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.