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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.
 Fall '08
 STAFF
 Math, Calculus

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