Unformatted text preview: serve that
inf H inf H x inf H
and so
inf H , inf H
H
and we have shown that inf H is close to H.
10. Prove that no open set can have a largest member.
Suppose that U is open and that x U. We need to show that x can’t be the largest member of U.
Using the fact that x must be an interior point of U we choose 0 such that
x , x
U.
The interval x, x is non empty and all its members must belong to U. Therefore x is not the
largest member of U.
11. Given a real number x and a set S of real numbers, prove that the following two conditions are equivalent:
a. The number x is an interior point of S. 98 b. It is possible to find an open set U such that x U S. Proof that condition a implies condition b: We assume that x is an interior point of S. Choose a
number 0 such that x , x
S and define U x , x . In this way we have found an
open set U such that x U S.
Proof that condition b implies condition a. We assume that condition b holds. Choose an open set
U such that x U S. Using the fact that x is an interior point of U we choose 0 such that
x , x
U. Since x , x
S we deduce that x is an interior point of S.
12. Prove that if S is any set of real numbers then the set of all interior points of S must be open.
Suppose that S is a set of real numbers and define U to be the set of interior points of S. To show
that U is open, suppose that x U. Choose 0 such that x , x
S. We deduce from the
results obtained in Exercise 11 that every number in the interval must be an interior point of S. In
other words, x , x
U. Therefore, since every member of U is an interior point of U the set
U must be open.
13. This exercise refers to the sum of two sets as it was defined in this exercise. Prove that if A is any set of real
numbers and U is an open set then the set A U must be 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 u , u
U. We shall show that x is an interior point of A U by showing
that
x , x
A U.
Suppose that t
x , x . Thus
t a u 
which gives us
 t a u
and so
ta
u , u
U.
Since
t a t a
we see that t A U and so
x , x
AU
as promised. Exercises on Closure
1. Suppose that
S 0, 1 Þ 1, 2 .
a. What is the set of interior points of S?
The set of interior points of S is 0, 1 Þ 1, 2 .
b. Given that U is the set of interior points of S, evaluate U.
0, 1 Þ 1, 2 0, 1 Þ 1, 2 S.
The purpose of parts a and b is to exhibit a set S such that, if U is the set of interior points of S
then U S.
c. Give an example of a set S of real numbers such that if U is the set of interior points of S then U S.
We could take S to be a singleton like 3 or it could be the set of all integers. It could also be
the set of all rational numbers between 0 and 1. 99 d. Give an example of a subset S of th...
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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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