Let $E$ and $F$ be two non-empty subsets of $\mathbb{R}^n$ for some $n \in \mathbb{N}$.
I'm wondering if there's a simple formula or bound for the Hausdorff dimension of $E \cap F$ in terms of the Hausdorff dimension of $E$ and the Hausdorff dimension of $F$.
My instinct tells me that we could write
$$\dim_H E \cap F \leq \min (\dim_H E,\ \dim_H F)$$
but this only comes from my (probably naive) intuition. For example, taking a line which intersects a plane, the Hausdorff dimension of the intersection can only be 0 (intersects at a point) or 1 (line is contained in the plane), both of which are less than or equal to 1, the Hausdorff dimension of the line. Clearly their intersection cannot be any more than 1 dimension, and I think this sort of reasoning can apply to general sets $E$ and $F$.
Does this actually hold for general subsets $E, F \subset \mathbb{R}^n$, and if so, how would I make the above argument rigorous?
Best Answer
Compare the following with the formal definition on Wikipedia (also linked on its nLab page): Let $X\subseteq Y$ be subsets of a metric space. Every countable open cover $(U_i)_{i\in \mathbb{N}}$ of $Y$ with $\operatorname{diam}(U_i)<\delta$ for every $i\in I$ is also a countable open cover of $X$, hence $H_\delta^d(X)\leq H_\delta^d(Y)$. Since limits preserve (weak) inequalities, we have $H^d(X)\leq H^d(Y)$. Since outer measures are always positive, we have $H^d(Y)=0\Rightarrow H^d(X)=0$ or $\{d\geq 0|H^d(Y)=0\}\subseteq\{d\geq 0|H^d(X)=0\}$. Applying the infimum yields $\dim_H(X)\leq\dim_H(Y)$. According to my comment above, this yields your formula, when being applied to the inclusions $E\cap F\subseteq E$ and $E\cap F\subseteq F$.