# Ideals of $C(X)$ with only finitely many number of zerosets

We denote the rings of all real valued continuous functions on compeletely regular Hausdorff space $X$ by $C(X)$.Let $I$ be a ring ideal of $C(X)$. define $$Z[I]:=\lbrace Z(f):\;f\in I\rbrace$$ where $Z(f):=\lbrace x \in X:\;f(x)=0\rbrace$. We are interested in knowing about the relation about algebraic properties of the ring $C(X)$ and topological properties of the space $X$. I have a question about the relation between the number of zero sets of $I$ in the finite case and it's relation with the topological space $X$. Let me pose my questions in the definite way.

Question1: for which positive integer $n$ we can have a ring ideal $I$, that $Z(I)$ contains only n elements?

Question2: If we have an ideal $I$, such that $|Z(I)|=n$ for some positive integer $n$, can we characterize the ideal $I$?

Question3: for which condition on the space $X$, we have an ideal $I$, with $|Z(I)|=n$, for some positive integer $n$?

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I've noticed you are asking a lot of questions and they are not getting much attention. There are several reasons. First, you should define $Z(f)$ in each one, since this is not standard notation. Also, you should use dollar signs to make the mathematics appear as mathematics. It wouldn't hurt to give more background, like what you have tried on the problem and why you care about the answer. Have you read the "How to Ask" page yet? Have you thought about asking on math.stackexchange? Spending time there for a while learning how to ask would not hurt you. –  David White Apr 30 '12 at 12:26
$Z(f)$ is standard notation, it is the zero set of $f$. As for the question: Have you considered principal ideals and nice spaces, say manifolds? –  Martin Brandenburg Apr 30 '12 at 18:11
Is actually $Z[I]$ equal to the intersection of all the sets $Z[f]$ with $f\in I$? –  Matthew Daws May 1 '12 at 20:38
Dear Matthew $Z(I):=${$Z(f):f\inC(X)$} i.e. every element of $Z(I)$ is a set of the form $Z(f)$ that $f$ belongs to the ideal $I$. Thank you for your comment –  Ali Reza May 1 '12 at 20:56
Shouldn't it be $Z(I)=\lbrace Z(f): f \in I\rbrace$ (instead of $f \in C(X)$) ? –  Ralph May 2 '12 at 0:22
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Edit: In the first version I had the additional assumption that each singleton in $X$ is a zero set. But as observed by AliReza Olfati the proof can be adapted to work in the general case. Therefore we have:

$Z(I)$ is finite if and only if the following holds:

1. There is $A \subseteq X$ closed and $x_1,...,x_m \in X\setminus A$ with $X=A\cup \lbrace x_1,...,x_m\rbrace$.
2. $Z(I) = \lbrace A \cup T \mid T \subseteq \lbrace x_1,...,x_m\rbrace\;\rbrace$. In particular $|Z(I)|=2^m$ is a power of $2$.
3. $I=(1-\delta_{A \cup T} \mid T \subseteq \lbrace x_1,...,x_m\rbrace)$

Proof: $(\Rightarrow)$ Let $Z(I) = \lbrace Z(f_1),...,Z(f_n)\rbrace$.

First note that $Z(f^2 + g^2) = Z(f) \cap Z(g)$. Hence $Z(I)$ is closed under $\cap$ and we have $$A := Z(f_1) \cap ... \cap Z(f_n) \in Z(I).$$

Let $A=Z(f_0)$.

Next let's show that $X \setminus A$ is finite. Suppose $X \setminus A$ is infinite. Since $X$ is Hausdorff and $A$ is closed, then there is a sequence of closed subsets $A \cup \lbrace x_1,...,x_k\rbrace$. Hence it's enough to show that there are only finitely many closed subsets $C \supseteq A$ in $X$. Since $X$ is completely regular, such a $C$ is the intersection of zero sets (Gillman-Jerison: Rings of continuous functions, Theorem 3.2). Thus we have a surjection $$\lbrace Z \subseteq X \mid Z \supseteq A \text{ zero set }\;\rbrace\to \lbrace C \subseteq X \mid C \supseteq A \text{ closed }\rbrace,\; \mathfrak{Z} \mapsto \bigcap_{Z \in \mathfrak{Z}}Z.$$ Let $Z=Z(h) \supseteq A$ be a zero set. Then $f_0h \in I$ with $Z(f_0h)=Z(f_0) \cup Z(h) = Z$, i.e. $Z \in Z(I)$. Hence the LHS of the map is just the finite set $Z(I)$ and consequently its image is also finite. Hence the finiteness of $X \setminus A$ and 1) are shown.

Let $X \setminus A = \lbrace x_1,...,x_m\rbrace$. Since $A$ is closed and $X$ is Hausdorff it follows that $A$, $\lbrace x_i\rbrace$ are both, open and closed. Hence a function $f: X \to \mathbb R$ is continuous iff $f|A$ is continuous. In particular we find continuous $h$ with $Z(h)=\lbrace x_i\rbrace$.

To see 2) let $f \in I$. By definition $A \subseteq Z(f)$. Thus $T := Z(f) \setminus A \subseteq \lbrace x_1,...,x_m\rbrace$ and $Z(f)=A \cup T$ is in the RHS. Conversely, if $T = \lbrace x_{p_1},...,x_{p_k}\rbrace$ then choose $h_j \in C(X)$ with $Z(h_j) = \lbrace x_{p_j}\rbrace$. Now $f := f_0h_1\cdots h_p \in I$ and $Z(f) = A \cup T \in Z(I)$.

3) Note that the functions used in the following are continuous by the remark preceding the proof of 2). Let $f \in I$ and let $X \setminus Z(f) = \lbrace x_{p_1},...,x_{p_k}\rbrace$. Then $f= \sum_j f(x_{p_j})(1-\delta_{X \setminus \lbrace x_{p_j} \rbrace})$ is contained in the RHS. Conversely, let $T \subseteq \lbrace x_1,...,x_m\rbrace$ be given. By 2) there is $f \in I$ with $Z(f) = A \cup T$. Define a continuous map $h$ by $h|Z(f) = 0$, $h(x) = 1/f(x)$ if $x \notin Z(f)$. Then $1-\delta_{A \cup T} = fh \in I$.

$(\Leftarrow)$ If $X$ is given by 1) and $I$ by 3) then $Z(I) = \lbrace A \cup T \mid T \subseteq \lbrace x_1,...,x_m\rbrace\;\rbrace$ follows easily. q.e.d.

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Thank you very much dear Ralph. you gave an interesting proof.Let me congratulate you for guessing the complete form of my Question. –  Ali Reza May 5 '12 at 8:10
But When I read your proof I found that you had a small wrong assumption in your interestisng proof. But I think this assumption can be modified. you assumed that if $x_1∈X∖A$ then there is a function $h∈C(X)$ with the property that $Z(h)={x_1}$. as you Know, if $x_1$ is not a $G_\delta$-point, we could not have this function. please modulate your assumption.thank you very much dear friend. –  Ali Reza May 5 '12 at 8:30
Have a look at the first sentence, where $X$ is supposed to have exactly this property. This surely doesn't cover all completely regular spaces, but a large class of spaces, including metric spaces (I've started a separate thread math.stackexchange.com/questions/141291/… to investigate these spaces). Also note that $(\Leftarrow)$ works for any Hausdorff space. –  Ralph May 5 '12 at 9:03
@Ralph Let me complete your proof.If we have A completely regular space $X$ with this property and $A$ defined as above in your proof, we can prove that $|X-A|<\infty$. this implies that every point of $X-A$ is a $G_\delta$ point.notice that every point of $X-A$ in this case is a zero-set. then we can resume the details of your proof. –  Ali Reza May 5 '12 at 19:18
Looks interesting, but there are two questions that came into my mind: 1) I don't see how finiteness of $X-A$ follows from the def. of complete regularity. Can you give some detail of your argument. 2) If $X-A$ is finite, don't you need to know that singletons are closed in order to have them as $G_\delta$'s ? (if so, one can assume $X$ to be Hausdorff - that's no great restriction and commonly used in conjunction with complete regularity). –  Ralph May 5 '12 at 20:32
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