Jordan Regions and the Inverse Function Theorem

Question

Let U be an open subset of $\mathbb{R}^2$ and let K be a compact subset of U. Suppose that f : U → R is a function of class $\mathbb{C}^1$(U) -continuously differentiable in U- and let $\Bbb E = \{(x, y)\in K \mid f(x, y) = 0 \}$ and that Df -jacobian of f- does not vanish on E. Investigate whether E is a Jordan region.

Since determinant of the jacobian matrix of f is not equal to zero in E, initially it came to my mind to use the inverse function theorem. Let's define F(x,y)=( f(x,y), 1) for any $(x,y) \in U$. F is continuously differentiable. Since for any $(x,y) \in E$ the jacobian of f is different than zero, then (by inverse function theorem) there is an open set W containing E such that F is injective on W and $F^{-1}$ is continuously differentiable on F(W). Thus E$ \subseteq$W, E=$F^{-1}((0,1))$ and (0,1) is a point of volume zero. Due to the fact that $F^{-1}$ is continuously differentiable on a compact set K, $E\subseteq K$, then for any region of volume zero its image under $F^{-1}$ will be zero, i.e., E=$F^{-1}((0,1))$ is of volume zero.

Is this proof sufficient? I have some doubts especially about showing that image of (0,1) under F$^{-1}$ is equal to E.

Answer 1

Note that the set $E:=\bigl\{(x,y)\in K\,\bigm|\,f(x,y)=0\bigr\}$ is not a two-dimensional region, but generically a curve. It seems that your question is the following: Can this curve bound a certain Jordan measurable region?

Note that your $f$ is a scalar function, hence the derivative $df(p)$ is a linear functional at each point $p\in U$, and should not be viewed as a "Jacobian". Usually $df(p)$ is realized as gradient vector $$\nabla f(p)=\bigl(f_x(p),f_y(p)\bigr)\ .$$ I'm interpreting the phrase "that Df -jacobian of f- does not vanish on E" in your question as assumption that $df(p)\ne0$ for all $p\in E$, which is the same thing as $\nabla f(p)\ne{\bf 0}$ for all $p\in E$. This implies that each point $p\in E$ is the center of a rectangular window $W$ such that $E\cap W$ is graph of a $C^1$ function $y=\phi(x)$ or $x=\psi(y)$. This, together with the compactness of $E$, allows to conclude that $E$ is the union of finitely many simply closed curves. It may be that $E$ bounds a Jordan region in your sense, but this is not necessarily the case. Consider the example $$f(x,y):=\cos(x^2+y^2)\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)$$ and $K$ the closed disc of radius $\sqrt{4\pi}$ centered at the origin.