Is the Whitney embedding theorem tight for all $n$? Whitney's embedding theorem states that any smooth $n$-manifold $M$ can be smoothly embedded into $\Bbb R^{2n}$. In dimension 1 this is tight: the circle cannot be embedded into $\Bbb R^1$. It is a theorem that if $M^n \hookrightarrow \Bbb R^{n+k}$ is a smooth embedding, and $w(M)$ the total Whitney class of the tangent bundle of $M$, then $(w(M)^{-1})_k = 0$. Now consider dimension $m = 2^n$; there $w(\Bbb{RP}^m) = 1+a+a^m$, and hence $w(\Bbb{RP}^m)^{-1} = 1+a+a^2+\dots + a^{m-1}$. Hence $\Bbb{RP}^m$ cannot be embedded into $\Bbb R^{2m-1}$, proving tightness of the bound in these dimensions. 
Question: Is Whitney's embedding theorem tight in all dimensions? (Let's restrict to compact manifolds.)
Note that the related immersion theorem, that there is an immersion $M^n \looparrowright \Bbb R^{2n-1}$, does not provide a tight bound for $n=3$, according to this page on the Manifold Atlas project: every compact $3$-manifold immerses into $\Bbb R^4$.

As referenced in the comments, this MathOverflow question gives the answer as no: every compact $3$-manifold embeds into $\Bbb R^5$, as proved by Wall. The Immersion theorem gives the optimal result for immersions: every compact $n$-manifold embeds into $\Bbb R^{2n-\alpha(n)}$, where $\alpha(n)$ is the number of $1$s in the $2$-adic expansion of $n$. $\Bbb{RP}^n$ cannot embed into any smaller Euclidean space, as seen by inspecting its Stiefel-Whitney classes.
