Clarifications on the least upper bound property and the irrational numbers

In summary, the least upper bound property does not hold for the set of rational numbers because the least upper bound is not rational.
  • #1
drobadur
2
0
Hello everyone.
I desperately need clarifications on the least upper bound property (as the title suggests). Here's the main question:

Why doesn't the set of rational numbers ℚ satisfy the least upper bound property?

Every textbook/website answer I have found uses this example:

Let S={x∈ℚ : x≤√2}. This has an upper bound in ℚ, but it has no supremum in ℚ, since √2 is irrational, therefore ℚ does not satisfy the least upper bound property.

Here's my major problem:
A supremum is not necessarily an element of the set it's being a least upper bound of.
By the definition of the least upper bound property:
(i)A real number x is called an upper bound for S if x ≥ s for all s ∈ S.
(ii)A real number x is the least upper bound (or supremum) for S if x is an upper bound for S and x ≤ y for every upper
bound of S.
Nowhere does this definition state that the supremum has to belong in that set.

It seems to me that √2 should perfectly qualify as the supremum of S by this very definition. After all x≤√2 for all x∈S.
But why do we dismiss it by saying it is not rational? Why does it HAVE to be? Why does the supremum of a subset of ℚ also HAS to be in ℚ, or else ℚ doesn't satisfy the least upper bound property?

I'm probably missing something and this question looks idiotic to some, but frankly I don't see it. I hope someone can help me out by giving a detailed, comprehensible answer. Thanks in advance.
 
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  • #2
You're using the wrong definition of supremum. Your definition only holds if the ambient set is the real numbers.

Given an ordered set [itex](E,\leq)[/itex] and let [itex]S\subseteq E[/itex].
We say that [itex]x\in E[/itex] is an upper bound of S if [itex]s\leq x[/itex] for all s in S.
We say that [itex]x\in E[/itex] is a supremum of S if it is an upper bound of S and if for each other upper bound y of S holds that [itex]x\leq y[/itex].

So, if you use this definition for [itex]E=\mathbb{Q}[/itex], then you see that the supremum must be in [itex]\mathbb{Q}[/itex].
 
  • #3
drobadur said:
Hello everyone.
I desperately need clarifications on the least upper bound property (as the title suggests). Here's the main question:

Why doesn't the set of rational numbers ℚ satisfy the least upper bound property?

Every textbook/website answer I have found uses this example:

Let S={x∈ℚ : x≤√2}. This has an upper bound in ℚ, but it has no supremum in ℚ, since √2 is irrational, therefore ℚ does not satisfy the least upper bound property.

Here's my major problem:
A supremum is not necessarily an element of the set it's being a least upper bound of.
By the definition of the least upper bound property:
(i)A real number x is called an upper bound for S if x ≥ s for all s ∈ S.
(ii)A real number x is the least upper bound (or supremum) for S if x is an upper bound for S and x ≤ y for every upper
bound of S.
Nowhere does this definition state that the supremum has to belong in that set.

It seems to me that √2 should perfectly qualify as the supremum of S by this very definition. After all x≤√2 for all x∈S.
But why do we dismiss it by saying it is not rational? Why does it HAVE to be? Why does the supremum of a subset of ℚ also HAS to be in ℚ, or else ℚ doesn't satisfy the least upper bound property?

I'm probably missing something and this question looks idiotic to some, but frankly I don't see it. I hope someone can help me out by giving a detailed, comprehensible answer. Thanks in advance.

It looks like a simple misunderstanding. A set has the least upper bound property if the least upper bound of any bounded subset is in the set. The rationals do not satisfy this property, since the least upper bound (as in the example) may not be rational.
 
  • #4
Thanks a lot!
 
  • #5


Dear reader,

Thank you for your question and for seeking clarifications on the least upper bound property and irrational numbers. I understand your confusion and I will try my best to provide a detailed and comprehensible answer.

First, let's define the least upper bound property. The least upper bound property, also known as the completeness property, states that every non-empty set of real numbers that is bounded above has a least upper bound (or supremum) that is also a real number. This means that for any set of real numbers, there is a smallest number that is greater than or equal to all the numbers in the set.

Now, let's look at the set of rational numbers, ℚ, and why it does not satisfy the least upper bound property. As you mentioned, the set S={x∈ℚ : x≤√2} has an upper bound in ℚ, but it does not have a supremum in ℚ. This is because the supremum of S is √2, which is an irrational number.

You are correct in saying that the definition of the supremum does not state that it has to be an element of the set it is being a least upper bound for. However, in the case of the set S, the supremum √2 is not a rational number, which means it is not an element of the set of rational numbers, ℚ. This is why we say that ℚ does not satisfy the least upper bound property.

To understand why the supremum of a subset of ℚ also has to be in ℚ, we need to understand the definition of a rational number. A rational number is any number that can be expressed as a fraction of two integers. In other words, it can be written in the form a/b, where a and b are integers and b≠0. Now, let's consider the number √2. This number cannot be expressed as a fraction of two integers. Therefore, it is not a rational number and it does not belong in the set of rational numbers, ℚ.

In conclusion, the reason why √2 cannot be the supremum of the set S in ℚ is because it is not a rational number. And since the set of rational numbers, ℚ, only includes rational numbers, it does not satisfy the least upper bound property. I hope this helps clarify your confusion. If you have any further questions, please do not hesitate to ask
 

1. What is the least upper bound property?

The least upper bound property, also known as the completeness property, states that every non-empty subset of the real numbers that is bounded above has a least upper bound. In other words, given a set of numbers, there will always be a largest number that is still smaller than all other upper bounds.

2. How does the least upper bound property relate to irrational numbers?

The least upper bound property guarantees that irrational numbers exist. This is because irrational numbers, such as pi or the square root of 2, cannot be expressed as a ratio of two integers and therefore, do not have an exact decimal representation. The least upper bound property ensures that there is always a real number that is slightly larger than these irrational numbers.

3. Why is the least upper bound property important in mathematics?

The least upper bound property is important because it allows us to define and work with real numbers. It is a fundamental property in the construction of the real number system and is essential in many mathematical proofs and calculations.

4. Is the least upper bound property unique to the real numbers?

Yes, the least upper bound property is a defining characteristic of the real numbers. It is not applicable to other number systems, such as the rational or irrational numbers, which do not have this property.

5. Can the least upper bound property be extended to other sets besides the real numbers?

Yes, the least upper bound property can be extended to other sets, such as the complex numbers or the set of continuous functions. However, the exact definition and application may vary depending on the specific set and its properties.

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