# For a continuous function $f:\Bbb R\to\Bbb R,$ let $Z(f)=\{x\in\Bbb R: f(x) = 0\}.$ Prove that $Z(f)$ is always closed.

For a continuous function $$f:\Bbb R\to\Bbb R,$$ let $$Z(f)=\{x\in\Bbb R: f(x) = 0\}.$$ Prove that $$Z(f)$$ is always closed.

I considered three cases.

The first case is when $$Z(f)$$ is empty, and then the claim is vacuously true.

The second case, is when $$Z(f)$$ has a finite number of elements and then also the claim is valid.

The third case, is when $$Z(f)$$ has an infinite number of elements. This is the part, where I am stuck. I did not have any idea on how to show that $$Z(f)$$ is closed.

Then searching through some books, I found a theorem as follows:

A function $$f:\Bbb R\to\Bbb R$$ is continuous on $$\Bbb R$$ if and only if for any $$F\subseteq \Bbb R$$ we have, $$f^{-1}(F)$$ is closed in whenever $$F$$ is closed in $$\Bbb R.$$

In this case, $$F=\{0\}\subseteq \Bbb R$$ is closed and hence, $$f^{-1}(F)=f^{-1}(\{0\})=Z(f)$$ is closed in $$\Bbb R$$ since, $$f$$ is continuous on $$\Bbb R.$$

Is my solution valid? I am having some doubts over it.

Yes, your solution is valid. In many intro analysis texts, it is standard to define $$f$$ as being a continuous map iff $$f^{-1}(U)$$ is open for every open set $$U$$ (I'll leave it to you to verify that this definition is equivalent to the ordinary $$\epsilon$$/$$\delta$$ definition). As a direct consequence, if $$F$$ is any closed set $$f^{-1}(F)$$ is closed.
Here is a hint for your third case, when $$Z(f)$$ has infinitely many elements: show that for any converging sequence $$\{x_n\}$$ of elements of $$Z(f)$$, then $$f(\lim_n x_n) = 0$$.