Hi,

How can I show that:

a) any zero set is closed

b) not every closed set is zero set?

Thanks for any help.

Results 1 to 6 of 6

- Mar 1st 2010, 10:05 AM #1

- Joined
- Oct 2009
- Posts
- 70

- Mar 1st 2010, 10:09 AM #2

- Mar 1st 2010, 10:20 AM #3

- Joined
- Oct 2009
- Posts
- 70

Here is definition which I used:

"Let X be a topological space and $\displaystyle f \in C(X)$, the ring of continuous function on X . The level set of f is defined at $\displaystyle r \in R$ is the set $\displaystyle f^{-1}:=\{x\in X : f(x)=r\}$. 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 $\displaystyle f \in C(X)$ .

- Mar 1st 2010, 10:47 AM #4

- Joined
- Feb 2010
- Posts
- 37

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

- Mar 1st 2010, 11:09 AM #5

- Joined
- Oct 2009
- Posts
- 70

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 $\displaystyle \tau=\{G \subset \mathbf{Z}:0 \in G \}$. Clearly $\displaystyle F=\mathbf{Z}^{\ast}=\mathbf{Z}\setminus 0$ is a closed subset but it is not a zero-set "because $\displaystyle C(X)=C_0(X)$".

This "" part in not clear for me.

- Mar 1st 2010, 12:38 PM #6
How about this. If $\displaystyle f:X\mapsto\mathbb{R}$ is continuous (where $\displaystyle X$ is a metric space) we define the zero set $\displaystyle \mathcal{Z}(f)=\left\{x\in X:f(x)=0\right\}$. Why is $\displaystyle \mathcal{Z}(f)$ closed? Because, $\displaystyle \{0\}$ is closed in $\displaystyle \mathbb{R}$ under the usual topology and since $\displaystyle f$ is continuous so is $\displaystyle f^{-1}(0)=\mathcal{Z}(f)$

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

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