I'm interested in a simple proof of the following fact:
- Let $V \subset \mathbb{R}^3$ be a bounded, open, connected, simply connected set. Let $\gamma$ be a piecewise-smooth simple closed curve in $V$. Then, there exists some smooth surface $S$ such that the boundary of $S$ is $\gamma$.
The reason why I want this fact, is to prove the following elementary fact in calculus:
- Let $V \subset \mathbb{R}^3$ be a bounded, open, connected, simply connected set. Let $F:V \to \mathbb{R}^3$ be a smooth vector field with curl zero. Then $F$ is a gradient of some function.
Since what I really want is a proof for 2, it will be also appreciated if you prove 2 avoiding the use of 1.
Currently, the proof of 2 in my mind is as follows:
Define $V$ to be a line integral of $F$. Showing that this definition does not depend on the choice of path suffices. We may take the path which is always parallel to one of the three axes. Then we can somehow(maybe by combinatorial argument, which I am struggling to make rigorous) split the 'rectangular' closed curve into sum of 2-d rectangles. Hence the claim follows by Stokes(or Green).
Any help will be appreciated.