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Unformatted text preview: ill hold whenever n N. Then, whenever n N we have
x ,x
x 5 ,x 5 .
xn
Now to prove that condition b implies condition a we assume that condition b holds. To prove that
condition a holds, suppose that 0. Using the fact that /5 is a positive number, we choose an
integer N such that the condition
xn
holds whenever n ,x 5
5
5
N we have x N. Thus for every n
xn x 5 5 5 ,x 5 5 x ,x . 9. Prove that
n 2 3n 1
2n 2 n 4 1
2 as n Ý.
We begin by observing that if n is any positive integer then
5n 5 .
n 2 3n 1
5n 2
1
2
4n
4n 2
2n 2 n 4
4n 2 2n 8
Now suppose that 0. The inequality
n 2 3n 1
1
2
2n 2 n 4
will hold when
5
4n
which requires that
n 5
4
Choose an integer N such that N 5/ 4 and observe that, whenever n N we have
5
5.
n 2 3n 1
1
2
4n
4N
2n 2 n 4
10. Given that x n is a sequence in a metric space X and that x X, prove that x n x as n Ý if and only if
d xn, x
0
The condition x n x as n Ý says that for every 0 there exists an integer N such that
whenever n N we have d x n , x .
The condition d x n , x
0 as n Ý says that for every 0 there exists an integer N such that
whenever n N we have d x n , x 0  .
These two conditions clearly say the same thing. as n Ý.
11. Suppose that x n is a sequence in a metric space X, that x X, and that y n is a sequence of real numbers
converging to 0. Suppose that the inequality
d x, x n
yn
holds for every n. Prove that x n x as n Ý.
Suppose that 0. Using the fact that y n 0 as n Ý, choose an integer N such that the
inequality y n will hold whenever n N. We see at once that the inequality d x n , x also holds
whenever n N. 159 12. Given that x n and y n are sequences of real numbers, that x n y n for each n and that y n Ý as n Ý,
prove that x n Ý as n Ý..
Suppose that w is a real number. Using the fact that y n Ý as n Ý we choose an integer N such
that the inequality y n w holds whenever n N. Then whenever n N we have x n y n w.
13. Suppose that x n and y n are sequences in a metric space X and that x and y are points of X.
a. Prove that for every n we have
d x, y d x, x n d x n , y n d y n , y and deduce that
d x, x n d y n , y .
d x, y d x n , y n
Then prove that for every n we have
d x, y
d x, x n d y n , y
d xn, yn
and deduce finally that
d x, y  d x, x n d y n , y .
d x n , y n
Given any n we have
d x, y
d x, x n d x n , y
d x, x n d x n , y n d y n , y
and the inequality
d x, x n d y n , y .
d x, y d x n , y n
follows at once. In the same way we can show that
d x, y
d x n , x d y, y n .
d xn, yn
Therefore
d x, y  d x, x n d y n , y .
d x n , y n
d x, y as n Ý.
b. Prove that if x n x and y n y as n Ý then d x n , y n
We assume that x n x and y n y as n Ý. Suppose that 0. Choose an integer N 1 such
that the inequality d x n , x 2 holds whenever n N 1 . Choose an integer N 2 such that the
inequality d y n , y 2 holds whenever n N 2 . We define N to be the larger of the integers N 1
and N 2 . Then whenever n N we...
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 Fall '08
 STAFF
 Math, Calculus

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