Hello I am self studying differential geometry. I am working on the following problem.
Let $\sigma:[0,1]\to \mathbb{R}^2$ a smooth simple closed curve on the plane; show that the image of $\sigma$ is the boundary of a regular domain.
I could not give a complete solution, my attempt is the following.
Since $\sigma$ is a regular curve there exists a finite family of open sets $\{ U_{i}\}_{i=1}^{m}$ such that $\sigma([0,1])\subseteq \bigcup_{i=1}^{n}U_{i}$ and on each open set $U_{i}$ there exists a diffeomorphism $\varphi_{i}$ such that $$ \varphi_{i}(U_{i} \cap \sigma([0,1])) \subseteq \{x_{2}=0\} $$ and $$ \varphi_{i}(U_{i} \cap \sigma([0,1])^{c} ) \subseteq \mathbb{R}^2 \setminus \{ 0 \}.$$ Then I observed that the family of open sets $ \{U_{i}\}_{i=1}^{m} \cup \{\mathbb{R}^{2}\setminus \sigma([0,1]) \}$ admits a partition of unity $\{\psi_{i}\}_{i=0}^{m}$ and I have defined the function $\sigma: \mathbb{R}^2 \to \mathbb{R}$ as $$ \sigma(p):=\sum_{i=1}^{m}\psi_{i}\, \varphi_{i}^{2},$$ where $\varphi_{i}^{2}$ is the $2-$nd component of the diffeomorphism $\varphi_{i}.$ So I defined the set $$ S=\{ p \in \mathbb{R}^{2}: \sigma(p)>0 \}$$ but I could not prove that $S$ is a regular domain, I suspect that I chose the wrong function.
What are the ground rules? Are you allowed to use the Jordan Curve Theorem? In your case of a smooth curve, the JCT can be proved from scratch, and the concepts involved in that proof might be what you need to finish your proof.
Suggestion: Google " JORDAN CURVE THEOREM FOR PIECEWISE SMOOTH CURVES" and you may find a helpful exposition by Pederson.