Showing $f:[0,1] \rightarrow \mathbb{R}$ has finitely many isolated zeros

Question

We say that $x_{0} \in \mathbb{R}$ is an isolated zero of $\mathbb{R}$-valued function $f$ $f(x_{0}) = 0$ and there exists an open set containing $x_{0}$ that contains no other zeros of $f$. Show that if $f : [0,1] \rightarrow \mathbb{R}$ is continuous and all of its zeros are isolated, then it only has finitely many isolated zeros.

Now I think I am on the right track here, but I need some help finishing this off. Let $\epsilon > 0$ and set $K : = [- \epsilon, \epsilon]$. Then $f^{-1}(K)$ contains all of the zeros of $f$ by construction and is closed since $f$ is continuous and $K$ is closed. Moreover, since $[0,1]$ is compact and $f^{-1}(K)$ is closed, we know that $f^{-1}(K)$ is compact.

From here, I think that I should say that if there were infinitely many isolated zeros then we could cook up an open cover without a finite subcover thereby contradicting compactness. To this end, let $\{ x_{\alpha} \}_{\alpha \in A}$ be the zeros of $f$ (which are all isolated by assumption) and let $\{U_{\alpha} \}_{\alpha \in A}$ be a corresponding collection of open sets such that $x_{\alpha} \in U_{\alpha}$ and $x_{\beta} \notin U_{\alpha}$ for $\alpha \neq \beta$, where $A$ is some infinite index set. If we let $U \subseteq [0,1]$ be an open set such that $f^{-1}(K) \setminus \bigcup_{\alpha \in A} U_{\alpha} \subseteq U$ and $U \cap \bigcup_{\alpha \in A} U_{\alpha} = \emptyset$. Then $U \cup \bigcup_{\alpha \in A} U_{\alpha}$ constitutes an open cover $f^{-1}(K)$ and thus we may refine it to a finite subcover. But any finite subcover will not contain infinitely many $U_{\alpha}$ and $x_{\alpha} \notin U_{\beta}$ for $\alpha \neq \beta$ and $x_{\alpha} \notin U$ by construction, which yields a contradiction.

Is this correct? Also is there a more direct way of showing this because I definitely feel like my argument could be improved upon.

Answer 1

I didn't find any error, but there is a simpler approach. Let $A$ be the set of non-zeros of $f$. For each zero $z$ of $f$, let $U_z$ be a neighborhood of $z$ with no other zero. So, if $Z$ is the set of zeros of $f$, $A$ together with the $U_z$'s form an open cover of $[0,1]$ and each element of this open cover has a zero, at most. So, if $Z$ was infinite, there would be no finite subcover.

Answer 2

If the set of zeros was not finite, then there would be some sequence of distinct zeros $x_n$ such that $x_n \to x$. By continuity, $f(x) = 0$, which contradicts the assumption that the zeros are isolated.

Answer 3

If the set of zeroes say $A$ is infinite then being bounded $A$ has a limit point in $[0,1]$ say $c$. Now we must have $f(c) =0$ and clearly $c$ would not be an isolated zero. To see why $f(c) =0$ just assume $f(c) \neq 0$ and by continuity there is a neighborhood of $c$ where $f$ is non-zero. Then the is neighborhood does not contain any point of $A$ contradicting that $c$ is a limit point of $A$.