Disconnecting a complex vector space

Question

Can a (complex) dimension $n$ subspace disconnect a (complex) dimension $n+1$ vector space ? If the answer is no, what if we replace "vector space" by "manifold" ?

Answer 1

A complex submanifold does not disconnect the complex space since the real codimension is $\ge 2$.

The analytic approach: Say inside $\mathbb{C}^n$ we have a subvariety $f=0$. Take $P$, $Q$ point with $f(P) \ne 0$, $f(Q) \ne 0$. Consider a complex affine line $L$ through $P$, $Q$ ( looks like the complex plane). The function $f$ restricted to $L$ is not identically zero and thus has a discrete zero set, that can be avoided by a path from $P$ to $Q$.

A geometric analysis approach: a subset $\Delta$ of the unit ball in $\mathbb{R}^N$ of Hausdorff dimension $< n-1$ does not disconnect the ball. Take $P$, $Q$ in the ball and project centrally from $P$ and from $Q$ to the boundary of the ball. Since the projection is Lipschitz map the images of $\Delta$ on the sphere do not cover the sphere. Take an uncovered point $R$. Then $PRQ$ is a broken line from $P$ to $Q$.