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Unformatted text preview: t every nonempty open set must contain a rational number. Suppose that U is
open and nonempty. Choose x U and choose 0 such that x , x
U. Since the interval
x , x must contain rational numbers, so must the set U. 5. Given that H is a closed subset of R 2 and that Q Q H, prove that H R 2 .
We need to show that the open set R 2 H is empty. To obtain a contradiction, suppose that
and choose a point a, b
R 2 H. Using the theorem on open subsets of R k we now
R2 H
choose 0 such that whenever a x a and b y b we have x, y
R 2 H.
Using the theorem on densesess of the set of rational numbers we choose rational numbers x and
y such that a x a and b y b and we observe that x, y
Q Q H,
contradiction our assumption that QQ H.
6. Given that U is an open subset of a metric space X and that H is closed, prove that the set U
that the set H U is closed.
Since
U HU
XH
which is the intersection of two open sets, the set U H is open. Since
H UH
XU
which is the intersection of two closed sets, the set H U is closed. H is open and 7. Prove that every nonempty closed set of real numbers that is bounded below must have a least member.
Suppose that H is a nonempty closed set of real numbers and that H is bounded below. We need
to show that inf H H. To obtain a contradiction, suppose that inf H belongs to the open set R H.
Choose 0 such that inf H , inf H
R H. Since no member of H can be less than inf H
and no member of H can lie in the interval inf H , inf H we see that the number inf H is a
lower bound of H, contradicting the fact that inf H is the greatest lower bound of H.
8. If A and B are subsets of the space R k for a given positive integer k, then the sum A B of A and B is defined,
by analogy with an earlier exercise. Prove that if A is any subset of R k and U is an open subset of R k then
the set A U is open. Solution: We need to show that every member of A U is an interior point of A U. Suppose that x A U. Choose a member a of A and a member u of U such that x a u. Using the fact that u U,
choose 0 such that B u,
U. We shall show that x is an interior point of A U by showing that
B x,
A U.
Suppose that t B x, . Thus
t au
which gives us
t a u
and so
t a B u,
U.
Since 118 t a t
we see that t a A U and so
B x, AU as promised.
9. Prove that the interval 1, 3 is not an open subset of the metric space R but that this interval is an open subset
of the metric space 0, 3 which is a subspace of R.
Given any positive number , the ball in the metric space R with center 3 and radius will contain
all of the numbers between 3 and 3 and will therefore fail to be included in the interval 1, 3 .
Thus, in the metric space R, the point 3 fails to be an interior point of the interval 1, 3 and
therefore this interval fails to be open in the metric space R. However, if 0 1 then, in the
metric space 0, 3 , the ball center 3 with radius is the interv...
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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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