Math Problem Statement

Use the PMI to prove the well ordering principle. Hint: let S ⊂ N be

nonempty, and define a property P (n) of natural numbers by setting P (n) to be true

exactly when 1, . . . , n̸ ∈ S.

Solution

To prove the Well-Ordering Principle (WOP) using Principle of Mathematical Induction (PMI), we'll follow the hint provided. The Well-Ordering Principle states that every non-empty subset SNS \subseteq \mathbb{N} of the natural numbers contains a least element.

Step-by-Step Proof:

1. Setup the Problem:

Let SNS \subseteq \mathbb{N} be a non-empty subset of the natural numbers. Our goal is to show that SS has a least element.

We define a property P(n)P(n) for a natural number nn as follows:

  • P(n)P(n) is true if and only if none of the numbers 1,2,,n1, 2, \dots, n are elements of SS, i.e., P(n):1,2,,nS.P(n): 1, 2, \dots, n \notin S. In other words, P(n)P(n) asserts that the first nn natural numbers do not belong to the set SS.

2. Base Case:

We use the contrapositive approach, where we assume SS is non-empty and attempt to derive a contradiction.

Since SNS \subseteq \mathbb{N} is non-empty, we know there must exist some natural number mSm \in S.

Suppose, for contradiction, that SS has no least element. This would mean that for all nNn \in \mathbb{N}, the first nn natural numbers do not belong to SS, implying that P(n)P(n) holds for all nNn \in \mathbb{N}.

Start with P(1)P(1), which states that the number 1 does not belong to SS. Since SS is non-empty, we know SS must contain some element, but P(1)P(1) implies 1S1 \notin S.

3. Induction Hypothesis:

Now, assume for some nNn \in \mathbb{N}, that P(n)P(n) is true. This means 1,2,,nS1, 2, \dots, n \notin S, i.e., none of the numbers from 1 to nn belong to SS.

4. Inductive Step:

We now prove that P(n+1)P(n+1) holds, assuming P(n)P(n) is true. By P(n)P(n), we know that none of the numbers 1,2,,n1, 2, \dots, n belong to SS.

We must show that n+1Sn+1 \notin S. If n+1Sn+1 \in S, then n+1n+1 would be a member of SS and would be the smallest member of SS, since by the induction hypothesis, no number less than n+1n+1 is in SS. But this contradicts the assumption that SS has no least element.

Therefore, n+1Sn+1 \notin S, and P(n+1)P(n+1) is true.

5. Conclusion by Induction:

By the principle of mathematical induction, P(n)P(n) is true for all nNn \in \mathbb{N}. This means that none of the natural numbers 1,2,3,1, 2, 3, \dots belong to SS.

However, this contradicts the fact that SS is non-empty (since SS must contain some natural number). Therefore, our original assumption that SS has no least element must be false.

6. Conclusion:

Hence, SS must have a least element. This proves the Well-Ordering Principle using the Principle of Mathematical Induction.


Would you like more details or have any questions on specific steps?

Here are 5 related questions:

  1. How does the contrapositive method support the proof of the well-ordering principle?
  2. What is the significance of the inductive hypothesis in this proof?
  3. Can you provide an example of a non-empty subset of natural numbers that has a least element?
  4. How does this proof compare with direct proofs of the well-ordering principle?
  5. How can the well-ordering principle be applied in other mathematical contexts, such as divisibility?

Tip: The Well-Ordering Principle and PMI are closely related; many proofs can be structured by converting one into the other. Understanding this connection helps in abstract problem-solving across different areas of mathematics.

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Math Problem Analysis

Mathematical Concepts

Principle of Mathematical Induction (PMI)
Well-Ordering Principle
Set Theory

Formulas

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Theorems

Well-Ordering Principle
Principle of Mathematical Induction

Suitable Grade Level

Undergraduate Mathematics or Advanced High School