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

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.

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No it is not two-sided because the complement of $RP^n$ in $RP^{n+1}$ is an $(n+1)$-cell.
well isn't $RP^1$ two sided in $RP^2$? since that 2-cell is wrapped twice around $RP^1=S^1$? – Hamton Jul 31 '14 at 14:54
Ah I see so any neighborhood we take of $RP^1$ is just a thickened up open border of the $n+1$-cell, which is def not disconnected – Hamton Jul 31 '14 at 15:03