Royden, Real Analysis 3rd ed.
Chapter 2
Problem 211
a.
If

x
n
−
l

≤
²
for
n
≥
N,
we get immediately

x
n
−
x
m

≤

x
n
−
l

+

x
m
−
l

≤
2
ε
(1)
for
n, m
≥
N.
b.
Equation (1) implies

x
n

≤

x
N

+ 2
ε
for all
n
≥
N.
Hence
{
x
n
}
is bounded by
max
1
≤
i
≤
N

x
i

+ 2
ε
.
c.
Assume
lim
x
n
m
=
l.
Then for any
ε
>
0
,
there is a number
L
such
that

x
n
m
−
l

≤
ε
for
m
≥
L.
(2)
Moreover, we may assume WLOG that
n
L
≥
N.
By (1) and (2),

x
n
−
l

≤

x
n
−
x
n
L

+

x
n
L
−
l

≤
2
ε
for
n
≥
n
L
.
Since
ε
can be arbitrarily small,
lim
x
n
=
l.
d.
By part
a
it is enough to know that every Cauchy sequence
{
x
n
}
converges.
By Problem 29.b
{
x
n
}
has a convergent subsequence and the
conclusion follows then from part
c
.
Problem 213
(i) only if part.
Let
l
=
lim
x
n
.
By de
fi
nition
l
= inf
a
n
= lim
a
n
,
where
a
n
=sup
k
≥
n
x
k
is a monotone sequence. Hence we have, for any
ε
>
0
,
l
≤
a
n
≤
l
+
ε
(1)
holds for
n
no less than a certain number
N.
By de
fi
nition of
a
N
,
x
k
≤
a
N
≤
l
+
ε
for
k
≥
N
and there exists
n
(
ε
)
≥
n
such that
x
n
(
ε
)
≥
a
n
−
ε
≥
l
−
ε
.
1
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Combining together, we have for any
ε
>
0
and
n,
l
−
ε
≤
x
m
≤
l
+
ε
holds for some
m
≥
n.
Now let
ε
vary like
2
−
1
,
3
−
1
, . . .
, we can get a sequence
{
x
n
k
}
such that
l
−
k
−
1
≤
x
n
k
≤
l
+
k
−
1
and
n
1
< n
2
< n
3
<
· · ·
,
which implies
lim
k
x
n
k
=
l.
(ii) if part.
The given conditions imply
l
−
ε
≤
a
n
≤
l
+
ε
holds for all
m
≥
n.
Hence
lim
x
n
= lim
a
m
=
l.
Problem 215
For each
n,
b
n
≡
inf
k
≥
n
x
k
≤
x
n
≤
sup
k
≥
n
x
n
≡
a
n
.
Letting
n
→ ∞
,
lim
x
n
= sup
n
b
n
= lim
b
n
≤
lim
a
n
= inf
n
a
n
=
lim
x
n
.
(1)
For any
ε
>
0
,
we knew from Problem 213 that
x
k
≤
l
+
ε
for
k
≥
N
1
.
(2)
Using the de
fi
nition of
lim
x
n
=
l,
we have similarly
x
k
≥
l
−
ε
for
k
≥
N
2
.
(3)
Combining together,
l
−
ε
≤
x
k
≤
l
+
ε
for
k
≥
N
=
N
1
∨
N
2
.
That is
lim
x
n
=
l.
Conversely, (2) and (3) will hold,
which imply
lim
x
n
≤
l
and
lim
x
n
≥
l.
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 Spring '08
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 Mathematical analysis, Extreme Value Theorem, Epigraph, lim xn, Semicontinuity, lim xm

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