How to prove the line exist?

Question

Let $n \geq 1$, and $M \subset \mathbb{R}^{n+2}$ be a smooth $n$-dimensional manifold which is closed in $\mathbb{R}^{n+2}$. Show that for each $m \in M$, there exists a line $L \subset \mathbb{R}^{n+2}$ such that $L\cap M = \{m\}$.

I tried to consider $m$ as the origin of $RP^{n+1}$ space and use Sard's Theorem, but it seems not to work.

Answer 1

Let $N= M\setminus \{m\}\times \bf R$ the product of $M\setminus \{m\}$ nad the real line, so that $N$ is a $n+1$ dimensional manifold. The map $N\to \bf R^{n+2}$ given by $f(x,t)= tm+(1-t)x$ is smooth, and its image is the set of points which are on a line which meets $L$ in $m$ and another point. This map is smooth and cannot be surjective (dimension). Let $p$ be a point not in the image of $f$. The line through $p$ and $m$ do not meet $M$ at another point.

Answer 2

Your approach is perfectly fine, but you messed up dimensions. The lines in $\Bbb R^{n+2}$ passing through $m\in M$ form an $\Bbb RP^{n+1}$. The chords of $M$ passing through $m$ form an $n$-dimensional submanifold thereof.