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Math Help - Zero set

  1. #1
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    Zero set

    Hi,
    How can I show that:
    a) any zero set is closed
    b) not every closed set is zero set?
    Thanks for any help.
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  2. #2
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    Quote Originally Posted by Arczi1984 View Post
    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?
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  3. #3
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    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) .
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  4. #4
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    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
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  5. #5
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    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.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Arczi1984 View Post
    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.
    Last edited by Drexel28; March 1st 2010 at 01:13 PM.
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