© 1997 Copyright by Sonia Sabogal. All rights reserved.

Question

The Sierpinski Triangle Sonia Sabogal

The Sierpinski Triangle is obtained as the residual set remaining when one begins with a triangle and applies the operations of dividing it into four triangles, connecting the middle points of the triangle's edges, and omitting the interior of the center one, then repeats this operation on each of the surviving 3 triangles, then repeats again on the surviving 9 triangles, and so on indefinitely. This space is homeomorphic to the unique nonempty compact set K of the complex plane that satisfies: K = w₁(K) \cup w₂(K) \cup w₃(K) where w₁, w₂, w₃ are maps of the complex plane defined by: w₁(z) = 0.5z, w₂(z) = 0.5z+0.5 and w₃(z) = 0.5z+0.5i (see W. J. Charatonik and A. Dilks; Topol. and its Appl.; 55(1994), 215-238, Example 2.7).

It is known that Cantor's space is the only totally disconnected, perfect compact metric space (up to homeomorphism), or that the unit interval is the only metric continuum with exactly two noncut points (up to homeomorphism) or that the circle S¹ is the only metric continuum K such that for any two points a and b, K-{a,b} is not connected (up to homeomorphism). G. T. Whyburn (see: G. T. Whyburn; Fund. Math; 45 (1958), 320-324) provides a topological characterization of the Sierpinski plane universal curve (this curve is obtained as the residual set remaining when one begins with a square and applies the operation of dividing it into nine equal squares and omitting the interior of the center one, then repeats this operation on each of the surviving 8 squares, then repeats again on the surviving 64 squares, and so on indefinitely) as the only plane locally connected 1-dimensional continuum S such that the boundary of each complementary domain of S is a simple closed curve and no two of these complementary domain boundaries intersect.

Question: Is there a known topological characterization of Sierpinski triangle?