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# The inductive process is complete the sequence c n

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The inductive process is complete; the sequence { C n } constructed. It being obvious that each C n is compact and that C n C n +1 , we conclude that the set E = \ n =0 C n is also compact. We also have because { C n } is a decreasing sequence of sets of finite measure m ( E ) = lim n →∞ m ( C n ) = lim n →∞ 2 n X k =1 m ( I ( n ) k ) = lim n →∞ 2 n γ = lim n →∞ (1 - α ) + α 2 3 n = 1 - α. Finally let us see that [0 , 1] \ E is dense in [0,1]. It suffices to see that if 0 a < b 1 then the interval ( a, b ) 6⊂ E ; i.e., E contains no non-empty intervals. In fact, in this case, if x [0 , 1], ² > 0, there must be y [0 , 1] \ E such that y ( x - ², x + ² ); i.e., there is y [0 , 1] \ E such that | x - y | < ² . So, for a contradiction, assume 0 a < b 1 and ( a, b ) E . In the case of the Cantor set, the contradiction was immediate, here it takes a few more steps. Since E = T n C n , we will have that ( a, b ) C n for all n N 0 . So for every n N 0 , ( a, b ) S 2 n k =1 I ( n ) k . Claim There is k such that ( a, b ) I ( n ) k . In fact, if not, then ( a, b ) has non empty intersection with at least two of these intervals, say with I ( n ) k 1 and I ( n ) k 2 , k 1 6 = k 2 . Let G be the union of all the intervals I ( n ) k with k 6 = k 1 , so I ( n ) k 2 G and I ( n ) k 1 G = . Since ( a, b ) I ( n ) k 1 G , I ( n ) k 1 and G are closed, and ( a, b ) I ( n ) k 1 6 = , ( a, b ) G 6 = , we have disconnected the connected set ( a, b ), and this cannot really be done. That is, we have reached a contradiction by assuming the claim to be false. The claim is established. Now ( a, b ) I ( n ) k implies b - a = m (( a, b )) m ( I ( n ) k ) = γ n = 1 - α 2 n + α 3 n . Since this is true for all n N 0 and the expression on the extreme right goes to 0 as n → ∞ , we either have b a or a contradiction. Since we assumed b > a , we have a contradiction, and we are done. 7
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• Spring '11
• Speinklo
• CN, Lebesgue measure, open intervals

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