Zeros of real analytic function 
Let $-\infty \le a<b\le \infty$ and $f:(a,b)\to \Bbb{R}$ be real analytic. Show that the set $\{x\in (a,b) : f(x)=0\}$ has no limit point in $(a,b)$.

One way I could think is to show that for non zero $f$ , the set is both open and closed that will imply that set is empty and thus has no limit point. But openness is not coming easily.

Let $Z = f^{-1} (\{0\})$, which is closed since $f$ is continuous. Hence
$Z$ contains any limit points of $Z$.
Suppose $\hat{x} \in Z$. Since $f$ is analytic, it has a power series representation on some open ball $B$ containing $x$. That is,
we have $f(x) = \sum_k a_k (x-\hat{x})^k$. If $a_k=0$ for all $k$ then $f$ is zero on $B$, otherwise there is a first $p$ such that $a_p \neq 0$,
and $f(x) = (x-\hat{x})^p \sum_{k \ge p} a_k (x-\hat{x})^{k-p} $, and so it follows that there is some open ball $B'$ such that $B' \cap Z = \{\hat{x}\}$.
Suppose $f$ is zero on $B$. Let $\bar{x} = \sup \{ x | [\hat{x},x] \subset Z \}$.
If $\bar{x} < b$, it follows from the previous paragraph that
that $f$ is zero on some open set containing $\bar{x}$,
which contradicts the definition of $\bar{x}$, hence $\bar{x} = \infty$.
A similar analysis for $x < \hat{x}$ shows that we have $Z = (a,b)$.
Hence if $f\neq 0$, then $Z$ is isolated. It follows that $Z$ has no limit
points.
