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Let f be a continuous real function on a metric space X. Let

  1. Nov 25, 2011 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    Definition of continuity on a metric space.

    3. The attempt at a solution

    Proof. We'll show that X/Z(f) = {p in X s.t. f(p) ≠ 0} is open. Choose p in X/Z(f). Since f is continuous, for every ε > 0 there exists a ∂ > 0 such that d(f(x),f(p)) < ε whenever d(x, p) < ∂.


    ..... Unfortunately, I suffered a brain hemorrhage before I could finish this. I think I was trying to show that p is an interior point of X/Z(f). Thoughts?
     
  2. jcsd
  3. Nov 25, 2011 #2

    Dick

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    I'd look into getting the brain hemorrhage treated. What might be confusing you is that you are using 'd' for the metric on R as well as on X. How about writing the metric on R as |f(x)-f(p)|<epsilon. Now pick epsilon=|f(p)|/2. Can f(x) be zero?
     
  4. Nov 25, 2011 #3

    Bacle2

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    Another approach: if f is continuous the inverse image of an open set is open, and the inverse image of a closed set is ....., and the subset {0} of the real numbers is ..... in the reals.
     
  5. Nov 28, 2011 #4
    Hold on a sec ...... Could you take a quick look at the solution on this UCLA Math website? http://www.math.ucla.edu/~elewis/Math230PDFs/Math%20230b%20HW1.pdf It's only 3 lines long and seems sketchy. How do we know that {0} is closed in our space X? If X = ℝ, then it would work out; but we're not given that.
     
  6. Nov 28, 2011 #5

    Dick

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    It does not say {0} is closed in X. It's says {0} is closed in R. Therefore f^(-1)({0}) is closed in X.
     
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