We know that the zero set of a Brownian motion $(B(t))$, $T:=\{s\in [0,1]:B(s)=0\}$, is almost surely homeomorphic to the Cantor set. I would like to prove that the zero set of the minimum the absolute value of $N$ Brownian motions, $S=\{s\in [0,1]:\min_i |B^i(s)|=0\}$, is also homeomorphic to the Cantor set.
My intuition for why this would be true is something along the following lines. $S=\bigcup_{i=1}^N S_i$, where $S_1,...,S_N$ are $N$ i.i.d copies of $T$. Then taking the union, by some scaling argument, $\bigcup S_i$ should also be homeomorphic to the Cantor set. I expect this because $B(t)$ has the same law $N^{-1/2}B(Nt)$, and the prefactor doesn't make a difference to the zero set, so since $B(Nt)$ is covering $N$ times 'more time' it should have a zero set which is like $N$ 'lots of' $T$, which is like $\bigcup S_i$.
This kind of maths is not my speciality so I would appreciate any help in making this arguments rigorous (if they are true!).
Simple topological or metric arguments, namely stability of some properties by finite unions gives the result. The independence of the $N$ Brownian motions is not required here.
Almost surely, $S$ is a closed set with $0$ Lebesgue measure and without isolated points since $S_1,\ldots,S_n$ are closed set with $0$ Lebesgue measure without isolated points. Therefore $S$ has an empty interior.