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i1 4. The Cantor intersection theorem depends upon the completeness of the real number system. Where in the
proof of the theorem is the completeness used?
We use the completeness to guarantee that a nonempty closed bounded set has a least member
and then we use the completeness again to guarantee that the sequence of all the least members
of the given sets has a supremum. Exercises on the BolzanoWeierstrass Theorem
1. The BolzanoWeierstrass theorem does not tell us that if a set S is bounded and infinite then at least one
member of S must be a limit point of S. Give an example of a bounded infinite set S such that no member of
S is a limit point of S.
The set
1
n Z
n
has only one limit point; the number 0, that does not belong to the set.
2. Prove that if H is a closed bounded infinite set then
H LH
.
Suppose that H is a closed bounded infinite set. We know from the BolzanoWeierstrass theorem
that H has a limit point. Since H is closed, every limit point of H must belong to H and so H must
have at least one limit point that belongs to H.
3. This exercise suggests a different proof of the BolzanoWeierstrass theorem:
a. Prove that if E is a nonempty bounded set with no least member then inf E is a limit point of E.
Suppose that E is a nonempty bounded set that has no least member. We know that the
number inf E is close to the set E but, since E has no least member, inf E does not belong to E.
As we know, a number that does not belong to a given set but is close to that S must be a limit
point of the set.
b. Prove that if a bounded set S has a nonempty subset that does not have a least member then S has a limit
point.
Suppose that S is a bounded set that has a nonnegative subset E that has no least member.
From part a we know that E has a limit point. Since any limit point of E must also be a limit point
of S we can conclude that S has a limit point. 151 c. Given that S is a bounded infinite set and that every nonempty subset of S has a least member, find an
example of a strictly increasing sequence in the set S. By considering the limit of this sequence, prove
that S must have a limit point.
Since S is a nonempty subset of itself, it has a least member. We define x 1 to be the least
member of S. Since S is infinite, the set S
x 1 is a nonempty subset of S. We define x 2 to be
the least member of S
x 1 . We note that x 1 x 2 . Since S is infinite, the set S
x 1 , x 2 is a
nonempty subset of S. We define x 3 to be the least member of S
x 1 , x 2 and note that
x 2 x 3 . Continuing in this way we obtain a strictly increasing sequence x n in the set S.
As we know, an increasing bounded sequence must converge. We define
x n Ý xn.
lim
0 we know that the condition
xn
x ,x
must hold for all sufficiently large integers n and, since the sequence x n is oneone we
deduce that the interval x , x must contain more than one member of S. Therefore x is a
limit point of S.
Given any number Some Further Exercises on Limit Points
The ex...
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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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