https://en.wikipedia.org/wiki/Transversality_(mathematics)

This problem may be a embarrassed one, but I even could not prove it for the 1 dimensional case.

Here is the problem:

>**Question 1** $latex M$ is a compact $latex n$-dimensional smooth manifold in $latex R^{n+1}$, take a point $p\notin M$. prove there is always a line $latex l_p$ pass $latex p$ and $latex l_p\cap M\neq \emptyset$, and $latex l_p$ intersect transversally with $latex M$.

You can naturally generated it to:

>**Qusetion 2** $latex M$ is a compact $n$-dimensional smooth manifold in $latex R^{n+m}$, take a point $latex p\notin M$. Prove $\forall 1\leq k\leq m$, there is always a hyperplane $latex P_p, dim(P_p)=k$ pass $p$ and $latex P_p\cap M\neq \emptyset$, and $latex P_p$ intersect transversally with $latex M$.

Thanking for Piotr pointed out, assuming “transverse” means “the tangent spaces intersect only at 0”.

We focus on question 1 for simplified.

Even in 1 dimension it is not easy at least for me, **warning**: a line $latex l$ pass $latex p$ may be intersect $M$ at several points combine a set $latex A_l$, $latex A_l$ could be finite, countable or even it is not countable (consider $latex M$ is induced by a smooth function for which the zeros set is Cantor set.)… And if there is one point $latex a\in A_l$, $l$ is tangent with the tangent line of $latex M$ at $latex a$, then $latex l$ is not intersect transversally with $latex M$.

**My attempt**:

I could use a dimensional argument and Sard’s theorem to establish a similar result but instead of a fix point $latex p$, we proof for generic point in $latex R^{n+1}$ which is not in $latex M$ we can choose such a line.

So it seems reasonable to develop the dimensional technique to attach the question 1, in 1 dimensional, it will relate to investigate the ordinary differential equation:

$latex \frac{f(x)-b}{x-a}=f'(x)$

Where $latex p=(a,b)$, $latex M$ have a parameterization $latex M=\{x,f(x)\}$. If there is a counterexample for the question 1, then there is another solution which satisfied the ODE in the sense:

at least for every line $latex l$ there is a intersection point $latex a_l\in l\cap M$, $latex f$ satisfied ODE at $latex a_l$.

This is just like the uniqueness of the solution of such a ODE is destroyed at some subspace of a line which have some special linear structure, I do not know if this point of view with be helpful.

I will appreciate for any useful answers and comments.

**Proof 1(provided by fedja)**

Area trick.(weakness:it seems we could not proof the transtivasally intersection point have positive measure by this way).

** Proof 2(provided by Piotr)**

#For the codimension 1 case.#

###Using Thom transversality theorem.###

Consider the maps $latex f_s:\mathbb{R} \to \mathbb{R}^n$ parametrized by $latex s \in S^{n-1}$ and given by $latex f_s(t) = p + t \cdot s$. The map $latex F(s,t) = f_s(t)$, $latex F:S^{n-1} \times \mathbb{R} \to \mathbb{R}^n$ is clearly transverse to $latex M$, thus Thom’s transversality says that $latex f_s$ is transverse to $latex M$ for almost all $latex s$. Now it suffices to prove that for an open set in $latex S^{n-1}$, the line given by $latex f_s$ intersects $latex M$. Proven below.

###Using Sard’s theorem directly.###

Thom’s transversality is usually proven using Sard’s theorem. Here is the idea.

Consider the projection $latex \Pi:\mathbb{R}^n \setminus \{p\}\to S^{n-1}_p$ onto a sphere centered at $latex p$. A line $latex l_p$ through $latex p$ intersects $latex M$ transversally if the two points $latex l_p \cap S^{n-1}_p$ are regular values of $latex \Pi$ (indeed, the critical points of $latex \Pi$ are exactly the points $latex x \in M$ at which the normal $latex \vec n_x$ is perpendicular to the radial direction (with respect to $p$)). By Sard’s theorem, the set of regular values is dense in $latex S^{n-1}_p$.

We need to choose any point $latex s$ on the sphere for which both $latex s$ and $latex -s$ are regular values, and the line $latex f_s$ through $latex p$ and $latex s$ actually intersects $latex M$. It suffices to prove that the set of points $latex s$ for which this line intersects $latex M$ contains an open set. We could now use the Jordan-Brouwer Separation Theorem and we would be done, but we can do it more directly (and in a way that seems to generalize).

###The set of points $latex s \in S$ for which $latex f_s$ intersects $latex s$ has nonempty interior.###

For each point $latex q \notin M$ the projection $latex \Pi:M \to S_{q,\varepsilon_q}^{n-1}$ onto the sphere centered at $latex q$, of radius $latex \varepsilon_q$ small enough so that the sphere does not intersect $latex M$, has some (topological) degree $latex d_q$. It is easy to check that if one takes any point $latex x \in M$ and considers the points $latex x \pm \delta \vec n_x$ for small $latex \delta$, the degrees of the corresponding maps differ by $latex 1$. It follows that we can find a point $latex q$ for which $latex d_q \neq d_p$, which guarantees that for every point $latex q’$ in a small open ball $latex B$ around $latex q$ (all these points have same degree $latex d_q$), the line joining $latex p$ and $latex q’$ intersects $latex M$. Projection of $latex B$ on $latex S_p^{n-1}$ is an open set which we sought.

#For the general case (partial solution).

I think a similar reasoning should work, however, notice that for $latex k < m$ we cannot make $latex P_p$ intersect transversally with $latex M$ because of dimensional reasons: the dimensions of $latex M$ and $latex P_p$ don’t add up to at least $latex n+m$. Recall that transversality implies Thus, either (1) you want to consider $latex k \geq m$, or (2) define “transversal intersection” for such manifolds saying that the tangent spaces have to intersect at an empty set.

Also, for $latex k>n$ we can just take any plane $latex P_p$ which works for $latex k=m$ and just extend it to a $latex k$-dimensional plane.

###Assuming $latex k = m$.###

A similar reasoning should work for $latex f_s:\mathbb{R}^m \to \mathbb{R}^{n+m}$ with $latex s = (s_1, \ldots, s_m)$ going over all families of pairwise perpendicular unit vectors, and $latex f_s(t_1,\ldots,t_m) = p+\sum_{j=1}^m t_i \cdot s_i$. Thom’s transversality says that for almost all choices of $latex s$, the plane $latex f_s$ is transverse to $latex M$.

### The nonempty interior issue. ###

The only thing left is to prove that the set of $latex s$ for which the intersection is nonempty has nonempty interior. Last time we proved that there is a zero-dimensional sphere containing $latex p$, namely $latex \{p,q\}$, which has nonzero linking number with $latex M$, and by deforming if to spheres $latex \{p,q’\}$ and taking lines through pairs $latex p,q’$, we got an open set of parameters for which the line intersects $latex M$.

Here should be able to do a similar trick by finding a $latex m-1$-dimensional sphere with nonzero linking number with $latex M$. The ball that bounds that sphere has to intersect $latex M$, thus the plane $latex P$ containing the sphere has to intersect $latex M$. By perturbing the sphere we get spheres with the same linking numbers, and get all the planes that lie in a neighbourhood of $latex P$; in particular, we get an open set of parameters $latex s$ for which $latex f_s$ intersects $latex M$.

Well, we don’t actually need a *round* sphere, but we do need a *smooth* sphere that lies in a $latex m$-dimensional plane. There’s some trickery needed to do this, but I am sure something like this can be done.

Maybe somebody else can do it better?

### For $latex k<m$ ###

I don’t really know how to attack this case, assuming “transverse” means “the tangent spaces intersect only at $latex 0$”.