Is $\mathbb{R}P^n$ "two-sided" in $\mathbb{R}P^{n+1}$

Question

i.e. Does $\mathbb{R}P^n$ have a tubular neighborhood $N$ such that $N-\mathbb{R}P^n$ is disconnected.

My guess is yes, but don't know how to show it convincingly ( or maybe only for $n$ odd, I'm using $\mathbb{R}P^1$ for inutition). Since $\mathbb{R}P^n$ is a smooth submanifold of $\mathbb{R}P^{n+1}$ I believe that guarantees us a tubular neighborhood.

Answer 1

No it is not two-sided because the complement of $RP^n$ in $RP^{n+1}$ is an $(n+1)$-cell.