the least upper bound property · archimedean property video: archimedean property here’s a cool...

LECTURE 5: THE COMPLETENESS AXIOM (II)

1. The Least Upper Bound Property

Video: Least Upper Bound Property

Last time: Defined the concept of a sup, which is a more relaxedversion of a max:

Sup

M = sup(S) if for all M1 < M there is s1 ∈ S such that s1 > M1

Example: sup[0, 4) = 4

Date: Wednesday, April 8, 2020.

1

https://youtu.be/OQ0HBjq8OWE

2 LECTURE 5: THE COMPLETENESS AXIOM (II)

Now why is this such an important concept? Because even though themax of S might not always exist the sup always exists1:

Least Upper Bound Property

If S is a nonempty subset of R that is bounded above, then S hasa least upper bound, that is sup(S) exists.

Note: Geometrically, this theorem is saying that R is complete, thatis it does not have any gaps/holes.

Non-Example: The property is NOT true for Q. Let:

S ={x ∈ Q | x2 < 2

}

Then S is bounded above by 3.2, but it doesn’t have a least upperbound in Q because sup(S) =

√2 but

√2 isn’t in Q.

In some sense, Q is broken: It has holes and gaps where the sup issupposed to be!

1The book calls it the Completeness Axiom2Because if x > 3, then x2 > 9 ≥ 2, so x cannot be in S, so by the contrapositive x ∈ S ⇒ x ≤ 3

LECTURE 5: THE COMPLETENESS AXIOM (II) 3

R doesn’t have that problem, it is complete, it has no holes, the supis exactly where it’s supposed to be!

Fun Fact: It’s always possible to fix a broken heart; it’s always pos-sible to complete a space with holes.

2. The Case of Inf

Video: inf(S) = − sup(−S)

From the above, we know that sup always exists, but what about inf?

Inf

m = inf(S) if for all m1 > m there is s1 ∈ S such that s1 < m1

https://youtu.be/x0ZweOLCWek


It turns out that inf and sup are related in a really elegant way!

Definition:

If S is any subset of R,then

−S = {−x | x ∈ S}

(In other words, reflect S across the origin)

Example: If S = [1, 3] then −S = [−3,−1]


Notice: In this example, inf(S) = 1, but also sup(−S) = −1, soinf(S) = 1 = −(−1) = − sup(−S), and in fact this is always true:

Important Fact (memorize!)

inf(S) = − sup(−S)

What this is saying is that anything that is true about sup is also truefor inf. We’ll see a consequence soon.

Proof of Important Fact:

Let m = − sup(−S) then: inf(S) = − sup(−S)⇔ inf(S) = m.

In order to show inf(S) = m, we need to show S is bounded below bym (skip3) and: If m1 > m then there is s1 ∈ S such that s1 < m1.

3Since sup(−S) = −m, −S is bounded above by −m, so for all (−s) ∈ −S, −s ≤ −m⇒ s ≥ mfor all s ∈ S


Now if m1 > m, then −m1 < −m. But −m = sup(−S), so since −m1

is smaller than the sup, there is s′ ∈ −S such that s′ > −m1.

But by definition of −S, s′ = −s1 for some s1 ∈ S.

Claim: This s1 works

Why? Because s′ > −m1 ⇒ −s1 > −m1 ⇒ s1 < m1 X �

Why useful? This basically says that you never have to prove state-ments with inf: Just prove the version with sup and use this theorem.In fact, let’s illustrate this with:

Greatest Lowest Bound Property

If S is a nonempty subset of R that is bounded below, then inf(S)exists.


Proof: Suppose S is a nonempty subset that is bounded below by m,then for all s ∈ S, s ≥ m > −∞, so for all s ∈ S, −s ≤ −m. Thissays that −S is bounded above by −m <∞.

By the Least Upper Bound Property, sup(−S) exists, and thereforeinf(S) exists because

inf(S) = − sup(−S)︸︷︷︸Exists

�

3. Archimedean Property

Video: Archimedean Property

Here’s a cool consequence of the least upper bound property:

Scenario: Suppose you go to the grocery store and your total comesdown to $100. Suppose you only have 1 dollar bills at hand (but asmany as you need). Can you pay your total? Yes! What if the totalis $1000 and you only have 1 cent coins? Still yes! This is the essence of:

Archimedean Property

If a > 0 and b > 0 are real numbers, then for some n ∈ N we havena > b

Interpretation: No matter how large the total b is and how smallyour currency a is, it is always possible to exceed b by using enough a

https://youtu.be/DssLr2xHWNg


Or, to quote the quote in the book, “Given enough time, one can emptya large bathtub with a small spoon.”

Proof: Assume a < b4 (Your total is bigger than your currency)

Suppose this is false, that is there are a > 0 (currency) and b > 0(total) such that na ≤ b for all n ∈ N:

In particular, this means if you let S = {na | n ∈ N · · · } (nonemptysince a ∈ S), then b is an upper bound for S. So by the Least UpperBound Property, M =: sup(S) exists.

4For the other cases: If a > b, n = 1 works, and if a = b, n = 2 works


Now consider M-a < M (since a > 0), so by the definition of sup(S),we have there is s1 ∈ S such that s1 > M − a

But by definition of S, s1 = n0a for some n0 ∈ N. Therefore:

s1 > M − a⇒ n0a > M − a⇒ n0a + a > M ⇒ (n0 + 1)a > M

On the other hand, since M = sup(S), for all n ∈ N, na ≤ M , hence(n0 + 1)a ≤M , but then (n0 + 1)a ≤M < (n0 + 1)a⇒⇐ �

4. Dense with me!

Video: Q is dense in R

Finally, using the Archimedean property, we can show the followingimportant fact about Q, It says that between two rational numbersthere always is a real number:

Theorem (Q is dense in R)

For any real numbers a and b with a < b there is r ∈ Q such thata < r < b

https://youtu.be/A9L6YVtoAsQ


Note: The point is that even if a and b are really close together, youcan always squeeze a rational number between them. Intuitively thisis saying that, even though Q has holes, it still fills up “most” of R,unlike Z for instance, which is pretty sparse.


Proof: Suppose a < b, WTF r = mn such that a < m

n < b.

STEP 1: Since b − a > 0, by the Archimedean property applied tob − a (your currency) and 1 (your total), there is n ∈ N such thatn(b− a) > 1, that is b− a > 1

n

Note: For the remainder of the proof, remember that n is fixed (ifyou want, think n = 3).


STEP 2: WLOG, assume a > 05

Main Idea: List all the fractions 0, 1n ,

2n ,

3n , · · · until you reach the last

one that is < b. This process has to stop because b is finite, and alsothat last fraction is guaranteed to be between a and b because a and bare at least 1

n apart.

Here are the details: Consider the following set:

S ={mn|m = 0, 1, 2, · · · and

m

n< b}

Then S is nonempty (0 ∈ S) and S is bounded above by b, so by theLeast Upper Bound Property, sup(S) = r exists.

Claim: r solves our problem

Why? First of all, S only has finitely many elements: By the Archimedeanproperty with 1

n (currency) and b (total), there is k ∈ N such that

k(

1n

)= k

n > b, so S has at most k elements (k because S includes 0

5For the other cases: If a < b < 0, then notice −a > −b > 0 and use this proof to find r rationalwith −b < r < −a and then −r does the trick. And if a = 0 then r = 1

n works since b− a > 1n and

if b = 0 then r = − 1n works. And if a < 0 < b, then let r = 0


but not kn)

In particular, since S is finite, sup(S) = max(S) (Think for exampleS = {1, 3, 5, 9}. You can compare all the elements of S one by one andpick the one that is is largest6). In particular, r = sup(S) = max(S) ∈S (by definition of max), so by definition of S, r is rational and r < b.

To show r > a, suppose r = mn ≤ a, but then

b− a >1

n⇒ b > a +

1

n≥ m

n+

1

n=

m + 1

n

Hence m+1n < b and so m+1

n is an element of S that is bigger thanr = m

n , which contradicts the fact that r = max(S)⇒⇐ �

6If you want to make this 100% rigorous, you can (but don’t have to) do induction on the sizeof S. Namely Pn would be the proposition “If S has n elements, then S has a max”

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