A differentiable one-parameter family of codimension 2 subspaces of $\mathbb{C}^n$ cannot fill $\mathbb{C}^n$, right? Suppose that $P(t)$ is a one-parameter family of rank 2 self-adjoint projections on $\mathbb{C}^n$ that vary analytically in the real parameter $t \in [0,1]$. I claim that there must exist a vector $x \in \mathbb{C}^n$ such that $P(t)x \neq 0$ for all $t$.    
In other words, I am hoping to prove that $\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t) \neq \mathbb{C}^n$. This seems like it is essentially a space-filling curve type argument, and hence the requirement that $P(t)$ be differentiable is probably important (in my example, $P(t)$ is analytic in $t$).  Does anyone know a reference that would provide a simple proof of this claim?  
 A: Let $Q(t):\mathbb C^n\to\mathbb C^n$ be the orthogonal projection onto $\text{ker}(P(t))$.
Then $t\mapsto Q(t)$ is as differentiable as $P$ was. Now the mapping $(t,x)\mapsto Q(t)(x)$ has rank at most $2n-2+1<2n$. Apply Sard's theorem: The set of regular values is Lebesgue nearly everything. Take a regular value $y$. It cannot be in the image, and thus is in no kernel of $P(t)$ for any $t$.
A: I assume $P$ varies analytically as in the body of the question, I do not know how to tackle the case where $P$ is just differentiable, I also read rank 2 as real rank though of course the problem is even easier if the rank is complex.
The map $\operatorname {ker}:A\rightarrow\operatorname {Gr}_{2n-2}\Bbb R^{2n}$ where $A\subseteq End(\Bbb R^{2n})$ are the endomorphisms of rank $2$ is $C^\infty$. Clearly the rank of $D(\operatorname {ker}\circ P)$ is less than or equal to 1. Consider the tautological bundle over the Grassmanian $T$ and the projection map $g:T\rightarrow \Bbb R^{2n}$. Since $T$ is locally trivial and second countable, we have $T$ is a countable union of spaces diffeomorphic to $\operatorname {Gr}_{n-2}\Bbb R^{2n}\times \Bbb R^{2n-2}=M_n$. Consider $f:\Bbb R\times\Bbb R^{2n-2} \rightarrow M_n$, $f_n(x,y)= (\operatorname {ker}\circ P(x),y)$. The rank of $D(g\circ f_n)$ is less than or equal to $2n-1$, so by Sard's theorem we have $\operatorname{img}(g\circ f_n)$
 is measure zero in $\Bbb R^{2n}$. So $\bigcup_n\operatorname{img}(g\circ f_n)=\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t)$ is a countable union of measure zero sets hence its complement is non-empty.
