[proofplan] We prove the contrapositive local-to-global mechanism behind the theorem. First, a local power-series argument shows that near any point of an [analytic function](/page/Analytic%20Function), either the function vanishes on a whole neighbourhood or its zero at that point is isolated. If a zero is not isolated, this local dichotomy forces the function to vanish on a neighbourhood of that zero. We then propagate this local vanishing across the connected set $U$ by showing that the set of points near which $f$ vanishes is both open and closed in $U$. [/proofplan]

[step:Establish the local dichotomy for zeros of a power series]Let $b \in U$. Since $U$ is open, choose $\rho>0$ such that $B(b,\rho) \subset U$. Since $f:U\to\mathbb{C}$ is analytic, there are coefficients $(c_n)_{n \ge 0}$ in $\mathbb{C}$ such that \begin{align*} f(z)=\sum_{n=0}^{\infty} c_n(z-b)^n \end{align*} for every $z \in B(b,\rho)$. If $c_n=0$ for every $n \ge 0$, then $f(z)=0$ for every $z \in B(b,\rho)$. Otherwise, let $m \ge 0$ be the least index such that $c_m \neq 0$. For $z \in B(b,\rho)$, define \begin{align*} g(z)=\sum_{k=0}^{\infty} c_{m+k}(z-b)^k. \end{align*} This defines a continuous map $g:B(b,\rho)\to\mathbb{C}$, because it is represented by a [power series](/page/Power%20Series) on $B(b,\rho)$. Moreover $g(b)=c_m \neq 0$, and \begin{align*} f(z)=(z-b)^m g(z) \end{align*} for every $z \in B(b,\rho)$. By continuity of $g$ at $b$, there exists $\delta \in (0,\rho)$ such that $g(z)\neq 0$ for every $z \in B(b,\delta)$. Hence, if $m=0$, then $f(z)\neq 0$ for every $z \in B(b,\delta)$; if $m\ge 1$, then \begin{align*} f(z)\neq 0 \quad \text{for every } z \in B(b,\delta)\setminus\{b\}. \end{align*} Thus, at each point $b\in U$, either $f$ vanishes on some neighbourhood of $b$, or the possible zero of $f$ at $b$ is isolated.[/step]

[guided]Fix a point $b \in U$. The point of using analyticity is that it turns the local behaviour of $f$ into a power-series question. Since $U$ is open, we may choose $\rho>0$ with $B(b,\rho)\subset U$. By the definition of analyticity, there are complex coefficients $(c_n)_{n\ge 0}$ such that \begin{align*} f(z)=\sum_{n=0}^{\infty} c_n(z-b)^n \end{align*} for every $z \in B(b,\rho)$. There are two possibilities. If every coefficient $c_n$ is zero, then the series is identically zero on $B(b,\rho)$, so $f$ vanishes on that whole neighbourhood. Now suppose not every coefficient is zero. By the [well-ordering principle](/theorems/721) applied to the nonempty set $\{n\ge 0:c_n\neq 0\}$, there is a least index $m\ge 0$ such that $c_m\neq 0$. We factor the first nonzero power out of the series. Define \begin{align*} g(z)=\sum_{k=0}^{\infty} c_{m+k}(z-b)^k \end{align*} for $z\in B(b,\rho)$. This is a continuous map $g:B(b,\rho)\to\mathbb{C}$, since it is represented by a power series on that ball. The value at the centre is \begin{align*} g(b)=c_m\neq 0. \end{align*} The original series factors as \begin{align*} f(z)=(z-b)^m g(z) \end{align*} for every $z\in B(b,\rho)$. Because $g(b)\neq 0$ and $g$ is continuous at $b$, there exists $\delta\in(0,\rho)$ such that $g(z)\neq 0$ whenever $z\in B(b,\delta)$. Therefore the only possible zero of $f$ in $B(b,\delta)$ comes from the factor $(z-b)^m$. If $m=0$, even that factor is absent and $f$ has no zeros in $B(b,\delta)$. If $m\ge 1$, then the factor $(z-b)^m$ vanishes only at $z=b$, so \begin{align*} f(z)\neq 0 \quad \text{for every } z\in B(b,\delta)\setminus\{b\}. \end{align*} This proves the local dichotomy: near any point, an analytic function is either identically zero or has at most one zero, namely the centre point itself.[/guided]

[step:Convert a non-isolated zero into local vanishing] Let $a \in U$ satisfy $f(a)=0$. Suppose, for contradiction, that $a$ is not isolated as a zero of $f$ relative to $U$. Since $U$ is open, choose $\rho_a>0$ such that $B(a,\rho_a)\subset U$. The assumption that $a$ is not isolated means that for every $r\in(0,\rho_a)$ there exists $z_r\in B(a,r)\setminus\{a\}$ such that $f(z_r)=0$. Apply the local dichotomy from the previous step at the point $a$. The alternative in which the zero at $a$ is isolated is incompatible with the existence of such points $z_r$ for every sufficiently small $r>0$. Therefore there exists $\varepsilon>0$ such that $B(a,\varepsilon)\subset U$ and \begin{align*} f(z)=0 \quad \text{for every } z\in B(a,\varepsilon). \end{align*} [/step]

[step:Propagate local vanishing through the connected domain] Define the local vanishing set \begin{align*} V=\{w\in U:\text{ there exists } r_w>0 \text{ such that } B(w,r_w)\subset U \text{ and } f(z)=0 \text{ for every } z\in B(w,r_w)\}. \end{align*} By the previous step, $a\in V$, so $V$ is nonempty. The definition immediately implies that $V$ is open in the relative topology of $U$. We prove that $U\setminus V$ is also open in $U$. Let $b\in U\setminus V$. Apply the local dichotomy at $b$. Since $b\notin V$, the alternative that $f$ vanishes on a neighbourhood of $b$ is impossible. Hence there exists $\delta_b>0$ such that $B(b,\delta_b)\subset U$ and, if $f(b)=0$, then \begin{align*} f(z)\neq 0 \quad \text{for every } z\in B(b,\delta_b)\setminus\{b\}; \end{align*} while if $f(b)\neq 0$, possibly shrinking $\delta_b$ by continuity of $f$ gives \begin{align*} f(z)\neq 0 \quad \text{for every } z\in B(b,\delta_b). \end{align*} In either case no point of $B(b,\delta_b)$ belongs to $V$, because every point of $V$ is a zero of $f$. Thus $B(b,\delta_b)\subset U\setminus V$, so $U\setminus V$ is open in $U$. Therefore $V$ is both open and closed in the connected set $U$. Since $V$ is nonempty, connectedness of $U$ gives $V=U$. Hence $f$ vanishes on a neighbourhood of every point of $U$, and in particular \begin{align*} f(z)=0 \quad \text{for every } z\in U. \end{align*} This says that $f$ is identically zero on $U$, contradicting the hypothesis. [/step]

[step:Conclude that every zero is isolated] The contradiction shows that no zero $a\in U$ of $f$ can fail to be isolated relative to $U$. Hence for every $a\in U$ with $f(a)=0$, there exists $r>0$ such that $B(a,r)\subset U$ and \begin{align*} f(z)\neq 0 \quad \text{for every } z\in B(a,r)\setminus\{a\}. \end{align*} Thus every zero of $f$ is isolated in $U$. [/step]