Consider any topological space $X$ and $\mathbb{R}$ be with usual topology. The set of all continuous functions from $X$ to $\mathbb{R}$, denoted by $C(X,\mathbb{R})$, is a commutative ring with unity under pointwise addition and pointwise multiplication. If we consider $X=[0,1]$ with usual topology, then $X$ is connected and $C(X,\mathbb{R})$ is not an integral domain. If $X$ is any set with indiscrete topology, then $X$ is connected and $C(X,\mathbb{R})$ is an integral domain. Again if we take $X$ with at least two different elements and if topology is taken as the discrete topology, then $X$ is disconnected and $C(X,\mathbb{R})$ is not an integral domain. So, from these facts I am unable to guess and prove anything about $C(X,\mathbb{R})$ when $X$ is arbitrary. Can anyone tell me about this fact? Does there exist any iff condition on $X$ to make $C(X,\mathbb{R})$ an integral domain?
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1An idea : If there is two closed set A and B such that $A \cup B = X$, then you can probably construct a non-zero continuous function f such that f(A) = 0 and a non-zero continuous function g such that g(B) = 0. If you can do this, $C(X, \mathbb{R})$ will not be an integral domain – Tryss Mar 05 '15 at 06:36
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A and B should be disjoint, right? This is true. So, if X is disconnected, then C(X,ℝ) is not an integral domain. But, converse of this fact is not true. I am seeking a necessary and sufficient condition. @ Tryss – Anirban Mar 05 '15 at 06:40
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Not disjoints, they can intersect. All that's important is that you can build a non-zero function on $A^C$ and $B^C$ – Tryss Mar 05 '15 at 06:42
3 Answers
The only case when $C(X)$ is a domain is when it is trivial, i.e. there are no nonconstant (real-valued) functions on $X$. Given a nonconstant function, you can manipulate it (by composing with maximum, addition of a constant and multiplication by $-1$) to obtain two continuous functions with disjoint supports. (Edited in according to Pete L. Clark's suggestion: more explicitly, if $f(x)=a<b=f(y)$ and $c=(b-a)/3$, consider $g_1=\max(0,f-a-2c)$ and $g_2=\max(0,b-f-2c)$. Then $g_1$ is nonzero where $f>b-c$ and $g_2$ is nonzero where $f<a+c$, so $g_1g_2=0$. Drawing a picture might help.)
I don't think there's a concise sufficient condition for that. It is clearly satisfied by the one-point space. It is clearly not satisfied by any nontrivial normal, or even completely regular space. On the other hand, there are many regular spaces with this property. For more information, see this related question, and this relevant paper.
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1Also: it is shown in Gillman and Jerison's Rings of Continuous Functions that for any topological space $X$, $C(X)$ is canonically isomorphic to $C(T(X))$, where $T(X)$ is the Tychonoff completion of $X$. Since essentially by definition the continuous functions separate points on a Tychonoff space, it seems that the necessary condition on a (nonempty!) space $X$ for $C(X) \cong \mathbb{R}$ is that the Tychonoff completion $T(X)$ be a one point space. – Pete L. Clark Mar 08 '15 at 04:24
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@PeteL.Clark: Funny, googling "Tychonoff completion" yields your commutative algebra notes as the first result. Shameless advertising? ;-) I thought that it is sufficiently obvious that as soon as I spell it out, anyone who cares should be able to see it immediately. Of course, these things can be misleading (it's easy to see why something is true post hoc), so I will be adding it. :) I erased the first version because I wasn't too sure about the algebra, I've just woken up. :) – tomasz Mar 08 '15 at 04:44
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Well, what is much more shameless is that I would like to add this as an exercise to my commutative algebra notes. May I? I will attribute it to you. – Pete L. Clark Mar 08 '15 at 04:49
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$C(X,\Bbb{R})$ is not an integral domain for any perfectly normal Hausdorff space $X$, in particular, it is not an integral domain for any metric space $X$. I didn't know about any iff condition
Even for the Tychonoff corkscrew you find the existence of two points not separable by a real-valued function. But this is not true for all pairs of points! To prove nondomainness, it is enough to find ONE pair of non real-valued-function-separable points.
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