1873_solutions

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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  0t xy and obser...
View Full Document

## 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.

Ask a homework question - tutors are online