Is the zero set of an analytic function closed nowhere dense?

Question

Given a non-constant analytic function $f(x)$ on a domain $D \subseteq \mathbb{R}^n$. I want to prove that

$\mathcal{Z} = \{x \in \mathcal{D} |f(x) = 0 \}$

is closed nowhere dense.

I originally wanted to prove this by contradiction as follows using the identity theorem.

"Assume that $\mathcal{Z}$ is not a closed nowhere dense set, then $\mathcal{Z}$ must contain a limit point", which shows that $f(x)$ is constant zero by the identity theorem and thus contradicts that $f(x)$ is non-constant.

However, I am not sure if the following statement is true:

"Assume that $\mathcal{Z}$ is not a closed nowhere dense set, then $\mathcal{Z}$ must contain a limit point"

1. If it is true, could you provide one reference?

2. If not, is $\mathcal{Z}$ closed nowhere dense, and how can I prove it?

Answer 1

Indeed, more is true. It is enough to assume that $\mathcal Z$ is not nowehere dense to show that it has a limit point.

To see it, recall that a set is nowhere dense, by definition, if and only if its closure has empty interior. Then if $\mathcal Z$ is not nowhere dense, its closure has nonempty interior. Any of these points is a limit point of $\mathcal Z$.

EDIT: As Conrad points out in reply to Shi James' comment, to complete the proof for the given $\mathcal Z$, which is a zero set of an analytic (hence continuous) function, we need to notice first that $\mathcal Z$ is necessarily closed.

Answer 2

We know $Z$is closed. So we want to show $Z$ is nowhere dense. Assume, to reach a contradiction, that the interior of $Z,$ denoted by $\text {int } Z,$ is nonempty.

Let $x\in \text {int } Z.$ Let $y\in D.$ Because $D$ is open and connected, $D$ is path connected. So there is a continuous path $\gamma:[0,1]\to D$ with $\gamma(0)=x, \gamma(1)=y.$

Let $E$ be the set of $t\in [0,1]$ such that $f=0$ in an open ball $B_{\gamma(t)}$ centered at $\gamma(t).$ Clearly $0\in E.$ By continuity of $\gamma,$ $\gamma^{-1}(B_{\gamma(t)})$ is an open neighborhood of $t$ in $[0,1].$ And if $s$ is in this neighborhood, $\gamma(s)\in B_{\gamma(t)},$ hence there is an open ball centered at $\gamma(s)$ where $f=0.$ It follows that $E$ is a nonempty open subset of $FE.$

Furthermore, since $f$ is real analytic in $D,$ $E$ is precisely the set of $t\in [0,1]$ where all derivatives of $f$ vanish at $\gamma(t).$ Thus $E$ is closed in $[0,1].$ It follows that $E$ is nonempty and clopen in $[0,1].$ Since $[0,1]$ is connected, $E=[0,1].$

This implies $f(y)=0.$ Since $y\in D$ was arbitrary, we have $f=0$ in $D.$ This is a contradiction, since $f$ was assumed to be nonconstant in $D.$ Therefore $\text {int } Z$ is empty, and $Z$ is nowhere dense as desired.