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Math Help - Proving the Zero Set of a Continuous Function is Closed

  1. #1
    Super Member redsoxfan325's Avatar
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    Proving the Zero Set of a Continuous Function is Closed

    Question: Let f be a continuous real function on a metric space X. Let Z(f) be the set of all p \in X such that f(p)=0. Prove that Z(f) is closed.

    My Thoughts: This statement is obvious if all the zeroes f are disjoint. Z has no limit points, and thus contains them all, so Z is closed. I don't understand how to prove that Z is closed if Z contains an entire interval's worth of zeroes though.

    Like, for instance, f = \{-x-2, x \in (-\infty, -2] | 0, x \in (-2,2) | x-2, x \in [2, \infty)\}. This function I believe is continuous but has zeroes on the entire interval [-2,2]. How would I prove a function similar to this has a closed Z(f).

    Are there any other cases I'm forgetting to consider?
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  2. #2
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    Quote Originally Posted by redsoxfan325 View Post
    Question: Let f be a continuous real function on a metric space X. Let Z(f) be the set of all p \in X such that f(p)=0. Prove that Z(f) is closed.
    I would show that X\backslash Z(f) is an open set.
    If f(a) \ne 0 then \left( {\exists \varepsilon  > 0} \right)\left[ {0 \notin \left( {f(a) - \varepsilon ,f(a) + \varepsilon } \right)} \right].
    Because f is continuous an inverse image of an open sets is an open set.
    Can you finish?
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  3. #3
    Super Member redsoxfan325's Avatar
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    Thanks for the hint, but I eventually solved it a different way, by showing that if any sequence \{x_n\} \in Z(f) converges to x, then x \in Z(f). (Thus Z(f) is closed.)
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