So first for n even, $RP^n$ is not orientable, hence can not be embedded in $\mathbb{R}^{n+1}$.
For odd n, $RP^{n}$ is orientable, hence the normal bundle is trivial. Now using stiefel-Whitney classes, one can prove when $n$ not of the form $2^k – 1$, it can not be embedded in $\mathbb{R}^{n+1}$. I would appreciate if someone can give an more elementary proof in this case.
Then for the left cases. $RP^3$ can not be embedded proving by homology thoery (Alexander sphere duality, lefschetz duality and a long exact sequence).
For the other cases, I do not know how to prove. I realized Don Davis has a table for the immersion and embedding of $RP^n$ (http://www.lehigh.edu/~dmd1/immtable). But the question I am asking is easier, hence there may be an answer and a proof I could follow.
Best Answer
I'll use cohomology with coefficients $\mathbb{Z}/2$ everywhere.
Suppose that the space $P=\mathbb{R}P^{n-1}$ embeds in $S^{n}$ (where $n>2$). Recall that $$ H^*(P)=(\mathbb{Z}/2)[x]/x^{n} = (\mathbb{Z}/2)\{1,x,\dotsc,x^{n-1}\} $$ By examining the top end of the long exact sequence of the pair $(S^{n},P)$ we find that $H^{n}(S^{n},P)$ has rank two. Lefschetz duality says that this group is isomorphic to $H_0(S^{n}\setminus P)$, so we see that $S^{n}\setminus P$ has two connected components. (I don't need any orientation conditions here as I am working mod 2.) Let $A$ and $B$ be the closures of these components, so $A\cap B=P$ and $A\cup B=S^{2n}$. Lefschetz duality also gives $H^{n}(A)\times H^{n}(B)=H^{n}(S^{n}\setminus P)=H_0(S^{n},P)=0$.
We now have a Mayer-Vietoris sequence relating the cohomology groups of $A$, $B$, $P$ and $S^{n}$. As $H^1(S^{n})=H^2(S^{n})=0$ this gives an isomorphism $H^1(A)\times H^1(B)\to H^1(P)=\{0,x\}$. After exchanging $A$ and $B$ if necessary, we can assume that $H^1(B)=0$ and that there is an element $a\in H^1(A)$ that maps to $x$ in $H^1(P)$. It follows that $a^{n-1}$ maps to $x^{n-1}$, which generates $H^{n-1}(P)$, so the Mayer-Vietoris connecting map $H^{n-1}(P)\to H^{n}(S^{n})=\mathbb{Z}/2$ must be zero. This contradicts exactness at the next stage, because $H^{n}(A)\times H^{n}(B)=0$.