A note on three dimensional good sets

Let X 1 , X 2 , . . . , X n be nonempty sets and let Ω = X 1 × X 2 × • • • × X n be their Cartesian product. We will write → x to denote a point (x 1 , x 2 , . . . , x n ) ∈ Ω.

For each 1 ≤ i ≤ n, Π i denotes the canonical projection of Ω onto X i .

A subset S ⊂ Ω is said to be good, if every complex valued function f on S is of the form:

for suitable functions u 1 , u 2 , . . . , u n on X 1 , X 2 , . . . , X n respectively ( [3], p. 181).

For a good set S, a subset B ⊂ n i=1 Π i S is said to be a boundary set of S, if for any complex valued function U on B and for any f : S -→ C the equation (1) subject to

admits a unique solution. For a good set there always exists a boundary set ( [3], p. 187).

A subset S ⊂ Ω is said to be full, if S is maximal good set in Π 1 S × Π 2 S × • x and → y . R is an equivalence relation, whose equivalence classes are called related components of S. The related components of S are full subsets of S (ref. [3]).

First we prove that when the dimension n = 3, a full component need not be a related component, by giving an example of a full set with infinitely many related components. Consider a countable set T which consists of the following points:

Call the first three points of T as D 0 and for n 1, let D n denote the first 3 + 5n points of S. Let A 0 = D 0 and for n ≥ 1 let

Then it is easy to see that every D n is good and has three point boundary. All the three points of the boundary of D n cannot come from the coordinates of points in D n-1 : because, if all of them occur as coordinates in D n-1 , they form a boundary for D n-1 . Given any function f on D n , there is a solution u 1 , u 2 , u 3 on D n-1 such that f (w 1 , w 2 , w 3 ) = u 1 (w 1 ) + u 2 (w 2 ) + u 3 (w 3 ), (w 1 , w 2 , w 3 ) ∈ D n-1 .

But then f ( → a 5n+3 ) fixes the value of u 3 (α 5n-2 ) by the following equation:

When we substitute this value of u 3 (α 5n-2 ) in the remaining four points of A n , we get a set of linearly dependent equations. This shows that the boundary of D n contains at least one of the five coordinates, α 5n-4 , α 5n-3 , α 5n-2 , α 5n-1 or α 5n , which are introduced in A n . One can observe the following properties of the points in the set A n : any k points of A n has at least k coordinates introduced in A n . (i.e, they do not occur as coordinates in D n-1 ). If we take a singleton

T is good as every finite subset of T is good. It cannot have a boundary B with more than two points: If |B| = 3, we can choose a n sufficiently large such that all the three points of b occur as coordinates in D n-1 . Then B is a boundary of D n which is not possible as observed above. If |B| > 3, we can choose n sufficiently large so that k = |B ∩ ∪ 3 i=1 Π i D n 4. Then these k points form a boundary of D n which is again not possible. So the boundary of T consists of only two points which shows that T is full.

We prove that no finite subset A of T other than singleton is full:

2, . . . l and k i = 0 for all other i. If k i1 > 1, as no subset other than singleton of A n is full, the set A ∩ A i1 is not full. When we add the points of A ∩ A i2 to A ∩ A i1 ( as we are adding k i2 points) we will be adding at least k i2 new coordinates. So the set A ∩ (A i1 ∪ A i2 ) is not full. Similarly when we keep adding A ∩ A ij to the set A ∩ (∪ k<j A i k ) the number of coordinates added is at least equal to the number of points added. So at each step A ∩ (∪ k≤j A i k ) is not full. In this way we get

in the first step when we add points of A ∩ A i2 to the singleton set A ∩ A i1 the new coordinates added is at least k i2 + 1. So A ∩ (A i1 ∪ A i2 ) is not full. In the remaining steps as we keep adding points from A ∩ A ij , the number of coordinates added is at least equal to the number of points added. So in the end we get A is not full. in F n is the whole set F n . To show that F n is full, consider the matrix M n whose rows correspond to the points

and columns correspond to the coordinates y 1 , y 2 , z 3 , α 1 , α 2 , …, α 5n .This is a 5n + 3 × 5n + 3 matrix:

1 1 0 0 0 0 0 0 . . 0 0 0 0 0 1 0 1 0 0 0 0 0 . . 0 0 0 0 0 0 0 0 1 1 1 0 0 . . 0 0 0 0 0 0 0 0 0 0 1 1 1 . . 0 0 0 0 0 0 0 1 1 0 0 0 1 . . 0 0 0 0 0 0 0 0 0 1 0 1 0 . . 0 0 0 0 0 0 1 0 0 0 1 0 0 . . 0 0 0 0 0 0 0 0 0 0 0 0 0 . . 1 1 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 0 0 0 0 . . 0 0 1 1 1 0 0 0 0 0 1 0 0 . . 1 0 0 0 1 0 0 0 0 0 0 0 0 . . 0 1 0 1 0 0 0 0 0 1 0 0 0 . . 0 0 1 0 0 0 1 1 0 0 0 0 0 . . 0 0 0 1 0

It has an inverse given by

-3 4 0 0 0 0 0 . . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

This shows that F n is full. To show that it is the geodesic between the points Let k = |F n | -|A|. As A is full there exists atleast k coordinates of points of F n which donot occur as coordinates in the points of A. (Because otherwise adding these k points we get F n and we will be adding less than k coord