17
$\begingroup$

A set $X\subset \mathbb{R}$ is called nice if, for every $\epsilon > 0$, there are a positive integer $k$ and some bounded intervals $I_1,I_2,...,I_k$ such that $X \subset I_1 \cup I_2 \cup \cdots \cup I_k$ and $\sum\limits_{j=1}^k |I_j|^{\epsilon} < \epsilon$.
Prove that there exist sets $X,Y \subset [0,1]$, both of them nice, such that $X+Y = [0,2]$, where $X+Y:=\{x+y\mid x\in X,y\in Y\}.$

This question is from an iberoamerican exam for undergraduate students (ciim 2010).

An attempt to solve this problem can be found at aops but it doesn't seem to be complete or correct. Any help is welcome.

$\endgroup$
  • 4
    $\begingroup$ Take $X=Y=C$, where $C$ is the middle-thirds Cantor set; there are elementary proofs here. $\endgroup$ – Brian M. Scott Jun 6 '15 at 16:40
  • 1
    $\begingroup$ Not zero-dimensional set, but set of measure zero! $\endgroup$ – Yiorgos S. Smyrlis Jun 6 '15 at 17:04
  • $\begingroup$ As @Yiorgos points out, the problem statement is wrong: the definition given in the problem is for sets of measure zero, not for zero-dimensional sets. The solution that I suggested in my comment is both zero-dimensional and of measure $0$. $\endgroup$ – Brian M. Scott Jun 6 '15 at 17:16
  • 1
    $\begingroup$ There is an $\varepsilon$ in the exponent as well, so this makes it stronger than being measure $0$, maybe? $\endgroup$ – Henno Brandsma Jun 6 '15 at 20:01
  • 2
    $\begingroup$ @jack: Ah, yes: I didn’t notice the exponent. I’ll leave the Cantor set suggestion there for now in case someone can use it as a starting point. $\endgroup$ – Brian M. Scott Jun 6 '15 at 21:03
1
$\begingroup$

The middle thirds Cantor Set doesn't work since at stage $N$ there are $2^N$ intervals of length $\frac{1}{3^N}$. So for some $\epsilon$ the Hausdorff dimension of this Cantor set, the series is more than $0$:

$$ \left(\frac{2}{3^{\epsilon }}\right)^N =1 \longrightarrow \epsilon = \frac{\log 2}{\log 3} $$

It is also the limit set of the two functions $T_1(x) = \frac{x}{3}$ and $T_2(x) = \frac{x}{3} + \frac{2}{3}$.

If we remove successively smaller percentages, we can get Cantor sets of positive Lebesgue measure. The Smith-Volterra-Cantor set is nowhere-dense and has Lebesgue measure $\frac{1}{2}$.

These disasters come up when you try to show $\int_a^b f'(x) \, dx = f(b) - f(a)$ for certain trig series.


Considering this is like the Putnam exam, let's try to take "aggressive" Cantor sets. Instead of removing the middle $\frac{1}{3}$ at each stage, let's remove the middle $1-\frac{1}{N}$. Now after stage $N$ there are $2^N$ sets, but these have measure:

$$ 2^N \cdot \left(\frac{1}{N!}\right)^\epsilon \approx \left(\frac{2 e^\epsilon}{N}\right)^N \to 0$$

and this is true for all $\epsilon$. Using iterated functions we are using $T_{1,N}(x) = \frac{x}{N}$ and $T_{2,N}(x) = \frac{x}{N} + (1-\frac{1}{N})$. Here are some notes on dimension theory.


Here a lot of work has gone into constructing nice sets, but we need two nice sets $X,Y\subset [0,1]$ with $X + Y = [0,2]$. Necessarily $0,1 \in X \cap Y$.

It is certainly true that $C + C = [0,2]$ I don't think this is true for the set I have constructed.

Another route might be to try building "Cantor sets" using continued fractions whose digits are bounded or avoid a certain number.

$$ A_k = \{ [a_1, \dots, a_k, \dots] : 0 \leq a_i < k \} \subset \mathbb{Q} $$

These sets may be sparse enough and have the sum property you desire.

$\endgroup$
  • $\begingroup$ It is indeed, not true at all for the Cantor set you have constructed. In fact, one can show for a large family of Cantor sets, there is a dichotomy between the sets that are "too large" (not nice) or "too small" (not satisfying $X+Y=[0,1]$). I'll write more about this if I have the time $\endgroup$ – Ewan Delanoy Jun 15 '15 at 16:42
  • 1
    $\begingroup$ @john Consider the continued fractions $A = \{ [0; a_1, a_2, \dots] : 1 \leq a_i \leq 4, 1 \leq i \}.$ Can you show that A is nice? If A is nice I can solve the problem. $\endgroup$ – jack Jun 17 '15 at 18:46
  • $\begingroup$ @Ewan Following the suggestion of John Mangual, do you know if the set of continued fractions $A = \{ [0; a_1, a_2, \dots] : 1 \leq a_i \leq 4, 1 \leq i \}$ is nice? I'm not sure if it's nice, but if the answer is positive I think I can solve the problem. $\endgroup$ – jack Jun 19 '15 at 14:54
  • $\begingroup$ @jack any reason in particular $4$? $\endgroup$ – cactus314 Jun 19 '15 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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