Sierpinski Relatives: Straight Lines

Using the relative FCF as an example I was shown a simple way to prove there was a vertical line through S1 (bottom right) and S3 (top right). In order for this to work the x-components had to match, and the y values had to go from [0,1/2] for F1 and [1/2,1] for F3.

F1F(-1/2x+1/2, 1/2y) F3(-1/2x+1/2, 1/2y +1/2)

As you can see these two maps satisfy both properties. Since F2 was not used in this proof at all, an relative with F as the map for F1 and F3 will have a straight line through S1 and S3.

We can use almost the same technique to prove there is a horizontal line through S1 and S2 (bottom left). Instead the Y components must match and the X values must got from [0,1/2] for F1 and [1/2,1] for F2. The relative with the form EE_ will have a horizontal line through S1 and S2.

F1E(1/2x, -1/2y+1/2) F2E(1/2x+1/2, -1/2y+1/2)

-1/2y+1/2 = -1/2y+1/2 The y values match

1/2x=x ... x=0
1/2x+1/2=x ... x=1, x values go from [0,1]