MAS221 Analysis, Semester 2 Solutions
Sarah Whitehouse
Chapter 5 Problems: Sequences in Rk
1. (a) If α = 0, then αxn = αx = 0 for all n, and the result is obvious. So
let α 6= 0.
ε
Let ε > 0. Then there is some N such that |xn − x| < |α|
whenever
n ≥ N.
Thus when n ≥ N , we have |αxn − αx| < ε, and the result follows.
(b) Let ε > 0. Then there are N1 and N2 such that:
– For n ≥ N1 , we have |xn − x| < 2ε .
– For n ≥ N2 , we have |y n − y| < 2ε .
Let N = max(N1 , N2 ). Then for n ≥ N , we have
|(xn + y n ) − (x + y)| ≤ |xn − x| + |y n − y| <
ε ε
+ = ε.
2 2
The result now follows.
2.
• The sequence (an ) does not converge, since the sequence of second
coordinates does not converge.
• The sequence (bn ) does not converge, since the sequence of first coordinates does not converge.
• The sequence (cn ) converges to (0, 0).
• The sequence (dn ) converges to (0, 0).
• The sequence (en ) converges to (0, 1).
• The sequence (f n ) does not converge, since the sequence of second
coordinates does not converge.
3. (a) The sequence is (1, 2), (2, 1), (1, 2), (2, 1), . . . so it does not converge.
(b) The sequence (ln n) does not converge, so neither does the sequence
(an ).
(c) Observe that u2 = 12 and v2 = − 12 . Also, since un+1 + vn+1 = un , we
have un+2 = un /2. And similarly, vn+2 = vn /2. From this, it follows
that (un , vn ) → (0, 0) as n → ∞.
(d) Observe that vn = 0 for n ≥ 2, and un = 1/2n−1 . So (un , vn ) → (0, 0)
as n → ∞.
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4. (a) We know that for any x ∈ R
x n
1+
→ ex
n
as n → ∞.
So the sequence (an ) converges to (e, e−1 ).
(b) We will prove that the sequence (xn ) is monotonic decreasing and
bounded below, and the sequence (yn ) is monotonic increasing and
bounded above.
First note that x1 ≥ 3 and y1 ≥ 3. Suppose xk ≥ 3 and yk ≥ 3.
Then
1 1
2
2
≤ + =
xk+1
3 3
3
√
and rearranging, xk+1 ≥ 3. From this it follows that yk+1 ≥ 3 · 3 =
3. Thus, for all n, xn ≥ 3 and yn ≥ 3 by induction.
In particular, the sequence xn is bounded below.
We now prove that (xn ) is monotonic decreasing, i.e. that xn+1 ≤ xn
for all n.
Using the fact that yn ≥ 3 for all n, we see that
2
3 + xn
1
1
≤
+ =
xn+1
xn
3
3xn
and so
3xn
xn+1
≤
2
3 + xn
and
xn+1 ≤
6xn
≤ xn
3 + xn
as xn ≥ 3. Thus the sequence (xn ) is monotonic decreasing as required.
√
In particular, xn ≤ 2 3 for all n.
We now show that yn ≤ xn+1 for all n. Observe that
1
2
1
1
=
−
≥
yn
xn+1
xn
xn+1
as the sequence (xn ) is monotonic decreasing. Taking reciprocals, it
follows that yn ≤ xn+1 .
The formula for yn+1 now tells us that
p
yn+1 ≥ yn2 = yn
so the sequence yn is monotonic increasing.
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Finally, we prove that the sequence (yn ) is bounded above. By the
above, we know that yn ≤ xn+1 for
sequence (xn ) is
√ all n. But the √
monotonic decreasing, with x1 = 2 3. Thus yn ≤ 2 3 for all n.
Putting all this together, we see that the sequences (xn ) and (yn )
both converge, so the sequence (xn , yn ) converges.
Note: We have
(x1 , y1 ) = (3.4642, 3.0000)
(x2 , y2 ) = (3.2154, 3.1057)
(x3 , y3 ) = (3.1596, 3.1325)
(x4 , y4 ) = (3.1460, 3.1392)
(x5 , y5 ) = (3.1426, 3.1408)
(x6 .y6 ) = (3.1418, 3.1414)
It looks like the sequence (xn , yn ) converges to (π, π).
5. Let (an ) be a bounded sequence
in R2 and write an = (xn , yn ). There is
p
some C such that |an | = x2n + yn2 ≤ C, so |xn | ≤ C and |yn | ≤ C for all
n.
Thus (xn ) is a bounded sequence of real numbers and so it has a convergent subsequence, say (xnk ). Consider the subsequence (ynk ) of (yn ).
This is again a bounded sequence of real numbers so it has a convergent
subsequence (ynkl ).
The sequence (xnkl ) is a subsequence of the convergent sequence (xnk ), so
it converges.
Thus the subsequence (ankl ) = ((xnkl , ynkl )) converges in each component,
so it converges.
6. (a) The interval [0, 1) in R is not open. Any open ball containing 0
contains elements less than 0, which do not belong to the set.
(b) The set {x ∈ R | x 6= 0} in R is open. To see this, let x ∈ R, x 6= 0.
Let δ = |x|. Then the ball B(x, δ) does not contain zero, so it is
contained in the set.
(c) The square (0, 1) × (0, 1) in R2 is open. To see this, let (x, y) ∈
(0, 1) × (0, 1). Let δ = min(x, 1 − x, y, 1 − y). Then B((x, y), δ) ⊆
(0, 1) × (0, 1).
(d) The line R × {0} in R2 is not open. Let (x, 0) ∈ R × {0}. Any open
ball around (x, 0) contains points (x, y) where y 6= 0.
7. (a) The set A is open. To see this, define a continuous map f : (x, y, z) →
R by f (x, y, z) = x2 + y 2 + z 2 .
Then A is the inverse image under f of the open interval (1, ∞).
Since f is continuous, A is open.
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(b) Let (x, y, z) ∈ B. Then x > 0, y > 0, z > 0. Set r = min(x, y, z).
Then if (x0 , y 0 , z 0 ) ∈ B((x, y, z), r) we have |x − x0 | < r ≤ x, |y − y 0 | <
r ≤ y, |z − z 0 | < r ≤ z. From this it follows that x0 > 0, y 0 > 0,
z 0 > 0.
Thus B((x, y, z), r) ⊆ B, and B is open.
(c) Observe (0, 0, 0) ∈ C. For any r > 0, the ball B((0, 0, 0), r) contains
elements (x, y, z) where x 6= 0. Thus the ball B((0, 0, 0), r) is not
contained in C, and C is not open.
(d) Certainly if (x, y, z) ∈ D, then |x| ≤ 1. Observe (1, 0, 0) ∈ D. But
for any r > 0, the ball B((1, 0, 0), r) contains elements (x, y, z) where
x > 1. Thus the ball B((1, 0, 0), r) is not contained in D, and D is
not open.
8. (a) If A = ∅, there is no sequence (an ) in A, and the condition required
for a closed set is vacuously true.
If A = Rk , then any sequence which converges in Rk has limit lying
in A, so Rk is closed.
(b) Let (an ) be a convergent sequence lying in {x}. Then an = x for all
n, and the sequence therefore converges to x, which of course lies in
{x}. Thus the set {x} is closed.
(c) Let (an ) be a convergent sequence of elements all lying in the square
[0, 1] × [0, 1]. Let an = (xn , yn ), and let (x, y) be the limit. Then
0 ≤ xn ≤ 1 and 0 ≤ yn ≤ 1 for all n, so by the sandwich rule,
0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. It follows that (x, y) ∈ [0, 1] × [0, 1].
Therefore [0, 1] × [0, 1] is closed.
9. (a) Let A ⊆ Rk be closed. We will show that U = Rk \A is open.
To see this, let a ∈ U = Rk \A. Suppose there is no r > 0 such that
B(a, r) ⊆ U = Rk \A.
Then for any n ∈ N, the ball B(a, 1/n) contains at least one element,
an , belonging to A.
Consider the sequence (an ). Observe that |an − a| < 1/n. It follows
that an → a as n → ∞. But A is closed, so a ∈ A, which is a
contradiction.
Thus we have some r > 0 such that B(a, r) ⊆ U , and U is open as
required.
Conversely, suppose U = Rk \A is open. Let a ∈ U . Let (an ) be a
sequence in A. There is some ε > 0 such that B(a, ε) ∩ A = ∅, so
|an − a| ≥ ε for all n. Thus, by the definition of convergence, (an )
does not converge to a.
Thus, if a sequence in A converges to a limit, that limit does not lie
in U , so it must lie in A. So A is closed.
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(b) Consider the interval [0, 1) in R. We have already seen in question 6
that this set is not open.
But the sequence (an ) given by an = 1 − 1/n lies in [0, 1) whereas
the limit is 1, which does not lie in [0, 1). Thus [0, 1) is not closed.
10. Observe that |(xn , yn )| = |z n | = |z|n .
If n < 1, |z|n → 0 as n → ∞, and if |z| > 1, the sequence (xn , yn ) is
unbounded.
Thus if |z| < 1, the sequence (xn , yn ) converges to (0, 0), and if |z| > 1,
the sequence does not converge.
11. (a) Let ε > 0. Then there exists N ∈ N such that |an − a| < ε whenever
n ≥ N.
Let (ank ) be a subsequence. Pick K such that nK ≥ N . Then if
k ≥ K, nk ≥ N , so |ank − a| < ε whenever k ≥ K.
Thus the subsequence (ank ) also converges to a, as required.
(b) In R, let an = (−1)n . Then (an ) does not converge. But the subsequence (a2n ) is constant, as a2n = 1 for all n. Thus a subsequence
can converge without the original sequence converging.
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