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I am trying to solve the problem related to the Sierspinski triangle. The triangle is shown as follow.

enter image description here Let $S$ be the intersection of all the finite stages

a). Show that $S$ is a nonempty compact set. b). Show that $S$ has no interior. c). Show that the boundaries of the triangles at the $n^th$ stage belong to $S$. Hence show that there is a path in $S$ from the top vertex of the original triangle that gets as close as desired (within $sigma$) to point in $S$.

Here is approach thinking process: ( I may be wrong) Answer for a): My approach is based on the Cantor's intersection Theoram and the Heine-Bohl Theorem. Based on the construction of the triangle, we have the decreasing sequence of sets $S(0)\subset S(1) \subset S(2) \subset S(3) \subset ... $ The set is not empty since some points remain in the sets when at most a countable number of points are removed from the sides of the original triangle. By Cantor intersection theorem, if $S_i$ are nonempty decreasing sequence of nonempty compact subsets of $\mathbb R^n$ then the intersection is not empty. By the Heine-Borel Theorem, a subset of $\mathbb R^n$ is compact iff it is closed and bounded. The original triangle is closed and bounded, hence it is compact. Hence $S$ is a nonempty compact set.

Answer b);

Note that the Cantor set has empty interior hence the set $S$ has interior.

Answer c).

I got stuck with c. To prove the boundaries of the triangles at the $n^th$ stage belong to $S$, i guess I could use induction. Since each triangle is bounded, all the triangle we get till $n^th$ stage is bounded. Note that $S-i \subset S$, hence boundaries of each $S_i$ at the $n^th$ stage belong $S$.

I don't know how to continue. Any help is appreciated and if you need more information about the problem let me know.

Thanks in advance!

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  • $\begingroup$ The notation $S(0)\subset S(1) \subset S(2) \subset S(3) \subset … $ indicates an increasing sequence of sets, not a decreasing one. $\endgroup$ – Lee Mosher Oct 31 '14 at 14:43
  • $\begingroup$ My bad it should be the other way. $\endgroup$ – needhelp Oct 31 '14 at 16:02

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