Cantor Sets Inside Cantor Sets

Question

(Or: "I heard you liked Cantor Sets...")

I'm working on a student project, and the following construction came up very naturally: If $C$ is the usual Cantor Set, build a countable union of copies of this set as follows: start from $C$, and add a one-third scale copy of $C$ in the interval $[1/3,2/3]$, add three one-ninth scale copies of $C$ in the intervals $[1/9, 2/9]$, $[4/9,5/9]$ and $[7/9, 8/9]$, etc.

To Clarify: basically, every time you remove an interval in the classical construction, in this version, you do not remove the whole interval but you replace it with an appropriately scaled copy of the Cantor set. But you also add Cantor sets in the "holes of the holes", so to speak. So that's why in the second stage, there is a copy of $C$ in $[4/9,5/9]$, that fills in the middle third gap that appears in the copy of $C$ that was added at the first stage.

I guess the best way to see it is that at step $n$, you add to the set previously constructed copies of $C$ scaled at $3^{-n}$ in every empty interval where such a copy will fit.

I have my reasons for wanting to look at this construction, but that got me wondering: it looks so natural that it may very well come up in several contexts.

So. Anyone knows where this construction first appeared? Is it especially notable? Does it illustrate any especially interesting property? I would hate to miss something good about it.

Answer 1

The set you are considering, obtained by "filling the gaps" of the Cantor set $C$ by rescaled copies of $C$, is a dense F_σ set of null Lebesgue measure. It may be equivalently described as the invariant superset of $C$ generated by the (non-commuting) mappings $x\mapsto x/3$, $x\mapsto x/3+1/3$, $x\mapsto x/3+2/3$. So, as Andreas says, it can also be described as the set of all points in the unit interval that admit a ternary expansion with finitely many 1's . The complement is a full-measure G_δ set (thus, of course, uncountable: there are uncountably many ways of having infinitely many 1's in one's ternary expansion).

Your construction is customary in the handicraft of examples and counter-examples in Topology and Measure Theory. For instance, a variant of it (with fat Cantor sets) produces a set $S$ that sub-divides any non-empty open set $A$ into two parts of positive measure: $|A\cap S| > 0$ and $|A\setminus S| > 0$.

Answer 2

The sets you describe should appear as blow-ups (rescaled limits) of the original Cantor set. I'll try to illustrate the basic idea at one of the endpoints of the Cantor set.

Let $E\subset\mathbb{R}^n$ be a closed set and let $x\in E$. We say that a closed set $F\subset\mathbb{R}^n$ is a blow-up of $E$ centered at $x$ if there exists a sequence of scales $r_i\rightarrow 0$ such that $$\lim_{i\rightarrow\infty} HD\left[\frac{E-x}{r_i} \cap B(0,t), F\cap B(0,t)\right]=0\quad\forall t>0$$ where the Hausdorff distance between compact sets $A$ and $B$ is given by $$HD[A,B]=\max\left(\sup_{a\in A} \mathrm{dist}(a,B),\ \sup_{b\in B} \mathrm{dist}(b,A)\right).$$ A closed set admits blow-ups at every point in the set by Blaschke's selection theorem. Cantor sets provide some basic examples where blow-ups of a set at a point are not unique (i.e. depend on the choice of scales).

Let $C\subset[0,1]$ denote the Cantor Middle-Thirds Set, i.e.\ $C=\bigcap_{i\geq 0} C_i$ where $C_0=[0,1]$ and $C_{i+1}$ is the closed set obtained by removing the ``middle-third'' of each connected component (interval) of $C_i$. The Cantor Middle-Thirds Set has a rich tangent structure. By self-similarity, $$ C=\frac{1}{3}C\cup\left(\frac{2}{3}+\frac{1}{3} C\right)=\frac{1}{3} C\cup \left(\frac{6}{9}+\frac{1}{9}C\right)\cup \left(\frac{8}{9}+\frac{1}{9}C\right).$$ Let $x=0$, let $r_i=3^{-i}$ for each $i\geq 0$ and let $s_i=(7/9)r_i$ for all $i\geq 0$. Then $C$ has distinct blow-ups centered at $0$ as $i\rightarrow\infty$ along the sequences $r_i$ and $s_i$ (see picture). To verify this assertion, simply note that $$ \frac{C}{r_i}\cap B_1=C\quad\text{ for all } i\geq 0,$$ while $$ \frac{C}{s_i}\cap B_1 =\frac{9}{7}\left[\frac{1}{3}C\cup \left(\frac{6}{9}+\frac{1}{9} C\right)\right]\neq C\quad\text{for all }i\geq 0.$$ This demonstrates that the blow-ups of a closed set at a point are not always unique.

Two blow-ups of Cantor set http://www.freeimagehosting.net/newuploads/7ukn3.png

The second blow-up of the Cantor set (along the scales $s_i$) have the effect of "filling in" a scaled copy of the Cantor set. My example is simpler than the example cited by the original poster. However, by blowing-up the Cantor set centered at other points of $C$, one should be able to get back to the original poster's example. I would argue that this means the original poster's example is quite natural, because it appears at part of the "blow-up" or "tangent structure" of the Cantor set.