Ex3-3

Chapter 3 Question 3

Question: Construct a two-dimensional Cantor set in the unit square $$\{(x,y):0\leq x,y\leq 1\}$$as follows. Subdivide the square into nine equal parts and keep only the four closed corner squares, removing the remaining region (which forms a cross). Then repeat this process n times for a suitably scaled version for the remaining squares, ad infinitum. Show that the resulting set is perfect, has plane measure zero, and equals $$C\times C$$.

Proof: Let $$C_1=[0,1/3]\cup [2/3,1]$$ as in the textbook for constructing Cantor set, let $$D_1$$be the union of four disjoint closed squares as described in the question, then we find that $$(x,y)\in D_1$$if and only if $$x\in[0,1/3]\cup [2/3,1]$$ and $$y\in[0,1/3]\cup[2/3,1]$$, which means precisely $$D_1=C_1\times C_1$$.

Now let $$D_2$$ be the result of subdividing each of the four squares, so it is a union of $$16$$disjoint squares. We choose $$C_2=[0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1]$$, as the same in the textbook, then by looking at the graph visually we may conclude $$D_2=C_2\times C_2$$. We claim that $$D_n=C_n\times C_n$$ for all $$n$$. For $$n=1$$ the assertion is already true. Assuming $$D_{n-1}=C_{n-1}\times C_{n-1}$$ for some $$n$$. From Exercise 2(a), we know both of $$x,y$$ satisfying $$(x,y)\in D_{n-1}$$ have some triadic expansion $$0.a_1\dots a_{n-1}a_{n}\dots$$ such that $$a_i$$ only takes value from $$0,2$$ for $$1\leq i\leq n-1$$. The process of cancelling crosses from $$D_{n-1}$$ to form $$D_n$$ means exactly we can further choose $$a_n$$ to be $$0, 2$$ only, hence $$D_n\subset C_n\times C_n$$. On the other hand if $$(x,y)\in C_n\times C_n$$, then $$x,y$$, up to their $$n$$th triadic expansions' integer can be expressed using only $$0,2$$, hence $$(x,y)$$ is not only contained in some square in $$D_{n-1}$$, but it also positioned at one of the four smaller corner squares, hence $$(x,y)\in D_n$$. This proves $$C_n\times C_n\subset D_n$$.

We define the 2-dimensional Cantor set as $$D=\bigcap_{n=1}^\infty D_n$$, and we claim that $$D=C\times C$$. For each $$n$$we have $$C_n\subset C$$, so $$D_n=C_n\times C_n\subset C\times C$$. Thus $$D\subset C\times C$$. On the other hand, let $$(x,y)\in C\times C$$, then $$x, y$$ are both contained in every $$C_n$$, thus $$(x,y)\in D_n$$ for each $$n$$, proving $$C\times C\subset D.$$

Now we observe some facts: when we perform each subdivision of squares from $$D_n$$ to $$D_{n+1}$$, it will make each square become 4 smaller squares, hence by noting $$D_1$$ has 4 disjoint squares, we deduce $$D_n$$ has $$4^n$$ disjoint squares. Moreover, each square in $$D_n$$has area 9 times the area of each square in $$D_{n+1}$$, hence by noting the area of each square in $$D_1$$is $$1/9$$, we deduce that each square in $$D_n$$has area $$1/9^n$$. Therefore, the total area, or, the outer measure of each $$D_n$$, is exactly $$(4/9)^n$$. We also have $$|D|_e\leq |D_n|_e=(4/9)^n$$ for all $$n$$, thus $$D$$ is measurable and has measure zero, by Example 2 above Theorem 3.12.

Finally, we prove $$D$$ is perfect by proving that each point $$(x,y)$$in $$D$$ is a limit point of $$D$$. For each square in $$D_n$$has side length $$1/3^n$$, so two points in a same square have a distance at most $$\sqrt2/3^n$$between them. Next, we observe that the four vertices of every square in $$D_n$$ are also points of $$D$$, since the subdivision operation never canceled out the four vertices of each square. Therefore, given $$u=(x,y)\in D$$, we find a square vertex $$u_1=(x_1,y_1)$$ from $$D_1$$such that $$u,u_1$$ lie in a same square in $$D_1$$. Then for each $$n$$ we find a square vertex $$u_n=(x_n,y_n)$$ such that $$u,u_n$$ lie in a same square in $$D_n$$. We then conclude

$$\lVert u-u_n\rVert \leq \dfrac{\sqrt2}{3^n}\quad\text{ for all }n=1,2,\ldots,$$

which means $$\{u_n\}$$ is a sequence in $$D$$ converges to $$u$$.