# How to prove the line exist?

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.

• I reformatted and slightly reworded your problem to make it more clear and legible; if I did so incorrectly please fix. Dec 23, 2016 at 14:28
• I don't have a solution, but I like your partial solution. Do I understand it right? If the statement is false, there exists a point $m\in M$ such that for all lines $L$ through $m$, there exist another point $m' \in M\cap L$. This gives a map $f \colon \mathbb{R}P^{n+1} \to M$ which is not surjective... Dec 23, 2016 at 14:49
• @MatthewLeingang,I get clear now.$M\setminus \left\{ m \right\} \rightarrow RP^{n+1}$ by x~y iff$x=λy$,and the map is smooth,as the image dimension is obvious less than $RP^{n+1}$,the rank will always less than $RP^{n+1}$ and its image in $RP^{n+1}$ is measure zero. So we find a point in $RP^{n+1}$ not in the image.
– gtx
Dec 24, 2016 at 0:58

## 2 Answers

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.

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.

• Please accept one of the answers so that the question will go off the "unanswered" list. :) Dec 24, 2016 at 1:18