18.100B/C: Fall 2008
Solutions to Practice Final Exam
1.
Suppose for sake of contradiction that
x >
0. Then
1
2
·
x >
0 because the product of two
positive quantities is positive. Thus
x
2
+ 0
<
x
2
+
x
2
(because
y < z
implies
x
+
y < x
+
z
for all
x
),
i.e.,
x
2
< x
. Also, since
x
2
>
0 we have by the givens that
x
≤
x
2
. But the results of the previous
two sentences are a contradiction with trichotomy. Thus
x
±
>
0. Since
x
≥
0, this implies
x
= 0, as
desired.
2.(a)
Let
s
k
=
∑
k
n
=1
a
n
. Then
s
k
=
a
1
2
+
a
k
2
+
k

1
X
n
=1
a
n
+1
+
a
n
2
>
k

1
X
n
=1
√
a
n
a
n
+1
,
where the last inequality follows because
a
n
+
a
n
+1
2

√
a
n
a
n
+1
=
1
2
(
√
a
n

√
a
n
+1
)
2
≥
0
.
But then
∑
∞
n
=1
a
n
is an upper bound on the partial sums of
∑
∞
n
=1
√
a
n
a
n
+1
. Thus these partial
sums are an increasing sequence bounded above and so they converge, as desired.
(b)
Deﬁne
a
0
=
a
1
and let
b
n
=
√
a
n
a
n

1
for
n
∈
N
. Then
∑
∞
n
=1
b
n
converges and
b
n
≥
a
n
because
a
n
is decreasing, so
∑
∞
n
=1
converges by the comparison test. (Challenge: can you ﬁnd a
counterexample to the converse when (
a
n
) is positive but not necessarily decreasing?)
3.(a)
Both
f
(
x
) = 4
x
(1

x
) and
f
(
x
) = 1
 
2
x

1

work nicely.
(b)
No function: continuous functions take connected sets to connected sets.
(c)
Deﬁne
f
(
x
) =
(
0
,
x
≤
1
,
1
,
x
≥
2
.
This function is continuous on [0
,
1]
∪
[2
,
3] and
f
([0
,
1]
∪
[2
,
3]) =
{
0
,
1
}
.
(d)
No function: suppose such a function
f
exists. There exists
x
1
for which
f
(
x
1
) = 1 and
x
2
for which
f
(
x
2
) = 2, so by the Intermediate Value Theorem there is
x
between
x
1
and
x
2
for which
f
(
x
) =
√
2, a contradiction. (Or, use connectedness again.)
(e)
No function: continuous functions take compact sets to compact sets.
4.
By uniform convergence of (
f
n
), we can choose some
N
∈
N
such taht
n
≥
N
implies

f
n
(
x
)

f
(
x
)

<
ε
3
for all
x
∈
E
. By uniform continuity of
f
N
, we can choose some
δ
such that
d
(
x, y
)
< δ
implies

f
N
(
x
)

f
N
(
y
)

<
ε
3
. Then for any
x, y
∈
E
such that
d
(
x, y
)
< δ
we have

f
(
x
)

f
(
y
)

=

f
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 Fall '10
 Prof.KatrinWehrheim
 Topology, Continuous function, Metric space, Riemann integral, Compact space

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