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 $\varepsilon$?

  • $\begingroup$ Some easy observations: Without the assumption that $A_c$ are closed, constructing such a set $A$ with Hausdorff-dimension $1$ would be easy. The more general case for $0 < \epsilon < 1$ is also easy (by noticing the trivial lower bound for the dimension of a connected set with at least two points). $\endgroup$ May 10, 2011 at 10:22
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    $\begingroup$ After giving it another thought, the set $A$ which I had in mind in the previous comment can easily be decomposed into closed and connected $A_c$ with Hausdorff-dimension $1$. However, I hesitate to give the example as the problem looks much like a homework problem. Arno, how did you end up considering this problem? $\endgroup$ May 10, 2011 at 11:18
  • $\begingroup$ I agree with Tapio (hi Tapio!). There are some easy constructions where A has dimension 1. Perhaps some background or motivation would help. $\endgroup$ May 10, 2011 at 11:26
  • $\begingroup$ I continue my monologue by noting that I was a bit too quick with thinking about the examples. I accidentally came up with a stronger example than needed. (Such that $A = \bigcup_{c \in \mathbb{R}} A_c$.) The original question only required $A_c \subset A$. For that it is easier to come up with an example. Hint: start from an uncountable closed set in $\mathbb{R}$ with Hausdorff-dimension $0$. $\endgroup$ May 10, 2011 at 11:31
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    $\begingroup$ Tapio and Pablo, despite easy counterexamples, I would encourage you to post your answers as answers rather than as comments. The MO system works best when answers are given as answers. $\endgroup$ May 10, 2011 at 13:22

1 Answer 1


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.

  • $\begingroup$ Thanks a lot for the answer! The case $\varepsilon > 1$ would only have been of interest to me if the base case would yield a lower bound $> 1$. To explaim my motivation: I am studying the problem of computing an element of a closed connected set from a description of the set. I conjectured that this is easier in dimension 2 than in 3+, and hoped that the Hausdorff-dimension could be used to prove that closed connected subsets of $\mathbb{R}^2$ are in some sense less complicated. The counterexample does not disprove my original conjecture, but demonstrates I'll need a different proof. $\endgroup$
    – Arno
    May 25, 2011 at 14:37

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