1873_solutions

a x is nonempty and since b x a x b x b x we deduce

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Unformatted text preview: But, since S is closed, every limit point of a subset of S must belong to S. 14. Suppose that S is a bounded set of real numbers and that L S  . Prove that every nonempty subset of S 130 must have both a largest and a smallest member. Can such a set S be infinite? The fact that every nonempty subset of S must have both a greatest and a least member follows at once from Exercise 12. To see that such a set S cannot be infinite we shall show that if S is an infinite set of real numbers and if every nonempty subset of S has a least member then S must have a subset that does not have a greatest member. We assume that S is an infinite set of real numbers and that every nonempty subset of S has a least member. Using the fact that S we define x 1 to be the least member of S. Using the fact that the set S x 1 is nonempty we define x 2 to be the least member of the set S x1 . Using the fact that the set S x 1 , x 2 is nonempty we define x 3 to be the least member of the set S x1, x2 . Continuing in this way we obtain a strictly increasing sequence x n in the set S and we see that the set x n n Z  does not have a greatest member. 15. Given any subset S of a metric space X, prove that the set L S must be closed. Solution: We shall show that any point that fails to belong to L S must fail to belong to L S . Suppose that x X L S . Choose a number   0 such that the ball B x,  contains only finitely many members of S. Given any point t in the ball B x,  , it follows from the fact that B x,  is a neighborhood of t and the fact that B x,  contains only finitely many members of S that t is not a limit point of S. Thus B x,  LS  and we have shown, as promised, that x does not belong to L S . 16. Prove that if U is an open subset of the metric space R k then L U  U. Of course L U U. Now suppose that x U. To show that x L U , suppose that   0. Using the fact that x U, choose a point y in the set U B x,  . Using the fact that the set U B x,  is open, choose  0 such that B y, U B x,  . We have now found more than one member of U that belongs to the ball B x,  and so we know that U B x,  x . 17. Suppose that S is a subset of a metric space, that L S , and that   0. Prove that there exist two different members x and y of S such that d x, y  . Choose a limit point u of the set S. Using the fact that the ball B u,  contains infinitely many 2 points of S, choose two different points x and y in S B u,  . We observe that 2 d x, y d x, u  d u, y      . 2 2 18. Prove that if S is a convex subset of R k and S contains more than one point then every member of S is a limit point of S. Suppose that S is a convex subset of R k and that S contains two different points x and y. In order to show that x is a limit point of S, suppose that   0. We observe that if t is a real number and if z  1 t x  ty then z x  1 t x  ty x  |t | x y and so the inequality z x   will hold as long as |t | x y  . We now choose a number t such that  0t xy and obser...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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