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Next, let i be continuous Let O be an open set in τ Then since i is continuous, i−1(O) is open in (X,τ0) Since i−1(O) = O, τ0 is finer than τ Proof B) Assume τ = τ0 Then τ0 is finer than τ By part A, i is continuous Since τ is also finer than τ0, i−1 is also continuous It is clear that i is both onetoone and ontoB n = O(log n) For a>1, this yields f (n)=ak f (1) c ak − 1 a − 1 = O(nlog b a) If n is not a power of b, then estimate with the next power of b to get the claimed bounds Title recurrence3key Author Andreas Klappenecker Created DateThe Euclidean algorithm to compute gcd ( a, b) can be described as follows If a b note that X a − 1 = ( X b − 1) X a − b ( X a − b − 1), hence every step of the Johnhcochrane Com ƒ~ƒbƒNƒX 1 "Ô ‰Âˆ¤‚¢ Œ¢
