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F'(t) 1~ K. 30 METRIC SPACES [ell. II (2) The set M = UkM k of all functions each of which satisfies a Lipschitz condition for some K is not closed. Since JJ1 contains the set of all polynomials, its closure is the entire space era, b]. e. let there be given two distinct metrics Pl(X, y) and P2(X, V). The metrics PI and P2 are said to be equivalent if there exist two positive constants a and b such that a < [PI (x, y) / P2(X, y)] < b for all x ~ y in R. If an arbitrary set M C X is closed (open) in the sense of the metric PI , then it is closed (open) in the sense of an arbitrary metric P2 which is equivalent to PI .

Necessity. :: R be the complete inverse image of the closed set M' ~ R' . We shall prove that M is closed. If x E [M], there exists a sequence {x n } of points in M which oonverges to x. But then, by Theorem 1, the sequence {f(x n )} converges to f(x). Since f(x n ) E M' and M' is closed, we have f(x) E M'; consequently x E M, which was to be proved. SUfficiency. Let x be an arbitrary point in R, y = f(x), and let O(y, e) be an arbitrary neighborhood of y. The set R' "" O(y, e) is closed (since it is the complement of an open set).

Proof. Let F = naF a, where the Fa are closed sets. Further, let x be a limit point of the set F. This means that an arbitrary neighborhood O(x, E) of x contains an infinite number of points of F. e. F is closed. Now let F be the sum of a finite number of closed sets: F = U~_l F i , and let x be a point not belonging to F. We shall show that x cannot be a limit point of F. In fact, x does not belong to any of the closed sets Fi and consequently it is not a limit point of any of them. Therefore for every i we can find a neighborhood O(x, Ei) of the point x which does not contain more than a finite number of points of F i .