Cumulative Gasket Pictures

Exercise Solutions

2. (ii) Suppose P1 lies in the largest removed triangle, hence inside the equilateral triangle S having vertices V1, V2, and V3.

Denote by T1, T2, and T3 the IFS rules equivalent to the rules of problem 1.

Then P2 = T1(P1), or P2 = T2(P1), or P2 = T3(P1), depending on the corner selected.

Consequently, P2 lies in one of the equilateral triangles T1(S), T2(S), and T3(S), labeled 1, 2, and 3 on the left above.

Say P2 = Ti(P1).

Note the three blue triangles on the right are T1, T2, and T3 of the green triangle, so P2 lies in one of the blue triangles.

Next, P3 = Tj(P2) = Tj(Ti(P1)), so P3 lies in one of the nine triangles 11 = T1(T1(S)), ..., 33 = T3(T3(S)) and so is not in one of the blue triangles.

All subsequent Pn = Tm(Tk( ... Tj(Ti(P1))...)) lie in Tm(Tk(S)) and so is not in one of the blue triangles.

Continuing in the obvious way shows each stage of removed triangles contains exactly one of the Pi.

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