Remmert–Stein theorem

Remarks

The question is local, so one may assume that $X$ is a neighbourhood of a point of $E$ in $\C^n$. The proof is by induction on the dimension of $E$ and uses projections. After a generic linear change of coordinates, the projection of $\overline{A}$ to $\C^k$ is proper near the point. Then $\overline{A}$ is described by the canonical defining polynomials of this branched covering. Their coefficients are bounded holomorphic functions outside a thin set, so they extend by the Riemann extension theorem. Example: $A = \set{z_2 = 0} \setminus \set{0}$ is an analytic subset of $\C^2 \setminus \set{0}$ of pure dimension $1$, and $E = \set{0}$ has dimension $0$, so $\overline{A}$, the $z_1$-axis, is analytic. The hypothesis $\dim E < k$ is necessary. Let $E = \set{z_1 = 0}$ in $\C^2$ and $A = \set{(z_1, e^{1/z_1}) : z_1 \ne 0}$, an analytic subset of $\C^2 \setminus E$ of pure dimension $1 = \dim E$. Since $e^{1/z_1}$ takes every nonzero value, and tends to $0$ along the negative real axis, in every neighbourhood of $z_1 = 0$, the closure $\overline{A}$ is $A \cup E$. It is not analytic: the components of an analytic curve other than $E$ would meet $E$ in a discrete set, but $A$ accumulates at every point of $E$.

Going deeper

For a student

The theorem is a basic extension result in complex analytic geometry, in the spirit of the Riemann extension theorem for functions. It gives a short proof of Chow's theorem, by extending the cone over an analytic subset of projective space across the origin. Bishop's theorem replaces the dimension condition on $E$ by a volume condition on $A$, and Shiffman's theorem by a Hausdorff measure condition on $E$.