Imaginary Hypercubes - Discrete and Computational Geometry and Graphs - page 183

From ( 3 ), the set of vertices of the polytope H n defined in Sect. 3.3 is the inter-
section of D n with the edges of C . Therefore, the convex hull of D n coincides
with H n . Similarly, we can show that the convex hull of T n
∞

is T n .

3 , H n
∞

and T n
∞

Theorem 12. For n are the only fractal imaginary cube

shapes obtained as the limit of an IFS that is composed of 3 n− 1

≥

homothetic

transformations with the scale 1 / 3 .

n satisfies # U =3 n− 1

Proof. Suppose that n ≥ 2 and that U ↂ{− 1 , 0 , 1 } and

Q n

n

∪{

a | a ∈

U } is an imaginary cube of C . We show that there exist b ∈{− 1 , 1 }

n

and r ∈{− 1 , 0 , 1 } such that U = U ( b ,r )for U ( b ,r )= { a ∈{− 1 , 0 , 1 } | a · b ≡ r

. It is clear that such a selection U is congruous to that of H n or T n .
We show this by induction on n , and it is true for n =2.
Note that we have U (

(mod 3) }

b ,r ) if and only if ( b ,r ). Since b ,r )= U ( b ,r )= ± (

simultaneous equations a 1 + a 2 ≡ r 1 , a 1 − a 2 ≡ r 2 (mod 3) always have a solution

± b , then we have

( a 1 ,a 2 )=2( r 1 + r 2 ,r 1 − r 2 ), one can also find that if b =

b ,r ) for any choice of r, r ∈{−

U ( b ,r ) ∩ U ( = ∅ 1 , 0 , 1 } .

Suppose that n ≥ 3. We divide U into three parts

U = U − 1 ×{− 1 }∪ U 0 ×{ 0 }∪ U 1 ×{ 1 } ,

n− 1

satisfies # U i =3 n− 2 Q n− 1

a

where U i ↂ{− 1 , 0 , 1 } and that ∪{ | a ∈ U i }

is an imaginary ( n − 1)-cube for i ∈{− 1 , 0 , 1 } . From the assumption, we can

put U i = U ( b i ,r i )for i ∈{− 1 , 0 , 1 } . Considering the projection in the n -th

direction, we have U i ∩ U j = ∅ for − 1 ≤ i<j ≤ 1. Therefore, we can assume

that b − 1 = b 0 = b 1 =( b 1 ,...,b n− 1 ), and we get { r − 1 ,r 0 ,r 1 } = {− 1 , 0 , 1 } .In

each case, there is b n ∈{− 1 , 1 } such that r 0 ≡ r − 1 − b n ≡ r 1 + b n (mod 3), and

hence we obtain U = U (( b 1 ,...,b n ) ,r 0 ).

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