I am reading the book Introduction to complex analytic geometry by Stanislaw Lojasiewicz. In Chapter II, we have that a subset $Z$ of a complex manifold $M$ is called an analytic subset of $M$ if every point of $M$ has an open neighbourhood $U$ such that the set $Z \cap U$ is a globally analytic subset of $U$ i.e. there exist finitely many holomorphic functions $f_1, \dotsc, f_k$ in $U$ such that
$Z \cap U =\{x \in U \ : \ f_1(x)=\dotsc =f_k(x)=0\}.$
On Page 207 of the book, we have the following corollary :
Corollary: Connected components of an analytic subset of a manifold $M$ are analytic subsets of $M$. Indeed, they are both open and closed in $M$.
Since any domain $\Omega$ in $\mathbb{C}^n$ is a complex domain (of dimension $n$), we can talk of analytic subset of $\Omega$. For a domain $\Omega$ in $\mathbb{C}^n (n>1)$ and a complex polynomial $p$ in $n$-variable, it follows that the subset $V=Z(p) \cap \Omega$ is an analytic subset of $\Omega$ given the zero set $Z(p)$ of polynomial $p$ has non-empty intersection with $\Omega$.
If $V$ is connected, then it is easy to see that $V$ is closed in $\Omega$ (in the subspace topology induced from $\mathbb{C}^2$ endowed with standard Euclidean topology) since $Z(p)$ is a closed subset of $\mathbb{C}^2$ (in the standard Euclidean topology) but as per the above mentioned corollary, $V$ must be open in $\Omega$. I am not able to see how $V$ can be open in $\Omega$.
My doubt: If $V$ is open in $\Omega$ which is an open subset of $\mathbb{C}^2$, then $V$ is an open subset of $\mathbb{C}^2$ which is not possible because an analytic subset must have empty interior.
Any hints or any correction to the statement of the corollary.