Existence of Zero Divisors in $C(X,\mathbb{R})$ 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?
 A: 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.
A: $C(X,\Bbb{R})$ 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
A: 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.
