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

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Homework Help Overview

The discussion revolves around the properties of a continuous real function defined on a metric space X, specifically focusing on the set Z(f) where the function equals zero. The original poster attempts to prove that Z(f) is closed by showing that its complement is open, but expresses confusion in their approach.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the continuity of the function and the implications for the set Z(f). There are suggestions to clarify the use of metrics and to consider the properties of inverse images of sets under continuous functions. Questions are raised about the closure of the set {0} in the context of the metric space X.

Discussion Status

The discussion is active, with participants providing guidance on clarifying the metric definitions and exploring the implications of continuity. There is an ongoing examination of the assumptions regarding the closure of sets in different spaces, and multiple interpretations of the problem are being explored.

Contextual Notes

There are concerns about the completeness of the original statement and the definitions being used, particularly regarding the nature of the metric space X and the properties of the real numbers. The original poster's attempt is noted to be incomplete, and there is a reference to external resources that may not align with the problem's requirements.

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?
 
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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?
 
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.
 
Dick said:
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.
 
Jamin2112 said:
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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