Is the image of a "$C^k$ open set" in $\mathbb{R}^n$ under a $C^k$ diffeomorphism also a $C^k$ open set? Let $U$ be an open subset of $\mathbb{R}^n$.
Suppose moreover that $U$ is $C^k$.   ($0 \leq k \leq \infty$).
This means that for each $x \in \partial U$, there exists an $r>0$ and a $C^k$ function defined on  $\mathbb{R}^{n-1}$ so that $B(x,r) \cap U = B(x,r)  \cap \{(x,y):y>f(x)\}$ and $B(x,r) \cap \partial U = B(x,r) \cap \{(x,y):y=f(x)\} $. You may "change the order of the coordinates" if needed.
Now, let $h:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a $C^k$ diffeomorphism.
Question: Is $h(U)$ also a $C^k$ open set?
Remark: Clearly, $\partial U$ is an embedded $C^k$ submanifold of $\mathbb{R}^n$ so that its image $h(\partial U)$ is also an embedded $C^k$ submanifold. Please note that this fact does not immediately solve the whole problem.
Any help will be fully appreciated.
 A: This is true. The difficulty in proving it seems to be to arrange for the correct ball radius $r$ in the ball $B(x,r)$. However, that difficulty can be sidestepped by rewording the definition of a $C^k$ open set using "open neighborhoods" instead of "open balls".

$U \subset \mathbb R^n$ is a $C^k$ open set if and only if for each $x \in \partial U$ there exists an open neighborhood $V \subset \mathbb R^n$ of $x$ and a $C^k$ function $f$ defined on $\mathbb R^{n-1}$, so that [everything holds as written except with $B(x,r)$ replaced by $V$].

The "open ball" definition clearly implies the "open neighborhood" definition. 
Conversely, suppose that the "open neighborhood" definition holds. Then for each $x$ with corresponding open neighborhood $V$, we simply choose $r>0$ so that $B(x,r) \subset V$, and now the "open ball" definition holds.
Now, assuming that $U$ is a $C^k$ open set and $h$ is a $C^k$ diffeomorphism, let's prove that $h(U)$ is a $C^k$ open set. 
For each $x \in \partial(h(U))$, consider $h^{-1}(x) \in \partial U$. Choose a function $f$ and an open neighborhood $V$ which witness that $U$ is a $C^k$ open set near the point $h^{-1}(x)$. Then the function $f h^{-1}$ and the open neighborhood $h(V)$ witnesses that $h(U)$ is a $C^k$ open set near $x$.
