Corollary 9.5.3F7B8B4ADF874

Proof. Let $X_1$ be a branch of $(X,x)$. If $X_1$ has dimension zero, it is the point $x$. Otherwise, suppose that $f-f(x)$ does not vanish identically on $X_1$, and apply proposition 9.5.2 to $X_1$ and its subgerm $X_1\cap \{ f=f(x)\} $. The holomorphic function $f\circ \gamma $ on the disc has a local maximum of its modulus at $0$, so it is constant by the maximum principle in one variable, and $f(\gamma (t))=f(x)$ for all $t$, contradicting the choice of $\gamma $.

For the second assertion, let $K$ be a compact analytic subset of an open set in $\mathbb C^ N$, and let $K_+$ be the set of its non-isolated points. The isolated points form an open subset of $K$, so $K_+$ is compact. Suppose that $K_+\neq \varnothing $, and let $y$ be a point of $K_+$ where $|z|^2$ attains its maximum on $K_+$. Since $y$ is not isolated, some branch of $(K,y)$ has positive dimension, and its small representatives, being pure-dimensional, consist of non-isolated points of $K$. So proposition 9.5.2, applied to this branch and the subgerm $\{ y\} $, gives a nonconstant holomorphic curve $\gamma $ in $K_+$ with $\gamma (0)=y$. The function $|\gamma (t)|^2$ is subharmonic, with Laplacian $4\sum _ j|\gamma _ j'(t)|^2$, and it has a local maximum at $0$; by the maximum principle (proposition 3.4.7) it is constant near $0$, so its Laplacian vanishes, all $\gamma _ j'$ vanish, and $\gamma $ is constant, a contradiction. Hence $K_+=\varnothing $, and $K$ is compact and discrete, hence finite. $\square$