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

  • Thread starter Jamin2112
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Homework Statement



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.

Homework Equations



Definition of continuity on a metric space.

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?
 

Answers and Replies

  • #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?
 
  • #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.
 
  • #4
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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?
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.
 
  • #5
Dick
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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.
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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