Intersection of lines with compact smooth manifolds everybody. I need a hint on this: I have to prove that a compact smooth submanifold of R^n intersects almost every one dimensional linear subspace (in R^n) in a finite set of points. I know I have to use transversality theorems, but I just don´t see how. Any ideas?
I know that suffices to prove it for a manifold without the origin and that it must be related to the identification of R^n that produces a projective space, but I can´t figure out how to join the pieces.
Thanks.
 A: Let $M$ be a compact submanifold of $R^n$; then $m=dim(M)<n$. I will use the identification of the space of lines (1-dimensional linear subspaces) in $R^n$ with $RP^{n-1}$. 
I will first give an argument in the case when $0\notin M$. Consider the projection $p: M\to RP^{n-1}$, which is the restriction of the projection $R^n \setminus 0 \to RP^{n-1}$. The map $p$ is smooth and, by Sard's theorem, almost every point $x\in RP^{n-1}$ is a regular value of $p$. Since $m\le dim (RP^{n-1})$, it follows that for almost every $x\in RP^{n-1}$, $dim( p^{-1}(x))\le 0$ and, hence, by compactness of $M$, the subset $p^{-1}(x)$ is finite. Therefore, $M$ intersects almost every line (represented by a point in $RP^{n-1}$) in a finite number of points. 
Consider now the case when $0\in M$. The above argument applied to $M \setminus 0$ shows that almost every line $l$ intersects $M$ in a discrete set of points. Since $M$ is compact, if $l\cap M$ is infinite, then this intersection contains a sequence converging to $0$. Thus, by smoothness of $M$, the line $l$ is tangent to $M$ at $0$. Therefore, almost every line not contained in $T_0M$ will intersect $M$ only in a finite subset. Since the set of lines contained in $T_0M$ has measure zero in $RP^{n-1}$ (as $m<n$), the assertion follows. qed 
