Fall 2016, Math 409, Section 502
Solutions to some problems from the fifth assignment
Exercise 2. Let A be a non-empty subset of R that is bounded above. Assume that sup A ∈
/ A. Prove that there exists a strictly increasing sequence
(xn )n of A so that limn xn = sup A.
Solution. Set s = sup A. We will recursively choose for each n ∈ N a number
xn ∈ A, so that x1 < · · · < xn and |xn − s| < 1/n. This will yield the desired
sequence.
By the approximation property for suprema, we may choose x1 ∈ A with
s − 1/n < x1 6 s. Assume that we have chosen x1 , . . . , xn satisfying the
desired properties. Since xn 6 s and s ∈
/ A, we have xn < s, i.e. s − xn > 0.
Set ε = min{1/(n + 1), s − xn }, which is positive. By the aproximation
property of suprema we may choose xn+1 ∈ A with s − ε < xn+1 6 s.
Since s − ε < xn+1 6 s < s + ε, by the fundamental theorem of absolute
values we conclude |xn+1 − s| < ε 6 1/(n + 1). Furthermore, observe that
xn = s − (s − xn ) 6 s − ε < xn+1 , therefore xn+1 satisfies the desired
properties.
Exercise 3. Let (xn )n be a real sequence and assume that xn ∈ Z for all
n ∈ N.
(i) If (xn )n is Cauchy, show that it is eventually constant (i.e. there
exists n0 ∈ N so that for all n > n0 we have xn = xn0 ).
(ii) If (xn )n converges to some a ∈ R, then a ∈ Z.
Solution. (i) Fix ε = 1/2 (or any other number in (0, 1)). As (xn )n is Cauchy,
there exists n0 ∈ N, so that for all m, n ∈ N with m > n > n0 we have
|xn − xm | < ε = 1/2. In particular, for all n > n0 we have |xn − xn0 | < 1/2.
For n ∈ N with n > n0 , as xn − xn0 is in Z, it is either zero or |xn − xn0 | > 1.
Since the second case is impossible, we conclude xn = xn0 .
(ii) If (xn )n converges to some a ∈ R, it is Cauchy. By (i), there exists
n0 so that xn = xn0 for all n > n0 . This yields limn xn = xn0 and hence
a = xn0 ∈ Z.
Exercise 5. Let (xn )n be a real sequence and assume that there exists a ∈ R
with 0 < a < 1, so that |xn − xn+1 | 6 an for all n ∈ N. Show that (xn )n
converges to some real number x.
Solution. We first observe that for n ∈ N,
n
X
1 − an+1
ak =
,
1−a
k=0
which yields that for n 6 m,
m
X
an − am+1
an
ak =
6
.
1−a
1−a
k=n
To show that (xn )n is convergent, it suffices to show that it is Cauchy and
to that end fix ε > 0. Since 0 < a < 1, we have limn an = 0 and 1 − a > 0.
Hence, there exists n0 ∈ N so that for all n > n0 we have an+1 < ε(1 − a).
We conclude that for m > n > n0
m
m
m
X
X
X
|xm − xn | = (xk − xk−1 ) 6
|xk − xk−1 | 6
ak−1
k=n+1
=
m−1
X
k=n+1
ak 6
k=n
k=n+1
an
< ε.
1−a
Exercise 7.
(i) Let (xn )n be a real sequence and r be a real number. Assume that
r < lim inf n xn . Show that there exists n0 ∈ N with r < xn for all
n > n0 .
(ii) Let (xn )n , (yn )n be real sequences so that yn 6= 0 for all n ∈ N and
0 < lim inf n yn < +∞. Show that
xn
lim supn xn
lim sup
6
.
y
lim inf n yn
n
n
Solution. (i) If the conclusion fails, then for all n0 ∈ N there exists n > n0
with xn 6 r, i.e. inf{xn : n > n0 } 6 r. We conclude
lim inf xn = lim inf{xk : k > n} 6 r,
n
n
which is absurd.
(ii) Note that for all m > n ∈ N we have
(∗)
xm 6 sup{xk : k > n} and ym > inf{yk : k > n}.
Choose any real number r with 0 < r < lim inf n yn (e.g. one can choose
r = (lim inf n yn )/2). By (i), there exists n0 ∈ N so that for all n > n0 we
have yn > r. This also in particular yields that for all n ∈ N, with n > n0 ,
inf{yk : k > n} > r > 0. Combining this with (∗), we obtain that for all
m > n > n0 we have
1
1
(∗∗)
6
.
ym
inf{yk : k > n}
Using the convention a · (+∞) = +∞ for all a > 0, the first part of (∗) and
(∗∗) yield that for all m > n > n0 we have
xm
sup{xk : k > n}
6
,
ym
inf{yk : k > n}
from which we deduce that for all n > n0
xk
sup{xk : k > n}
(∗ ∗ ∗)
sup
:k>n 6
.
yk
inf{yk : k > n}
In conclusion, using (∗ ∗ ∗) we calculate
xn
xk
sup{xk : k > n}
lim sup
= lim sup
: k > n 6 lim
n
n inf{yk : k > n}
yn
yk
n
lim supn xn
limn sup{xk : k > n}
.
=
=
limn inf{yk : k > n}
lim inf n yn