why are you sure that no additional conditions are needed? $\endgroup$
    – Anatoly Kochubei
    Commented Jun 15, 2012 at 19:13
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In general the answer to both questions is no: there are pseudocompact zero-dimensional spaces (see, e.g., this answer) whose Cech-Stone compactifications are not zero-dimensional. For an $X$ like that one has $C(X)=C_b(X)$ and this ring is isomorphic to $C(\beta X)$.

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