FractionalGeometry-Chap2

Have we found a logical inconsistency that will bring

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: middle ninth, leaving eight 1/9 × 1/9 squares, then removing the middle ninth of these, and continuing. (a) Find IFS rules to generate the Sierpinsk carpet. (b) Find the minimum ǫ for which Oǫ is simply-connected, that is, has no holes. (c) Find the minimum ǫ for which Oǫ has exactly 1 hole. (d) Find the minimum ǫ for which Oǫ has exactly 1 + 8 holes. (e) For every nonnegative integer k , find the minimum ǫ for which Oǫ has exactly 1 + 8 + 82 + · · · + 8k holes. Figure 2.20: The Sierpinski carpet, subject of Prxs 2.2.2. Practice problem solutions PrxsSol 2.2.1 (a) For every ǫ > 0, and for each rational q , there is an irrational within ǫ of q , and for each irrational i, there is a rational within ǫ of i. (This is a simple consequence of the fact that both the rationals and the irrationals are densein [0, 1].) That is, for every ǫ > 0 Q ⊆ Iǫ and I ⊆ Qǫ and so h(Q, I ) ≤ ǫ for all ǫ > 0. That is, h(Q, I ) = 0. (b) The result of (a) certainly appears to contradict positive-definiteness. We have two sets, Q and I , which...
View Full Document

This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

Ask a homework question - tutors are online