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_1be the union of four disjoint closed squares as described in the question, then we find that (x,y)\in D_1if 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 16disjoint 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 nth 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 nwe 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_nhas area 9 times the area of each square in D_{n+1}, hence by noting the area of each square in D_1is 1/9, we deduce that each square in D_nhas 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_nhas side length 1/3^n, so two points in a same square have a distance at most \sqrt2/3^nbetween 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_1such 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.