We have shown that the theorem holds in all cases.
Proof: The last assertion was shown in theorem 7.7, and the second statement is clear, and it is also clear that if .
Suppose that .
I will show that
To prove (7.65),
let be a precision function for
. and hence , which is what we wanted to show.
Proof: We saw in theorem 3.71 that for all . If , . This sequence diverges, since it is not bounded. If , then by the previous theorem , so
area , and side . Note that an equilateral triangle with side has area . Starting with , we will now construct a sequence of polygons. will have sides, all having length . We let (so has sides of length ). To construct from we attach an equilateral triangle with side of length to the middle third of each side of .
A horizontal side of might be replaced by . Each side of is replaced by 4 sides of length , so will have sides of length . The figure shows some of these polygons. I will call the polygons snowflake polygons. We have for all . The snowflake is the union of all of the sets ; i.e., a point is in if and only if it is in for some .
Find the area of (in terms of the area of ), for example