# Thread: Proving the Zero Set of a Continuous Function is Closed

1. ## 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 $\displaystyle p \in X$ 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, $\displaystyle f = \{-x-2, x \in (-\infty, -2] | 0, x \in (-2,2) | x-2, x \in [2, \infty)\}$. 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?

2. 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 $\displaystyle p \in X$ such that f(p)=0. Prove that Z(f) is closed.
I would show that $\displaystyle X\backslash Z(f)$ is an open set.
If $\displaystyle f(a) \ne 0$ then $\displaystyle \left( {\exists \varepsilon > 0} \right)\left[ {0 \notin \left( {f(a) - \varepsilon ,f(a) + \varepsilon } \right)} \right]$.
Because $\displaystyle f$ is continuous an inverse image of an open sets is an open set.
Can you finish?

3. Thanks for the hint, but I eventually solved it a different way, by showing that if any sequence $\displaystyle \{x_n\} \in Z(f)$ converges to $\displaystyle x$, then $\displaystyle x \in Z(f)$. (Thus $\displaystyle Z(f)$ is closed.)

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# for a continuous function the set of all zeros is a closed set??

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