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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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@NajibIdrissi thanks, just fixed it –  Pilo Jul 31 at 14:55

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up vote 3 down vote accepted

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

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well isn't $RP^1$ two sided in $RP^2$? since that 2-cell is wrapped twice around $RP^1=S^1$? –  Pilo Jul 31 at 14:54
    
It's not the 2-cell that's wrapped but rather its boundary. If you remove it you are left with an open 2-cell. –  user72694 Jul 31 at 14:56
    
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 –  Pilo Jul 31 at 15:03

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