Show that $\{v: (M+v)\cap N= \emptyset\}$ is dense 
Given $X\subset \mathbb{R}^p$ and $v\in \mathbb{R}^p$, let $X + v =
 \{x+v; x\in X\}$. Let $M, N\subset \mathbb{R}^p$ be surfaces of class
  $C^1$ such that $\dim M + \dim N <p$. Show that $\{v: (M+v)\cap N= \emptyset\}$  is dense in $\mathbb{R}^p.$
Suggestion: Prove that the set of points $x-y\in\mathbb{R}^p$, where $x\in M$ and $y\in N$ has measure zero

Why proving that the vector from $M$ to $N$ has measure zero will help? I truly have no clue on this one, as I see no connection from the guess to the exercise. 
 A: If we suppose the hint is true, then you just have to observe that 
$M+v\cap N\neq\emptyset\ \Longleftrightarrow v\in N-M, $
where $N-M=\{ n-m; n\in N,m\in M\}$. This means that the complement of the set in question has measure zero, so it must be dense.
Now for the hint, define the morphism $\phi:M\times N \to \mathbb{R}^p$ defined by $(m,n)\mapsto n-m$. We want to show that the image of this morphism haz measure zero. We will use the following lemma:
Let $f:U\subseteq \mathbb{R}^q\to \mathbb{R}^n$ be a $\mathcal{C}^1$ morphism such that $U$ is open and $q<n$. Then $f(U)$ has measure zero.
You can find a basic proof of this here: 
Smooth image of a null set has measure zero (I know it says "smooth" but the exactly same proof holds for $\mathcal{C}^1$ functions).
Set $q:=\dim(M)+\dim(N)$. Since $M\times N$ is a variety of dimension $q$, we can get a family of $\mathcal{C}^1$ homeomorpshisms $\{\psi_i:V_i\subseteq \mathcal{R}^q \to U_i\subseteq M\times N\}_{i\in\mathbb{N}} $ such that the $V_i$ are open and $\{U_i\}$ is an open cover of $M\times N$.
Observe that $Im(\phi)=\bigcup_i Im(\phi \circ \psi_i)$. Applying the lemma to each of these functions, we get
$|N-M|\leq \sum_i |\phi\circ\psi_i|=0.$
A: I'll use the following well-known result: Suppose $j<k,$ and $f: [0,1]^j \to \mathbb R^k$ is Lipschitz. Then $\mu_k(f([0,1]^j)) = 0.$ (Here $\mu_k$ is Lebesgue measure on $\mathbb R^k.$) The proof of this is not difficult. I'll leave it for now; ask if you have questions.
Now suppose $m,n,p$ are positive integers such that $m+n < p.$ Suppose $f: [0,1]^m \to \mathbb R^p, g: [0,1]^n \to \mathbb R^p,$ and both $f,g$ are Lipschitz. Define
$$F(x,y) = g(y)-f(x),\,\, (x,y) \in [0,1]^m \times [0,1]^n.$$
Then $\mu_p(F([0,1]^m \times [0,1]^n)) = 0.$ This is because $F$ is Lipschitz on $[0,1]^m \times [0,1]^n$ and $m+n< p.$ Apply the first paragraph.
Continuing with this example, set $A=f([0,1]^m), B = g([0,1]^n).$ Then $F([0,1]^m \times [0,1]^n)$ is precisely the set of $v\in \mathbb R^p$ such that $(A+v)\cap B\ne \emptyset.$ Proof: Let $(x,y) \in [0,1]^m \times [0,1]^n.$ Then $g(y) = f(x) + F(x,y).$ Since $g(y)\in B, f(x) \in A,$ we see that with $v=F(x,y)$ the set $(A+v)\cap B\ne \emptyset$ as desired. I'll omit the inclusion in the other direction; it's just as easy.
Back to our $C^1$ surfaces $M,N.$ Let $m,n$ be the dimensions of $M,N.$ Now $M = \cup_{i=1}^{\infty}A_i,$ where each $A_i$ is the image of a $C^1$ map defined on $[0,1]^m$ with bounded derivatives on that set. Such maps are certainly Lipschitz. The same thing can be said for $N.$
So we have $M = \cup_{i=1}^{\infty}A_i$ and $N = \cup_{j=1}^{\infty} B_j.$ For each pair $A_i,B_j$ we are in the situation discussed above. Thus for each pair $i,j,$ the set $E_{i,j}$ of vectors $v$ such that $(A_i+v)\cap B_j\ne \emptyset$ has $\mu_p$-measure $0.$ Thus the union of all $E_{i,j}$ has $\mu_p$-measure $0.$ The complement of this union is then a dense set of vectors $v$ such that $M+v$ and $N$ are disjoint.
A: If we prove the suggestion, then the set of points $x - y \in \mathbb{R}^p$ with $x \in M$ and $y \in N$ has empty interior, since any set of measure zero has empty interior. Then, this implies the complement (which is the set of vectors $v$ such that $M + v$ and $N$ are disjoint) is dense in $\mathbb{R}^p$.
Now let's prove the suggestion.
Let $m = dim \ M$. Since $M$ is a surface of class $C^1$ in $\mathbb{R}^p$ then, by definition, it can be covered by an enumerable collection of open sets $U^{i}_{M} \subset \mathbb{R}^p$ (we can suppose the collection enumerable by Lindelof's Theorem) such that each $V^{i}_{M} = U^{i}_{M} \cap M$ admits a parametrization $\phi_{M}^i: V_{0M}^i \rightarrow V^{i}_{M}$ of class $C^1$, where $V_{0M}^i$ is an open set in $\mathbb{R}^m$.
We repeat this notation for $N$. Let $n = dim \ N$. Since $N$ is a surface of class $C^1$ in $\mathbb{R}^p$ then, by definition, it can be covered by an enumerable collection of open sets $U^{i}_{N} \subset \mathbb{R}^p$ (we can suppose the collection enumerable by Lindelof's Theorem) such that each $V^{i}_{N} = U^{i}_{N} \cap N$ admits a parametrization $\phi_{N}^i: V_{0N}^i \rightarrow V^{i}_{N}$ of class $C^1$, where $V_{0N}^i$ is an open set in $\mathbb{R}^n$.
Then, consider the enumerable collection of parametrizations:
\begin{align}
\phi^{i, j}: V_{0M}^i \times V_{0N}^j &\rightarrow \mathbb{R}^p \\
                         (x_0, y_0)        &\mapsto \phi_{M}^i(x_0) - \phi_{N}^j(y_0)
\end{align}
Notice that the set of points $x - y \in \mathbb{R}^p$, where $x \in M$ and $y \in N$ is equal to $\bigcup_{i, j} Im(\phi^{i, j})$. Therefore, it suffices to prove that $Im(\phi^{i, j})$ has measure zero.
Finally, $Im(\phi^{i, j})$ has measure zero because $V_{0M}^i \times V_{0N}^j$ has measure zero in $\mathbb{R}^p$ (since $V_{0M}^i \times V_{0N}^j \subset \mathbb{R}^m \times \mathbb{R}^n$ and $m + n = dim \ M + dim \ N  < p$) and $\phi^{i, j}$ is of class $C^1$ (remember that measure zero sets are invariant by applications of class $C^1$).
PS: Although there are already answers to this questions, I decided to write this one since I saw it on an analysis book I am studying ("Curso de análise 2 - vol2" by Elon Lages Lima) and in this book the concept of variety has not yet been defined.
