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.


1 Answer 1


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 $f(\lim_n x_n) = 0$.


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