I am thinking about the following: https://users.math.yale.edu/public_html/People/frame/Fractals/FracAndDim/DimAlg/intersection.html states that most placements of two fractals whose dimensions add to more than the dimension of the whole space are so that the two have nonempty intersection. But let's think about two $\alpha$-stable subordinators, one started in zero, one started in one. Their range is of dimension $\alpha$, so for $\alpha>\frac12$ the intersection of their ranges should be of dimension $2\alpha -1$. But how does one proof that this indeed happens with probability 1? Is there some elementary geometric argument?
All the best and thx in advance!
Edit My basic idea is the following: if a set $F$ has dimension $\alpha$ and is embedded in $\mathbb R^1$ its codimension is $1-\alpha$, but $\mathbb R^1 \backslash F$ still contains sets of dimension $1-\alpha+ \varepsilon$. Or just think of a plane $P$ in $\mathbb R^3$ which has codimension $1$ but obviously $\mathbb R^3\backslash P$ still contains sets of dimension $2$, namely planes. And obviuosly $\mathbb R^3$ contains more sets of dimension 2 than $\mathbb R^3\backslash P$. But one has to measure this somehow with respect to the law of $\alpha$-stable subordinators I suspect.