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Unformatted text preview: Section 6 Econ 204, GSI: Hui Zheng Key words Extreme Value Theorem, Intermediate Value Theorem, Monotonically Increasing, Completeness, Contraction Mapping Theorem, Cauchy Sequence Section 6.1 Properties of Continuous Functions & Lecture 5 Theorem 1 (Extreme Value Theorem): Let f be a continuous real-valued function on [ a;b ] . Then f assumes its minimum and maximum on [ a;b ] . In particular, f is bounded above and below. & Lecture 5 Theorem 2 (Intermediate Value Theorem): Suppose f : [ a;b ] ! R is continuous, and f ( a ) < d < f ( b ) . Then there exists c 2 ( a;b ) such that f ( c ) = d . Example 6.1.1 If f is continuous real-valued function on [ a;b ] , then f ([ a;b ]) is a closed interval. Solution: The Extreme Value Theorem shows that the range of f is bounded, and the extrema are attained. Thus there are points c and d in [ a;b ] such that f ( c ) = m := inf x 2 [ a;b ] f ( x ) f ( d ) = M := sup x 2 [ a;b ] f ( x ) Suppose that c d (the case c > d may be handled similarly by considering the function f ). Pick any arbitrary point y in ( m;M ) . Since f ( x ) is continuous on the interval [ c;d ] . Thus by the Intermediate Value Theorem, there is a point x 2 ( c;d ) such that f ( x ) = y . This is true for every point y 2 ( m;M ) , as well as two end points. Therefore, f ([ a;b ]) = [ m;M ] . Section 6.2 Cauchy Sequence & Lecture 5 De&nition 6: A sequence f x g in a metric space ( X;d ) is Cauchy if 8 " > 9 N ( " ) n;m > N ( " ) ) d ( x n ; x m ) < " & Lecture 5 Theorem 7: Every convergent sequence in a metric space is Cauchy. Example 6.2.1 Show that the sequence f x n g = ( & 1) n n is Cauchy with Euclidean metric. Solution 1: Pick an " > , choose N > 2 " , for every m;n > N , j x m x n j < j ( & 1) m m ( & 1) n n j j 1 m + 1 n j < 2 N < " . So it is Cauchy. Solution 2: We know that f x n g ! . All convergent sequences are Cauchy (Theorem 7). So f x n g is Cauchy. Example 6.2.2 Show that if x n and y n are Cauchy sequences from a metric space X , then d ( x n ;y n ) con- verges. Solution: Because X is not necessarily complete, we cannot rely on the convergence of x n and y n ....
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- Summer '08