Proving the Zero Set of a Continuous Function is Closed

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March 2nd 2009, 01:09 PM
redsoxfan325

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?
March 2nd 2009, 02:11 PM
Plato

Quote:

Originally Posted by redsoxfan325

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?
March 2nd 2009, 08:31 PM
redsoxfan325

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.)