Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\varepsilon$?

share|improve this question
    
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). –  Tapio Rajala May 10 '11 at 10:22
1  
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? –  Tapio Rajala May 10 '11 at 11:18
    
I agree with Tapio (hi Tapio!). There are some easy constructions where A has dimension 1. Perhaps some background or motivation would help. –  Pablo Shmerkin May 10 '11 at 11:26
    
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 $0$. –  Tapio Rajala May 10 '11 at 11:31
1  
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. –  Joel David Hamkins May 10 '11 at 13:22
show 1 more comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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. –  Arno May 25 '11 at 14:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.