By H. A Lauwerier

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It follows that for Ixl :::; r, Iyl :::; r we have I[f(y) - f(x)](y - X)-l - g(x) I :::; 2: co n=2 lanllY - xl 2n2 r n- 2 ly - xl- 1 co = Iy - xl 2: n2lanlr n- 2 = Kly n=2 xl, K constant. Thus f'(x) = g(x). 5. 11) Ix - xol < R. 2: n(n co n=k I)(n - 2)··· (n - k + I)an(x Proof. 4 by induction on k. 6) are determined uniquely by the function f (provided the radius of convergence is positive). Exercises 1. Find the function defined for Ixl < 1 by f(x) = 2::'=1 xnjn. ) 2. 6), then If x Xo co = n~o (n + l)-lan(X - xo)n+1.

Suppose (S, d) is a metric space. A subset A c S is open if and only if its complement is closed. Proof Let B be the complement of A. Suppose B is closed, and suppose x E A. Then x is not a limit point of B, so for some r > 0 we have Br(x) n B = 0. Thus Br(x) c A, and A is a neighborhood of x. Conversely, suppose A is open and suppose x ¢ B. Then x E A, so for some r > 0 we have Br(x) c A. Then Br(x) n B = 0, and x is not a limit point of B. It follows that every limit point of B is in B. 0 The set of limit points of a subset A c S is called the closure of A; we shall denote it by A - .

Now consider the neighborhood N(x). This contains an interval (x - e, X + e). If we choose m so large that bm - am < e, then since am ~ x ~ bm this implies [am, bm] c N(x). But this means that [am, bm] is nice. This contradiction proves the theorem for the case n = 1. The same method of proof works in IRn, where instead of intervals we use squares, cubes, or their higher dimensional analogues. For example, when n = 2 we choose M so large that A is contained in the square with corners ( ±M, ±M).