Suppose $X=\bigcup_i X_i$ is a countable union. I'm trying to prove a statement which wikipedia says follows directly from the definition of Hausdorff Dimension: http://en.wikipedia.org/wiki/Hausdorff_dimension#Behaviour_under_unions_and_products
$\text{Hdim}(\bigcup_i X_i) = \inf\lbrace d\ge 0:\inf \lbrace \sum_i r_i^d:\text{there is a cover of} \bigcup_i X_i \text{by balls of radii}\, r_i>0 \rbrace =0\rbrace $
Can someone explain why/how a supremum would arise here? Does it have anything to do with the criticial value of $d$?
Try this formula on simple planar sets. If you take the union of a line segment (dimension 1) and a square (dimension 2), what will be the dimension of the union? In general, it should be more or less intuitive that the dimension is determined by the "thickest" part of the set.
Since the dimension is monotone (larger set $\implies$ bigger dimension), we immediately get $\operatorname{Hdim}(\bigcup X_i)\ge \operatorname{Hdim}(X_i)$ for every $i$, hence $$\operatorname{Hdim}\left(\bigcup X_i\right)\ge \sup_i\operatorname{Hdim}(X_i)$$ The reverse implication takes a bit of work. Let $s=\sup_i\operatorname{Hdim}(X_i)$. It suffices to show (why?) that $\mathcal H^{s+\epsilon}(\bigcup X_i)=0$ for every $\epsilon>0$. Fix $\delta>0$. For each $i$ we can cover $X_i$ by balls such that the sum of their radii to power $s+\epsilon$ is less than $2^{-i}\delta$ (why?). The union of all such balls over $i$ covers $\bigcup X_i$. The conclusion follows (how?).