Let $S_0\subset \mathbb R^N$ be given, where we assume that $S_0$ is a connected $C^1$ hyper surface and $\mathcal H^{N-1}(S_0)<\infty$. Here $S_0$ may not be able to represented as a graph. (up to rotation)
Now assume in some way I remove some points from $S_0$ and have a new set $S$, the new set $S$, which is embedded in $S_0$, can be represented as a graph and also every point in $S$ has density 1. (Now $S$ may not be connected)
My question: can I find a new connected $C^1$ hyper surface $S^\ast$ such that $S^\ast$ contains $S$ and can be represented as a graph? If yes, I can transfer $S^\ast$ into a hyperplane, so does $S$ itself.
My goal is to find a way to flatten $S$...i.e., transfer it into a hyperplane.
My try: Let the collection $\mathcal S:=\{S_n\}$ be connected $C^1$ hyper surface and contains $S$, i.e., $S\subset S_n$ for each $n$. Then I know the collection $\mathcal S$ is not empty since at least $S_0$ is inside. Then I can have a ascending chain $S_i$ such that $$ S\subset\cdots\subset S_n\subset S_{n-1}\subset\cdots\subset S_1 $$ and hence $\bar S:=\bigcap S_n$ is the upper bound for this chain and hence by Zorn's lemma we have there exists at least one maximal element, we call it $C^\ast$.
In the comment, I realize that $\bar S$ may not be $C^1$. Then what regularity condition should I add?
I think $C^*$ is the surface I want. But I have trouble to prove it is actually can be represented as a graph. If I can, then by local chart and usual trick we also do when we study Sobolev space, I could flatten $C^\ast$, i.e., flatten $S$ as I wanted.
Any help is really welcome!