Area Measure

Strangely, the total area of the parts removed in the construction of the gasket is equal to the area of the equilateral triangle we started with. In the first step we removed 1/4 of the area. In the next step we took out three triangular regions each of which had 1/42 the area of the original. Continuing this series, we find that the total fraction of the area removed was

1/4 + 3(1/42) + 32(1/43) + . . . + 3(n-1)(1/4n) + . . .

This infinite series converges quite quickly to 1, as shown by the following cell which produces a table of partial sums s[n] for n=10,20,...,80.

Input := 

Clear[n,a,k]
a[k_]:=3^(k-1)/(4)^k;1/(1-x);  
s[n_]:=Sum[a[k],{k,1,n}]//N;
Table[{n,s[n]},{n,10,80,10}]//TableForm

Output =

10   0.943686

20   0.996829

30   0.999821

40   0.99999

50   0.999999

60   1.

70   1.

80   1.

This sum can also be found analytically using the fact that

1/(1-x) = 1 + x + x2 + ..., if |x|<1.

Since the area removed is equal to the area of the original triangle, the remaining set, the Sierpinski Gasket, has zero area measure. Thus, S is a perfect gasket because it divides its complement (the white regions in the pictures above) into infinitely many triangular regions without using any thickness to do so.