Zero set

**Arczi1984** · March 1st 2010, 10:05 AM

Hi,
How can I show that:
a) any zero set is closed
b) not every closed set is zero set?
Thanks for any help.

**Plato** · March 1st 2010, 10:09 AM

Originally Posted by Arczi1984

How can I show that:
a) any zero set is closed b) not every closed set is zero set?

What do you mean by a zero set?

**Arczi1984** · March 1st 2010, 10:20 AM

Here is definition which I used:
"Let X be a topological space and $f \in C(X)$ , the ring of continuous function on X . The level set of f is defined at $r \in R$ is the set $f^{-1}:=\{x\in X : f(x)=r\}$ . The zero set of f is defined to be the level set of f at 0 . The zero set of f is denoted by Z(f) . A subset A of X is called a zero set of X if A=Z(f) for some $f \in C(X)$ .

**facenian** · March 1st 2010, 10:47 AM

Let $x \notin Z(f)$ then $f(x) \neq 0$ then there exists $\epsilon>0$ such that $0 \notin (f(x)-\epsilon,f(x)+\epsilon)$ and because of continuity there exist a neibourhood U_x of x such that $f(U_x)\subset (f(x)-\epsilon,f(x)+\epsilon)$ ,ie,Z(f) is closed because its complement is open

**Arczi1984** · March 1st 2010, 11:09 AM

Thanks.
I tried to solve the second part. I found some example but I do not understand the last part. Here is this example:
Let us consider X=Z and $\tau=\{G \subset \mathbf{Z}:0 \in G \}$ . Clearly $F=\mathbf{Z}^{\ast}=\mathbf{Z}\setminus 0$ is a closed subset but it is not a zero-set "because $C(X)=C_0(X)$ ".
This "" part in not clear for me.

**Drexel28** · March 1st 2010, 12:38 PM

Originally Posted by Arczi1984

Hi,
How can I show that:
a) any zero set is closed
b) not every closed set is zero set?
Thanks for any help.

How about this. If $f:X\mapsto\mathbb{R}$ is continuous (where $X$ is a metric space) we define the zero set $\mathcal{Z}(f)=\left\{x\in X:f(x)=0\right\}$ . Why is $\mathcal{Z}(f)$ closed? Because, $\{0\}$ is closed in $\mathbb{R}$ under the usual topology and since $f$ is continuous so is $f^{-1}(0)=\mathcal{Z}(f)$

Alternatively, let $x\in\overline{\mathcal{Z}(f)}$ . If $x\in\mathcal{Z}(f)$ we're done, so assume not. Then, $x$ is a limit point of $\mathcal{Z}(f)$ and so there exists a sequence of points $\{z_n\}_{n\in\mathbb{N}}\subseteq\mathcal{Z}(f)$ such that $z_n\to x$ . But, since $f$ is continuous we know that $f(x)=\lim\text{ }f(z_n)=\lim\text{ }0=0$ and so $x\in\mathcal{Z}(f)$ . It follows that $\mathcal{Z}(f)=\overline{\mathcal{Z}(f)}$ which finishes the argument.

What about $f:\mathbb{R}\mapsto\mathbb{R}$ which is the identity function. $\{3\}$ is closed in the domain but it's not the zero set.

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