VSFCF  CVM 1.1 
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 C_{1}×C_{1} 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, C_{2}×C_{2} 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 C_{k}×C_{k} except for (0,0)) the projection is onetoone.
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=.a_{1}a_{2}a_{3}... and b=.b_{1}b_{2}b_{3}... ) using only 0's and 3's. We map (a,b) to c=2a+b=(2a_{1}+b_{1})/4+ (2a_{2}+b_{2})/16+(2a_{3}+b_{3})/64+ ···. Write the base 4 expansion of c/3=.c_{1}c_{2}c_{3}.... Note that in each of the four possible cases for (a_{k},b_{k}) we get 2a_{k}+b_{k} divisible by 3 and so:
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.)

