(New page: <math>\text{Show: Given r } \in [-1,1] \text{, show there exist elements in the Cantor set } x,y \text{ such that } x-y=r.</math> <math>\text{Proof: Let } \mathcal{C} \text{ denote the Can...)
 
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=Solution for HW2.17, MA598, Weigel, Summer 2009=
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<math>\text{Show: Given r } \in [-1,1] \text{, show there exist elements in the Cantor set } x,y \text{ such that } x-y=r.</math>
 
<math>\text{Show: Given r } \in [-1,1] \text{, show there exist elements in the Cantor set } x,y \text{ such that } x-y=r.</math>
 
<math>\text{Proof: Let } \mathcal{C} \text{ denote the Cantor set.  Define } f: </math> <math> \mathcal{C} \times \mathcal{C} \rightarrow [0,1] \text{ by } (x,y) </math> <math> \mapsto \frac{x+y}{2}. </math>  
 
<math>\text{Proof: Let } \mathcal{C} \text{ denote the Cantor set.  Define } f: </math> <math> \mathcal{C} \times \mathcal{C} \rightarrow [0,1] \text{ by } (x,y) </math> <math> \mapsto \frac{x+y}{2}. </math>  
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<math>\exists x, y \in \mathcal{C} \text{ s.t. } r+1 = x+y \Rightarrow r=x-(1-y).  </math>
 
<math>\exists x, y \in \mathcal{C} \text{ s.t. } r+1 = x+y \Rightarrow r=x-(1-y).  </math>
 
<math>\text{  Since } 1-y \in \mathcal{C} \text{ by symmetry, } \square. </math>
 
<math>\text{  Since } 1-y \in \mathcal{C} \text{ by symmetry, } \square. </math>
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Revision as of 05:09, 11 June 2013


Solution for HW2.17, MA598, Weigel, Summer 2009

$ \text{Show: Given r } \in [-1,1] \text{, show there exist elements in the Cantor set } x,y \text{ such that } x-y=r. $ $ \text{Proof: Let } \mathcal{C} \text{ denote the Cantor set. Define } f: $ $ \mathcal{C} \times \mathcal{C} \rightarrow [0,1] \text{ by } (x,y) $ $ \mapsto \frac{x+y}{2}. $ $ \text{ Now f is clearly onto by examining the ternary representation of an element of } [0,1]. \text{ Given } r \in [-1,1], \frac{r+1}{2} \in [0,1] \Rightarrow $ $ \exists x, y \in \mathcal{C} \text{ s.t. } r+1 = x+y \Rightarrow r=x-(1-y). $ $ \text{ Since } 1-y \in \mathcal{C} \text{ by symmetry, } \square. $



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