Least Upper Bounds
where we prove Theorems 1 and 2
from the last chapter.
Definitions
A set A of real numbers is bounded above, if
there exists a number x, such that x a, for all
aA.
x is an upper bound for A.
x is a least upper bound, supremum, sup(A) for
A, if
(i) x is an upper bound for A, and
(ii) if y is an upper bound for A, then x y.
Similarly, z = inf A = infimum of A
= greatest lower bound of A
Note: neither the supremum nor the infimum
have to be in A.
Exercise
Find the least upper bound u of A = [0,1) ∩ ℚ
in ℝ.
Is u in A?
Exercise
Find the least upper bound u of A = [0,1) ∩ ℚ
in ℚ.
Exercise
Consider the sequence defined recursively by
x1 = 2
xn+1 = 2 + xn
Show, by induction that xn < 2, for all n.
Exercise
Consider the sequence defined recursively by
x1 = 2
xn+1 = 2 + xn
Show, by induction that xn < xn+1, for all n.
Note:
Example of a set that is bounded above but
doesn’t have a least upper bound: Ø, the
empty set.
Any real number is an upper bound.
The Least Upper bound Property of
Real Numbers
If A is a set of real numbers, A ≠ Ø, and A
bounded above, then A has a least upper
bound.
Note:
As we have seen, ℚ does not have the least
upper bound property.
Exercise
Suppose A, B are nonempty sets of real
numbers, such that a b, for all aA, bB.
(i) Prove sup A b for all bB.
(ii) Prove sup A inf B.
Proof of Theorem 7.1
If f is continuous on [a,b] and f(a) < 0 < f(b),
then there is an x in (a,b) such that f(x) = 0.
Proof idea:
find the smallest x[a,b], such that f(x) = 0.
Question
In the proof of Theorem 7.1, we showed that
there exists a smallest x in (a,b) with f(x) = 0.
Is there necessarily a second smallest x in (a,b)
with f(x) = 0?
Exercise
Show that there is a largest x in (a,b) with
f(x) = 0.
Exercise
Given f continuous, nonnegative on [a,b],
f(a) = f(b) = 0 and there exists an x0, such that
f(x0) > 0.
Then there exist c, d, such that
a c < x0 < d b and
f(c) = f(d) = 0 and for all x in (c,d): f(x) > 0.
Theorem 8.1
If f is continuous at a, then there exists >0
such that f is bounded above on (a-, a+).
Proof: this follows from the continuity of f at a.
Try it.
Similarly:
• f bounded below on a suitable (a-, a+).
• f bounded above and below if a is the
endpoint of a closed interval and f
continuous at a.
Theorem 8.2
Archimedian Property
ℕ is not bounded above.
Proof: (by contradiction)
Theorem 8.2
Archimedian Property
ℕ is not bounded above.
Proof: (by contradiction) Suppose not.
Since ℕ is not empty and bounded above, there is
a least upper bound .
So n, for all nℕ, in particular
n+1. (n+1ℕ, if nℕ)
Thus, -1 n, for all nℕ and is not a least
upper bound. Contradiction
Corollary
For every ε>0, there exists an nℕ, such that
1
< ε.
𝑛
Proof: by contradiction.
Try it.
We have actually used this many times before,
when dealing with limits.
Definition
A set of real numbers A is dense, if every open
interval contains an element of A.
Exercise
ℚ is dense in ℝ.
Proof: Consider the open interval (x,y), x < y.
Need to show that there exists qℚ, such that
x < q < y.
Try it.
Nested Interval Theorem
Consider a sequence of closed intervals
I1=[a1,b1], I2=[a2,b2],…, where
a1 … an … bm … b1.
Prove that there is a point x which is in every In.
Also show that this conclusion is false for open
intervals.
Exercise
Suppose f is continuous on [a,b], then either
𝑎+𝑏
• 𝑓 =0
2
𝑎+𝑏
• 𝑓 𝑎 <0<𝑓 ∗
2
𝑎+𝑏
• f < 0 < 𝑓(𝑏)
2
Suppose (*). Apply the same argument to
𝑎+𝑏
[a, ].
2
Claim: This process will lead to x[a,b] such that f(x)=0.
Definition
The point x is an accumulation point of the set
A, if for any open set O containing x, there are
infinitely many points from A in O.
x doesn’t have to be in A.
Exercises
Show: Every point in [0,1] is an accumulation
point.
Show: No point at all is an accumulation point of
the set of natural numbers ℕ.
Show: every point on the real line, both rational
and irrational, is an accumulation point of the
set ℚ.
Exercise (additional)
Show: very point of a closed interval [a, b] is an
accumulation point of (a, b). No point outside
can be.
Definitons
Suppose A is a set of real numbers.
The limit superior = lim A = largest
accumulation point of A.
The limit inferior = lim A = smallest
accumulation point of A.
Execises
Find the limits superior and inferior, minima,
maxima, if they exist, of:
1
1. 𝐴 = 𝑛
𝑛 ∈ ℕ}
2. 𝐴 = 𝑥 0 𝑥 2, 𝑥 rational}
1
3. 𝐴 = 𝑛
+ −1 𝑛 𝑛 ∈ ℕ}
Uniform continuity
this measures how much a function oscillates
i.e. how fast a function grows locally (ultimately
its slope).
Example
f(x) = x2 is continuous, i.e it is continuous at each 𝑎 :
𝑎 ∈ ℝ, ε > 0, δ𝑎ε > 0,
such that 𝑥 satisfying 𝑥 − 𝑎 < δ𝑎ε
|f(x) − f(𝑎)|< ε
here the δ depends on 𝑎 and on ε.
Draw picture.
We aim for a function where the δ depends only on
on ε, and not on 𝑎.
Definition
f is uniformly continuous on an interval I, if for
every ε > 0, there is a δε > 0,
such that for all x,y ∈ I, if |x − y| < δ, then
|f(x) – f(y)| < ε.
Are they uniformly continuous?
1
1. f(x) = on (0,1)
𝑥
1 1
2. f(x) = on ( ,1)
𝑥 𝑛
3. f(x) = sin x on ℝ
4. f(x) = x3 on [-100,100]
5. f(x) = x
6. f(x) = ex
7. f(x) = e-x on ℝ+
Lemma
If f is uniformly continuous on [a,b] and on
[b,c], then f is uniformly continuous on [a,c].
Technically:
Lemma
Let a < b < c and f continuous on [a,c].
Let ε > 0 and suppose
(i) δ1 such that if 𝑥, 𝑦 ∈ 𝑎, 𝑏 and
𝑥 − 𝑦 < δ1, then |f(x) – f(y)| < ε
(ii) δ2 such that if 𝑥, 𝑦 ∈ 𝑏, 𝑐 and
𝑥 − 𝑦 < δ2, then |f(x) – f(y)| < ε
then
δ > 0 such that if 𝑥, 𝑦 ∈ 𝑎, 𝑐 and
𝑥 − 𝑦 < δ, then |f(x) – f(y)| < ε.
Theorem 8.3
If f is continuous on [a,b], then f is uniformly
continuous on [a,b].
Theorem 7.2
If f is continuous on [a,b], then f is bounded on [a,b].
Proof: Take ε = 1, then there exists a δ>0, such that for
any x, y satisfying |x – y|< δ, then
|f(x) – f(y)|<1. Cover [a,b] with overlapping
subintervals
δ 3δ (2n+1)δ
[a, a + δ), (a + , a+ ), … ,(a+ , b]
2 2 2
There are n such intervals. Over each of them f varies by
at most 1, so altogether |f(x) – f(y)|< n
for any x, y in [a,b].
Theorem 7.3
If f is continuous on [a,b], then there are
numbers y, z [a,b] such that
for all x [a,b]: 𝑓(𝑥) ≤ 𝑓(𝑦) and
for all x [a,b]: 𝑓(𝑥) ≥ 𝑓(𝑧)
Proof: see text
