Solutions to exercise sheet 3: Supremum and infimum
1. (a) Let x, y ∈ R and suppose that for all ε ∈ (0, ∞) we have that
x − y < ε. We proof this by contradiction. Suppose that x > y.
In that case x − y ∈ (0, ∞) and so x − y < x − y which is clearly
a contradiction. So x ≤ y.
(b) We have that for all ε > 0, |x−y| ≤ ε and thus −ε < x−y < ε. So
by the first part of the question x ≤ y. Moreover by multiplying
through by −1 we can see that for all ε > 0 we have y − x < ε
and so by the first part of the question y ≤ x. Since y ≤ x and
x ≤ y we must have x = y.
2. Let A, B ⊆ R be non-empty and bounded with a ⊆ B. To show that
sup A ≤ sup B we let a ∈ A which also means that a ∈ B. Since
a ∈ B and B is bounded (so sup B 6= ∞) we have that a ≤ sup B and
so sup B is an upper bound for A which means that sup A ≤ sup B.
To show that inf A ≥ inf B is almost identical. Let a ∈ A and note
this means that a ∈ B. Since a ∈ B and B is bounded we know that
a ≥ inf B. So inf B is a lower bound for A and thus inf A ≥ inf B.
3. Let A ⊆ R be bounded and non-empty and B = {|a| : a ∈ A}. We
first consider the case when | sup A| ≥ | inf A| in this case sup B =
sup A. Note that in this case we must have that since sup A ≥ inf A,
sup A ≥ 0. Let b ∈ B then b = |a| for some a ∈ A, if b = a then
b ≤ sup A. On the otherhand if b = −a then a ≤ 0 and inf A ≤. We
have that a ≥ inf A and so −a ≥ − inf A = | inf A| ≤ sup A. So sup A
is an upper bound for B.
Now let y < sup A (we need to show y cannot be an upper bound
for B). We know that there exists a ∈ A such that a > y and so
|a| ≥ a > y and since |a| ∈ B y cannot be an upperbound for B.
Therefore when | sup A| ≥ | inf A| we have that sup B = sup A.
We now consider the case when | sup A| < | inf A|, in this case sup B =
− inf A. This means that inf A < 0. So if we let b ∈ B then b = |a| for
some a ∈ A. If b = a then b ≤ sup A ≤ | inf A| = − inf A. If b = −a
then a ≥ inf A and so b ≤ − inf A.
Now let y < − inf A (we need to show y cannot be an upper bound
for B). Thus we have that −y > inf A and so there exists a ∈ A with
a < −y and thus −a > y. Thus |a| ∈ B and |a| ≥ −a > y and so y
cannot be an upper bound for B. Therefore when | sup A| < | inf A|
we have that sup B = − inf A.
4. (a) Claim: inf A1 = 1 and sup A1 = 3
Justification: If we let x ∈ A1 then 1 < x < 2 or 2 < x < 3
and so 1 < x < 3. Thus inf A1 ≥ 1 and sup A1 ≤ 3. To show
1
that inf A1 = 1. Let 3/2 > y > 1 and note that in this case
3/2 > y > y − (y − 1)/2 > 1 and so y − (y − 1)/2 ∈ A1 which
means y cannot be a lower bound for A1 . So inf A < y and since
this holds for all 3/2 > y > 1 we can conclude that inf A ≤ 1 and
so inf A = 1.
To show that sup A1 = 3. We let 2 < y < 3 and note that then
2 < y < y + (3 − y)/2 < y and so y + (3 − y)/2 ∈ A and thus
sup A ≥ y. Since this holds for all 2 < y < 3 we can conclude
that sup A ≥ 3 and so sup A = 3.
(b) Claim: inf A2 = 0 and sup A2 = 1
Justification: For any n ∈ N we have that
n−1
1
= 1 − ≤ 1.
n
n
So sup A2 ≤ 1 and inf A2 ≥ 0. If we take n = 1 then we can see
that 0 ∈ A2 and so 0 ≤ inf A2 . Thus inf A2 = 0. Let 0 < y < 1.
(Note: to show that sup A2 = 1 we still need to show that for any
0 < y < 1 we can find z ∈ A2 with z > y). By the Archimedean
1
principle we can find n ∈ N with n ≥ 1−y
> 0 (since we wanted
n
>
y).
We
then
have
n−1
0≤
1
1
n−1
= 1 − > 1 − 1 = 1 − (1 − y) = y.
n
n
1−y
Thus for all y < 1 we can find a ∈ A2 with a > y. Therefore
sup A2 ≥ 1 and since we already have that sup A2 ≤ 1 we can
conclude sup A2 = 1.
(c) Claim: inf A3 = −∞ and sup A3 = ∞.
2
n2
Justification: We fix x ∈ R, if n ∈ N then nn−+1
1 ≥ n = n. By
2
the Archimedean principle we can find n ∈ N with n > x and
then
n2 + 1
≥ n > x.
n − 12
Thus we have shown that A3 is not bounded above and so sup A3 =
∞.
On the other hand for n ∈ N then
n2 + 1
n2
n
(−n)2 + 1
1 =−
1 ≤ − 2n = − 2 .
−n − 2
n+ 2
So if we fix x ∈ R and by the Archimedean principle choose n ∈ N
such that n > 2x then we have that
n
(−n)2 + 1
1 ≤ − 2 < .x
−n − 2
2
This means that for all x ∈ R we can find a ∈ A3 with a < x and
so A3 is not bounded below and so inf A3 = −∞.
(d) Claim: inf A4 = −∞ and sup A4 = 1.
x
Justification: We have that for all x ≥ 0, x−1
x ≤ x = 1. So
1
.
sup A4 ≤ 1. If 0 < y < 1 then we can take z ∈ R with z > 1−y
z−1
In this case z ∈ A4 and
z−1
1
= 1 − > 1 − (1 − y) = y.
z
z
So sup A4 ≥ 1 and thus sup A4 = 1.
On the other hand if y ∈ R and y < 0 then if 0 < x <
have that
x−1
1
= 1 − < 1 − (1 − y) = y.
x
x
Thus A4 is not bounded below and so inf A4 = −∞.
1
1−y
we
(e) Claim: inf A5 = −1 and sup A5 = 7/3.
Justification: If n ∈ N is even then (−1)n = 1 and we have
n(−1)n + 5
n+5
4
=
=1+
.
n+1
n+1
n+1
If n ∈ N is odd we have
n(−1)n + 5
−n + 5
6
=
= −1 +
.
n+1
n+1
n+1
So when n is even
1≤
n(−1)n + 5
4
7
≤1+ =
n+1
3
3
and when n is odd
−1 ≤ −1 +
6
≤ 2.
n+1
So inf A5 ≥ −1 and sup A5 ≤ 7/3. If we take n = 2 we have
n +5
that n(−1)
= 7/3 and so sup A5 = 7/3. Now let y > −1. By
n+1
the Archimedean principle we can find n ∈ N which is odd and
5−y
6
y+1 < n. Since n is odd we will have that −1 + n+1 ∈ A5 and
−1 +
6
< −1 +
n+1
Thus inf A5 = −1.
3
6
6
= −1 + 6 = y.
+1
y+1
5−y
y+1
5. (a) Suppose that sup A = ∞. This means that for all x ∈ R we can
find a ∈ A such that a > x. To show that sup A − B = ∞ we
need to show for all x ∈ R there exists z ∈ A − B such that z > x.
To do this we fix x ∈ R and b ∈ B (note we can do this since B is
non-empty). Since sup A = ∞ we can find a ∈ A where a > x+b.
Thus a − b ∈ A − B and a − b > x. So sup A − B = ∞.
(b) Now suppose that both A and B are bounded and non-empty
and so both have supremum and infimum in the real numbers.
If we let z ∈ A − B then we can find a ∈ A and b ∈ B where
z = a − b. We then have that a ≤ sup A and b ≥ inf B. So
z = a − b ≤ sup A − inf B.
Thus sup(A − B) ≤ sup A − inf B. Now we need to show that if
y < sup A − inf B then we can find z > y with z ∈ A − B. To do
this let y = sup A − inf B − ε where ε > 0. We can find a ∈ A
with a > sup A − ε/2 and b ∈ B with b < inf B + ε/2. Thus
a − b ∈ (A − B) and
(a − b) > sup A − ε/2 − (sup B − ε/2) = sup A − inf B − ε = y.
6. Firstly suppose that inf A = 0. We wish to show sup B = ∞. We let
x ∈ R (we want to find b ∈ B with b > x). If x ≤ 0 this is easy
since we just take a ∈ A (A is non-empty) and observe that a−1 ∈ B
with a−1 > 0. If x > 0 then since inf A = 0 we can find a ∈ A with
a < x−1 and so ax < 1 and thus a−1 > x and since a−1 ∈ B we have
found b ∈ B with b > x.
Now suppose that inf A 6= 0. Since A ⊆ (0, ∞) and is non-empty
this means that inf A ∈ (0, ∞). Now let b ∈ B. By the definition of
B we can find a ∈ A with b = a−1 . Since a ≥ inf A we have that
a(inf A)−1 ≥ 1 and so b = a−1 ≤ (inf A)−1 . Thus sup B ≤ (inf A)−1 .
Now let y < (inf A)−1 . We can conclude that y −1 > inf A and so we
can find a ∈ A with a < y −1 and thus a−1 > y. Since a−1 ∈ B we can
conclude that sup B ≥ y and so sup B = (inf A)−1 .
2
7. (a) Let x, y ∈ R we have that x2 + xy + y 2 = (x + y/2)2 + 3y4 ≥ 0.
Moreover when x = y = 0 we have x2 + xy + y 2 = 0 and to have
(x + y/2)2 + 3/4y 2 = 0 we need that y 2 = 0 and x = −y/2 which
gives that x = y = 0.
(b) If x, y ∈ R and x3 < y 3 then x3 − y 3 < 0 and so by factorising we
have
(x − y)(x2 + xy + y 2 ) < 0.
By the first part we know that x2 +xy+y 2 > 0 and so multiplying
both sides by (x2 + xy + y 2 )−1 yields that x − y > 0 and thus
x < y.
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(c) If we let a ∈ A then a3 < 2 < y 3 and so since a3 < y 3 by the
previous part of the question we must have that a < y.
(d) Let y ∈ R with y 3 > 2 and let 1 > ε > 0. We have that
(y − ε)3 = y 3 − 3εy 2 + 3yε2 − ε3 .
Since y > 0 and ε < 1 we have
(y − ε)3 ≥ y 3 − 3εy 2 − ε = y 3 − ε(3y 2 + 1).
We want to find ε > 0 where (y − ε) > 2 and so to do this we
y 3 −2
take ε < 3y
2 +1 . This gives that
(y − ε)3 > y 3 − (3y 2 + 1)
y3 − 2
= y 3 − y 3 + 2 = 2.
3y 2 + 1
Since (y − ε) > 2 we know that for all a ∈ A a ≤ (y − ε) and so
(y − ε) is an upper bound for A and thus y > sup A.
(e) Now suppose that y ∈ R with y 3 < 2. We want to find z ∈ R
with z > y and z 3 < 2 in which case we know that z ∈ A and
so sup A > y. Again we want to find ε sufficient small so that
(y + ε)3 < 2. However it helps to first deal with the case when
y < 1. In this case since 1 ∈ A we know that y ≤ sup A. So now
let 1 ≤ y with y 3 < 2.
For 1 > ε > 0 we have that
(y + ε)3 = y 3 + 3y 2 ε + 3yε2 + ε3 ≤ y 3 + 7ε.
So if we take 0 < ε <
2−y 3
7
we have that
(y + ε)3 ≤ y 3 + 7
2 − y3
= 2.
7
(f) By the previous two parts we cannot have (sup A)3 > 2 or sup A <
2 and so we must have that (sup A)3 = 2.
8. (a) For A1 = (0, 1] we know that sup A1 = 1 and 1 ∈ A1 . So A1 has
a maximum value.
(b) For A2 = n−1
n : n ∈ N we know by question 4 (b) that sup A2 =
1. However for all n ∈ N
n−1
1
=1− <1
n
n
and so 1 ∈
/ A2 and we can conclude that A2 does not have a
maximum value.
5
(c) For A3 =
n
1
|n+ 21 |
o
: n ∈ Z we know that if x ∈ A3 then x =
for some n ∈ Z and since |n + 12 | ≥ 12 | for all n ∈ Z we have
1
|n+ 12 |
1
≤ 2.
|n + 12 |
So sup A3 ≤ 2 but if we take n = 0 we have that
1
|0+ 12 |
= 2 ∈ A3
and so sup A3 = 2 and A3 does have a maximum value.
6