Are two compact $n$-dim submanifolds with boundary of $\mathbb{R}^n$ with identical boundaries coincide? Let $U,V$ be the interiors of codimension $0$ compact embedded submanifolds with boundary of $\mathbb{R}^n$. Suppose that $\partial U=\partial V$.
Is it true that $U=V$?
(In other words, I am asking if two compact $n$-dim submanifolds with boundary of $\mathbb{R}^n$ with the same boundary coincide).
Note that in general, it is not true that two open subsets of $\mathbb{R}^n$ with the same boundary, are identical.
(e.g the left and right half-planes in $\mathbb{R}^2$).
 A: Yes, this is true.  
To prove it, consider the compact $n-1$-dimensional submanifold $M = \partial U = \partial V$. Note that each of $U,V$ must be unions of components of $\mathbb{R}^n-M$.
The case where $M$ is connected is called the Jordan-Brouwer separation theorem, which says: $\mathbb{R}^n-M$ has two components, one bounded and the other nonbounded; and the closure of each component is a codimension 0 submanifold with boundary $M$. It follows that $U$ and $V$ are both equal to the bounded component of $\mathbb{R}^n-M$.
For the general case, suppose that $M$ has $k$ components, $M = M_1 \cup \cdots \cup M_k$. By applying the Jordan-Brouwer theorem together with an induction argument, one can prove that the complement $\mathbb{R}^n-M$ has $k+1$ components $C_0,C_1,...,C_{k}$, exactly one of them is unbounded -- let's say $C_0$ is unbounded -- and each $M_i$ is contained in the closure of exactly two of the $C_j$'s. Furthermore there exists a unique function
$$f : \{C_0,...,C_{k}\} \to \{0,1\}
$$
such that $f(C_0)=0$, and such that for each $M_i$, if $C_a,C_b$ are the two $C$'s whose closures contain $M_i$ then $f$ takes one of $C_a,C_b$ to $0$ and the other to $1$. It follows that $U,V$ are both equal to the union of those $C_a$ such that $f(C_a)=1$.
