Let A be a nonempty set of real numbers. A number m ∈ A is a least element of A if x ≥ m
for every x ∈ A. For example, every finite nonempty set of real numbers and N have a least
element, while Z and the open interval (0, 1) of real numbers do not have a least element. A
nonempty set S of real numbers is said to be well-ordered if every nonempty subset of S has
a least element.
(a) Show that if S is a nonempty set of real numbers and S does not have a least element,
then S is not well-ordered.
(b) Show that the closed interval [0, 1] of real numbers is not well-ordered (thereby showing
that a set with a least element may not be well-ordered).

### Part (a): Show that if \( S \) is a nonempty set of real numbers and \( S \) does not have a least element, then \( S \) is not well-ordered.

#### Proof:

**Assume** that \( S \) is a nonempty set of real numbers that does **not** have a least element.

By the definition of a well-ordered set, every nonempty subset of \( S \) must have a least element for \( S \) to be well-ordered. We aim to show that \( S \) is not well-ordered.

Since \( S \) does not have a least element by assumption, the entire set \( S \) (which is nonempty) itself does not have a least element. This contradicts the definition of a well-ordered set, as \( S \) is a nonempty subset of itself, and hence, it must have a least element if \( S \) were well-ordered.

Thus, \( S \) cannot be well-ordered.

**Conclusion**: If \( S \) does not have a least element, it cannot be well-ordered.

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### Part (b): Show that the closed interval \([0, 1]\) of real numbers is not well-ordered.

#### Proof:

Consider the closed interval \( [0, 1] \), which is the set of all real numbers \( x \) such that \( 0 \leq x \leq 1 \). This set does have a least element, specifically \( 0 \), because \( 0 \in [0, 1] \) and for all \( x \in [0, 1] \), \( x \geq 0 \).

However, we need to check if \( [0, 1] \) is well-ordered, which requires that **every** nonempty subset of \( [0, 1] \) must have a least element.

Let's consider a specific nonempty subset of \( [0, 1] \), such as the open interval \( (0, 1] \), which is the set of all real numbers strictly between \( 0 \) and \( 1 \), including \( 1 \) but excluding \( 0 \).

- This subset \( (0, 1] \) **does not** have a least element. For any \( x \in (0, 1] \), there is always a smaller number that is still in \( (0, 1] \), because real numbers are dense. Therefore, there is no smallest element in \( (0, 1] \), despite the fact that it is a nonempty subset of \( [0, 1] \).

Since \( (0, 1] \subseteq [0, 1] \) is a nonempty subset that does not have a least element, this shows that the closed interval \( [0, 1] \) is **not** well-ordered.

**Conclusion**: The closed interval \( [0, 1] \) is not well-ordered, even though it has a least element. This example demonstrates that having a least element does not guarantee that a set is well-ordered.

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### Do you want details or have any questions?

Here are 5 questions to further explore these concepts:

1. Can a set with more than one least element be well-ordered?
2. How does the concept of density in real numbers affect the existence of least elements in subsets?
3. What is an example of a set of real numbers that is well-ordered?
4. How does well-ordering relate to the Axiom of Choice?
5. Can you think of a subset of \( [0, 1] \) that is well-ordered?

**Tip**: When proving that a set is well-ordered, always verify that **every** nonempty subset has a least element, not just the set itself.

Let A be a nonempty set of real numbers. A number m ∈ A is a least element of A if x ≥ m for every x ∈ A. For example, every finite nonempty set of real numbers and N have a least element, while Z and the open interval (0, 1) of real numbers do not have a least element. A nonempty set S of real numbers is said to be well-ordered if every nonempty subset of S has a least element. (a) Show that if S is a nonempty set of real numbers and S does not have a least element, then S is not well-ordered. (b) Show that the closed interval [0, 1] of real numbers is not well-ordered (thereby showing that a set with a least element may not be well-ordered).

Learn how to prove that the closed interval [0, 1] of real numbers is not well-ordered, despite having a least element. This proof involves the concepts of well-ordering, least elements, and the density of real numbers. Suitable for undergraduate-level math students.

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