Introduction to Mathematical Analysis I - Second Edition

81 In the first case, set a 2 = a 1 and b 1 = c 1 . In the second case, set a 2 = c 1 and b 2 = b 1 . Now set I 2 = [ a 2 , b 2 ] . Note that in either case, f ( a 2 ) < γ < f ( b 2 ) . Set c 2 = ( a 2 + b 2 ) / 2. If f ( c 2 ) = γ , again we are done. Otherwise, either f ( c 2 ) > γ or f ( c 2 ) < γ . In the first case, set a 3 = a 2 and b 3 = c 2 . In the second case, set a 3 = c 2 and b 3 = b 2 . Now set I 3 = [ a 3 , b 3 ] . Note that in either case, f ( a 3 ) < γ < f ( b 3 ) . Proceeding in this way, either we find some c n 0 such that f ( c n 0 ) = γ and, hence, the proof is complete, or we construct a sequence of closed bounded intervals { I n } with I n = [ a n , b n ] such that for all n , (i) I n ⊃ I n + 1 , (ii) b n − a n = ( b − a ) / 2 n − 1 , and (iii) f ( a n ) < γ < f ( b n ) . In this case, we proceed as follows. Condition (ii) implies that lim n → ∞ ( b n − a n ) = 0. By the Nested Intervals Theorem (Theorem 2.3.3 , part (b)), there exists c ∈ [ a , b ] such that T ∞ n = 1 I n = { c } . Moreover, as we see from the proof of that theorem, a n → c and b n → c as n → ∞ . By the continuity of f , we get lim n → ∞ f ( a n ) = f ( c ) and lim n → ∞ f ( b n ) = f ( c ) . Since f ( a n ) < γ < f ( b n ) for all n , condition (iii) above and Theorem 2.1.5 give f ( c ) ≤ γ and f ( c ) ≥ γ . It follows that f ( c ) = γ . Note that, since f ( a ) < γ < f ( b ) , then c ∈ ( a , b ) . The proof is now complete. Now we are going to discuss the continuity of the inverse function. For a function f : D → E , where E is a subset of R , we can define the new function f : D → R by the same function notation. The function f : D → E is said to be continuous at a point ¯ x ∈ D if the corresponding function f : D → R is continuous at ¯ x . Theorem 3.4.10 Let f : [ a , b ] → R be strictly increasing and continuous on [ a , b ] . Let c = f ( a ) and d = f ( b ) . Then f is one-to-one, f ([ a , b ]) = [ c , d ] , and the inverse function f − 1 defined on [ c , d ] by f − 1 ( f ( x )) = x where x ∈ [ a , b ] , is a continuous function from [ c , d ] onto [ a , b ] .