Show that for any Cantor set $C \subset R^n$ there exists a submanifold $M \subset R^{n+1}$ such that $M \cap (R^n \times \{0\}) = C$.
I think $M$ should not be transverse to $R^n \times \{0\},$ otherwise the intersection $C$ would be a manifold. This suggests that for each $p \in C \subset (R^n \times \{0\})$, the candidate manifold will only hit $p$ but not pass thorough it. For example, if $n=1$. Then for $p \in C$, $M$ on the neighborhood of $p$ would be some curves such that its vertex is $p$. However, I don't know how to rigorously construct such manifold (and am also quite dubious about its validity). Any comments will be much appreciated.