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(ii) ( a
) is monotonic increasing and bounded above by b
.
( b
) is monotonic decreasing and bounded below by a
.
(iii) Let ( a
)
a and ( b
)
b . Then by the inequality rule (3.76) a b .
By 3.80 (b) a
a and by analogy b
b
for all n ,so
a
a
b
b
. Thus any c such that a
c
b , lies in all the
].
(iv) Suppose c was in all the intervals. Consider n
[ a
, b
1/ c .
43
1], even terms constant, so this subsequence is
convergent.
(ii) [
(i) [
1,
2,
2], subsequence of even terms (1
1/2 n )
1.
44 Convergent
bounded, but not conversely by qn 43.
45 All terms within [0, 1]. 0.1, 0.10, 0.100, etc.
0.1; 0.1, 0.11, 0.111,
0.1111, etc.
.
47 The infinite decimal sequence for
2 is a bounded sequence of
terminating decimals, that is, of rational numbers. From qn 3.80, if any
subsequence is convergent, it is convergent to 2. So if the
completeness principle is not adopted, qn 46 can fail.
48 Choosepositive
, say
.
49
1/ n
(2 n
7)/7 n
. Take
.
50
,
and
areclustr points.
and
arenot.
51 No. Can take
shortest distance between two points in the set.
52
Z.
53
(i) Chooseany a
in A . Now choose a
in A but different from a
.
Construct the sequence by choosing a new point in A for each
successive term.
U .
(iii) By qn 46.
(iv) a need not be in A , but L a U by the closed interval
property (3.78).
(v) Given
(ii) L a
0, there is an N such that n
N a
a
.
54 Each term in the infinite decimal sequence for
2 lies in the interval
[1, 2]. If is a rational number in [1, 2] and we choose
such that
0
-neighbourhood of q contains at most a
finite number of terms of the sequence.
q
2
then an
55 If n N a
a
/2, then also
a
a
/2.
56 d
. d
d
... d
d
. d
d
... d
0.00 . . . 0 d
d
... d
(1/10
)(1
1/10
), so
a
a
1/10
.
 
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