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Unformatted text preview: show that the conclusion
x n y n 0 can fail if both x n and y n are unbounded. Hint: Make the observation that both of the sequences x n and y n must be bounded. Explain
carefully how you arrive at this conclusion. Now make use of the fact that 166 x n y n  y n  x n
yn 1 for every n.
14. Suppose that x n and y n are sequences of real numbers, that y n 1 and that for each n we have
z n x n y n . Prove that the sequences x n and z n have the same set of partial limits.
It follows from the part of Exercise 3 for which a solution is provided above that if x is an partial limit
of x n then the number x 1 x is a partial limit of the sequence z n . At the same time, since
x n z n /y n for all sufficiently large n we know that if x is any partial limit of z n then the number
x/1 x is a partial limit of the sequence x n .
15. Given that x n is a sequence in R k and that for every point x in R k we have x n x 0 as n Ý, prove that
the sequence x n converges to the point O.
Taking x 1, 0, , 0 we see that the sequence of first coordinates of the points x n converges to
zero. We can repeat this argument for each of the other coordinates. Exercises on Sequences and the Topology
1. Prove that a set S of real numbers is unbounded below if and only if there exists a sequence x n in S such
that x n
Ý. Solution: We want to show that, for a given set S of real numbers, the following conditions are
equivalent:
a. The set S is unbounded below.
b. There exists a sequence x n in the set S such that x n Ý as n Ý. To show that condition b implies condition a, assume that condition b holds and choose a sequence x n in
S such that x n
Ý as n Ý. If w is any real number then it follows at once from the fact that the
sequence x n is eventually in the interval Ý, w that w is not a lower bound of S. Therefore S is not
bounded above.
Now to show that condition a implies condition b, assume that condition a holds. For each positive integer
n we use the fact that the number n is not a lower bound of S to choose a number that we shall call x n
such that x n n. The sequence x n that we have made in this way is in the set S and it is clear that
xn
Ý.
2. Suppose that S is a nonempty set of real numbers and that is an upper bound of S. Prove that the following
conditions are equivalent:
a. We have sup S.
b. There exists a sequence x n in S such that x n as n Ý.
In Exercise 7 of the exercises on closure we saw that an upper bound of set S is close to S
if and only if is equal to sup S. We also saw in a recent theorem that a number is close to a
set S if and only if there exists a sequence in S that converges to . The present exercise
follows at once from these two facts. Of course, we can write this exercise out directly if we
wish.
4. Given a sequence x n that is frequently in a subset S of a metric space X, and given a partial limit x of the
sequence x n , is it necessarily true that x S?
No, this statement need not be true. If we define x n ...
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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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