Hausdorff-dimension of connected closed subsets of R^2 Let $A \subseteq \mathbb{R}^2$ be closed and connected, and for any $c \in \mathbb{R}$, let $A_c \subseteq A$ be a closed and connected subset of $A$, s.t. for $c \neq d$ we have $A_c \cap A_d = \emptyset$.
If the Hausdorff-dimension of each $A_c$ is at least 1, what lower bound can be given for the Hausdorff-dimension of $A$? More generally, if each $A_c$ has Hausdorff-dimension $\varepsilon$, is there a lower bound for the Hausdorff-dimension of $A$ better than $\varepsilon$?
 A: As Joel pointed out, the MO system works better with answers (or hints) as answers rather than as comments. Even though Joel did not suggest giving a more complete answer, I will do so.
Let $C$ be a Cantor set with Hausdorff-dimension $0$. Then $A = (C \times \mathbb{R}) \cup ([0,1]\times\{0\})$ has Hausdorff-dimension $1$ and it can be written as a disjoint union
$$
A = \bigcup_{c \in \mathbb{R}}A_c,
$$
where $A_c$ are of the form 
$$(\{a\} \times \mathbb{R}) \cup (\{b\} \times \mathbb{R}) \cup ([a,b] \times \{0\})$$
if $a$ and $b$ are the boundary points of an interval which gets removed in the construction of the Cantor set, or of the form
$$\{a\} \times \mathbb{R}$$
if $a$ is a point in the Cantor set which is not such a boundary point. Parametrization for the collection $A_c$ can be obtained for example by relating a dyadic decomposition of $\mathbb{R}$ with the construction of $C$.
As for the more general case. First of all there are no closed connected subsets of $\mathbb{R}^2$ with Hausdorff-dimension strictly between $0$ and $1$. (Assume that a connected set has at least two points. Take a line such that the orthogonal projection of the set to the line is not a singleton. The projected set must also be connected, so it is an interval. The projection can only decrease the Hausdorff-dimension so the original set has dimension at least $1$.) If $\epsilon = 0$, then $A$ has dimension at least $1$ (because it has at least $2$ points). 
So, we are left with the case $\epsilon > 1$ for which I do not have an answer yet.
