Corollary 9.3.16725857B32C0E

Proof. Fix $x\in A$, let $A_1,\ldots ,A_ p$ be representatives of the branches of $(A,x)$, and for each $i$ choose a parametrization polydisc $P_ i$ of $(A_ i,x)$ so small that $A\cap P_ i=\bigcup _ j A_ j\cap P_ i$. By corollary 9.3.13(1), $A_{\mathrm{sing}}\cap A_ i$ is an analytic subset of $A_ i$ whose germ at $x$ differs from $(A_ i,x)$. Hence $A_{\mathrm{sing}}$ meets the connected covering locus of $A_ i$ in a proper analytic subset, and by lemma 9.2.9 and corollary 9.3.6, the covering locus minus $A_{\mathrm{sing}}$ is connected and dense in $A_ i\cap P_ i$. The set $R_ i=A_ i\cap P_ i\cap A_{\mathrm{reg}}$ lies between this connected set and its closure, so it is connected and dense in $A_ i\cap P_ i$. In particular $A_{\mathrm{reg}}$ is dense in $A$.

Let $\{ S_\alpha \} $ be the connected components of the manifold $A_{\mathrm{reg}}$, and $Z_\alpha =\overline{S_\alpha }$. Each $R_ i$ above is connected, so it lies in a single component $S_{\alpha (i)}$. Let $U=\bigcap _ iP_ i$. Then $S_\alpha \cap U$ is contained in the union of the $A_ i$ with $\alpha (i)=\alpha $, because points of $A_{\mathrm{reg}}\cap A_ i\cap U$ lie in $R_ i$; conversely $A_ i\cap U\subset \overline{R_ i}\subset Z_{\alpha (i)}$. Hence

Thus each $Z_\alpha $ is analytic, only finitely many of them meet $U$, and their union is $A$. Since $A_{\mathrm{reg}}$ is open in $A$ and its connected components are open in it, $S_\alpha $ is open in $A$. So every point of $S_\alpha $ has a neighborhood that meets no other $S_\beta $, and therefore does not lie in $Z_\beta =\overline{S_\beta }$ for $\beta \neq \alpha $. Thus $S_\alpha \cap Z_\beta =\varnothing $ for $\beta \neq \alpha $, and no $Z_\alpha $ is contained in the union of the others. The same remark shows that $S_\alpha \subset (Z_\alpha )_{\mathrm{reg}}$: near a point of $S_\alpha $, the set $Z_\alpha $ coincides with the manifold $A$. Hence $(Z_\alpha )_{\mathrm{reg}}$ lies between the connected set $S_\alpha $ and its closure, and is connected.

If $Z_\alpha $ were the union of two analytic subsets $Z'$, $Z''$ different from it, then $S_\alpha \cap Z'$ and $S_\alpha \cap Z''$ would be analytic subsets of the connected manifold $S_\alpha $ different from $S_\alpha $, since otherwise their closure would be $Z_\alpha $. As proper analytic subsets of a connected manifold, they would be closed and nowhere dense in $S_\alpha $ (lemma 9.2.8), and so would their union, which is $S_\alpha $. This contradiction shows that $Z_\alpha $ is irreducible.

If $M$ has a countable basis, so does $A_{\mathrm{reg}}$; each $S_\alpha $ is open in $A_{\mathrm{reg}}$ and contains a basic open set, and disjoint components contain different ones, so there are at most countably many components. Finally, a holomorphic function vanishing on a nonempty open subset of the connected manifold $(Z_\alpha )_{\mathrm{reg}}$ vanishes on all of it by the identity theorem, and hence on its closure $Z_\alpha $.

It remains to prove uniqueness. Let $A=\bigcup _\beta Y_\beta $ be a locally finite decomposition into irreducible analytic subsets, none contained in another. A connected manifold is not a locally finite union of closed nowhere dense subsets, since such a union is closed and nowhere dense. Since proper analytic subsets of the connected manifold $S_\alpha $ are nowhere dense (lemma 9.2.8), each $S_\alpha $ lies in some $Y_\beta $, and then $Z_\alpha \subset Y_\beta $. Conversely, fix $\beta $. By what we have proved, applied to the analytic set $Y_\beta $, the set $Y_\beta $ is the union of its irreducible components, and it has only one, for otherwise it would be the union of one of them and of the union of the others, two analytic subsets different from $Y_\beta $. Hence $(Y_\beta )_{\mathrm{reg}}$ is a connected manifold, dense in $Y_\beta $ and covered by the locally finite family of closed analytic subsets $(Y_\beta )_{\mathrm{reg}}\cap Z_\alpha $; as before, $Y_\beta \subset Z_\alpha $ for some $\alpha $. Choosing $\beta '$ with $Z_\alpha \subset Y_{\beta '}$, we get $Y_\beta \subset Y_{\beta '}$, so $\beta '=\beta $ and $Y_\beta =Z_\alpha $. Thus every $Y_\beta $ is an irreducible component of $A$; and every component $Z_\alpha $, being contained in some $Y_\beta =Z_{\alpha '}$, equals it, because $S_\alpha \cap Z_{\alpha '}=\varnothing $ for $\alpha '\neq \alpha $. $\square$