Show $M_1\cap M_2$ submanifold iff $N_x(M_1)\cap N_x(M_2) = \{0\}$ and dimensions Let $M_1,M_2 \subseteq \mathbb{R}^3$ two-dimendional submanifolds of $\mathbb{R}^3$ such that for every point $x\in M_1\cap M_2$
$$N_x(M_1)\cap N_x(M_2) = \{0\}.$$
Show $M_1\cap M_2$ is an one-dimensional submanifold of $\mathbb{R}^3$.
Let now $M_1,M_2\subseteq\mathbb{R}^n$ the submanifolds with dimensions $k$ and $l$ with $k+l \gt n$. Show that if for every point $x\in M_1\cap M_2$
$$N_x(M_1)\cap N_x(M_2) = \{0\}$$ holds, then $M_1\cap M_2$ is a submanifold. What is its dimension?
 A: If you zoom into the point of intersection $x \in M_1 \cap M_2$, you have that $M_1$ and $M_2$ are approximately planes in 3-space intersecting at a point.  Then you have their normals pointing in different directions. 
The intersection should have dimension $2+2-3=1$ or something like that.  Your job is for formalize this - or another - argument with open sets, tangent spaces, etc.

For your $\mathbb{R}^n$ example, $M_1$ and $M_2$ can also be approximated by linear spaces, so this becomes an instance of the rank-nullity theorem.
A: I use the following fact, that if $M\subseteq \mathbb{R}^3$ is a two dimensional submanifold, then for every $x \in M$ there is a differentiable function $f$ defined on an open neighbourhood $U$ of $x$ (in $\mathbb{R}$) such that 
1) $M\cap U$ is the set of zeros of $f$.
2)$f$ has non zero differential, meaning that the map $df: T_x(\mathbb{R}^3) \rightarrow T_{f(x)}(\mathbb{R})$ is non zero. In the language on elementary calculus this means that the gradient of $f$ is non zero. To elaborate, since this will come up latter, there are directions for which the directional derivative is non zero. We see however that since $df$ is a map from a three dimensional space to a one dimensional space, the kernel is two dimensional. So there will be two independent directions with zero directional derivative.
Since the gradient is normal to the surface, the derivative in the normal direction is non zero and the derivative in the two tangent directions will be zero.
I also use the converse that the set of zeros of such a function with nonzero derivative will be a submanifold. 
These facts are a consequence of the inverse function theorem and are proved by establishing a coordinate system such that the submanifold is defined by setting one coordinate equal to zero.
Now say we have two manifolds as in the question $M_1, M_2 \subseteq \mathbb{R}$. Let 
$x \in M_1 \cap M_2$ then there are differentiable functions $f_1$ and $f_2$ as above such that $M_i$ is the set of zeros of $f_i$ in a neighbourhood of $x$.
Now consider the function $f_2$ defined on $M_1$. The set of zero directions for $df_2$ is the tangent space for $M_2$ as above. Now since by assumption the intersection of the two tangent spaces of $M_1$ and $M_2$ is one dimensional, there must be a direction on the tangent space of $M_1$ for which $f_2$ has a non zero directional derivative. This means that if we consider $f_2 \restriction M_1$ we have a function with a non zero derivative. The zero set of $f_2 \restriction M_1$ is $M_1 \cap M_2$. This shows that the intersection is a submanifold of $M_1$ and therefore of $\mathbb{R}$.
That was the case of codimension $1$, in the general case of $M_1,M_2\subseteq \mathbb{R}^n$ of dimensions $k$ and $l$ respectively, the manifolds have codimensions, $n-k$ and $n-l$. So $M_2$ will be defined (locally) as the set of zeros of $n-l$, differentiable functions or better a function, $f_2:U\rightarrow \mathbb{R}^{n-l}$ with surjective derivative $df_2$ (and $U \subseteq \mathbb{R}$). The kernel of $df$ is the tangent space to $M_2$ which is $l$ dimensional.
If we now look at $f_2 \restriction M_1$. Then $d(f_2 \restriction M_1):T_x(M_1)\rightarrow \mathbb{R}^{n-l}$ is again surjective since the kernel of $df_2$ is the tangent space to $M_2$. (The intersection of the two tangent spaces has dimension $k+l-n$ so rank nullity theorem gives the dimension of the image as $k-(k+l-n)=n-l$) Thus the zero set of $f_2 \restriction M_1$ is the a submanifold of $M_1$ and is equal to $M_1\cap M_2$ the dimension $d$ is given by $d+n-l=k$ so 
$d=k+l-n$. 
