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# Some Dimension One Cantor Sets

If we produce a Cantor set by deleting middle halves, we get C a Cantor set of dimension -1/log2(1/4)=1/2. C consists of points with base four expansions that use only 0's and 3's. Thus, C×C in R2, is a Cantor set of dimension one (see the first frame in Figure 8). Sandefur ([San, p. 120]) pointed out how one can intuitively see this by shifting the pieces of C×C to fill up the unit interval. His approach is animated in Figure 8.

Figure 8. A Dimension One Cantor Set shifting onto an interval by Sandefur's method. Frames 1 and 9 of 17 are shown. Also available is an animation (58 kb) and a speed controlled version, The coloring should help the viewer to follow certain points. Source code is available.

Here's another way to see this. If we project the points parallel to the line 2x+y=0 note that the first approximation given by C1×C1 has image equal to the interval [0,1.5]. There are only three pairs of points from different components that get mapped to the same point, for example (0,3/4) and (1/4,1/4). At the next level, C2×C2 is also projected onto [0,1.5] with only finitely many pairs of points in different components mapping to the same point. And for C×C we project onto [0,1.5] and if we delete countably many points (e.g., all the "lower left" corners from every stage Ck×Ck except for (0,0)) the projection is one-to-one.

Figure 9. A Dimension One Cantor Set with a projection onto an interval. The green slanted lines indicate the direction of projection. Shown are the starting frame and a partially projected frame. Also available is an animation (113kb) and a speed controlled version. Use the browser "back" button to return. Source code is available.

Analytically, one can describe this projection as follows. The points of C×C are the points (a,b) with both coordinates base 4 expansions (say a=.a1a2a3... and b=.b1b2b3... ) using only 0's and 3's. We map (a,b) to c=2a+b=(2a1+b1)/4+ (2a2+b2)/16+(2a3+b3)/64+ ···. Write the base 4 expansion of c/3=.c1c2c3.... Note that in each of the four possible cases for (ak,bk) we get 2ak+bk divisible by 3 and so:

• ck= 0, if (ak,bk)=(0,0)
• ck= 1, if (ak,bk)=(0,3)
• ck= 2, if (ak,bk)=(3,0)
• ck= 3, if (ak,bk)=(3,3)
Thus f(x,y)=2x+y maps C×C onto the interval [0,3]. Since f/2 is the desired projection, the projection maps onto the interval [0,1.5].

As Falconer points out ([Fa, p. 75]) projections cannot increase dimension. Thus the dimension of C×C must be at least one. (We already know that it equals one, but this gives extra corroboration that this meager set has positive dimension.)

Communications in Visual Mathematics, vol 1, no 1, August 1998.