University:
The University Of Western Ontario
Course:
MATH 3020A | Introduction to Abstract Algebra
Academic year:

2024
Views:

266

Pages:

2
Author:

Kyuwon J.

Practice problem Zorn’s Lemma Problem 1: Prove that every vector space has a basis using Zorn’s Lemma. Solution: 1. Consider the set of all linearly independent subsets of the vector space V . 2. Order these subsets by inclusion. 3. Any chain (totally ordered subset) of linearly independent sets has an upper bound, which is their union. 4. By Zorn’s Lemma, the set has a maximal element. 5. This maximal element must span V , because if it does not, we can add a vector to it, contradicting maximality. 6. Hence, this maximal element is a basis for V . Problem 2: Using the Axiom of Choice, prove that every set can be well-ordered. Solution: 1. Let X be a set. 2. Define a choice function f such that for any non-empty subset A ⊆ X, f (A) ∈ A. 3. Use f to construct a well-ordering by transfinite recursion: choose the least element in each subset. 4. This process defines a well-ordering of X, demonstrating the claim. Problem 3: Prove that the Axiom of Choice implies Zorn’s Lemma. Solution: 1. Let S be a partially ordered set where every chain has an upper bound. 2. Define a choice function on the collection of all subsets of S. 3. Use transfinite recursion to construct a maximal element: start with an arbitrary element and keep adding upper bounds of chains. 4. This process terminates at a maximal element, proving Zorn’s Lemma. Problem 4: Show that the Well-Ordering Theorem implies the Axiom of Choice. Solution: 1. Assume every set can be well-ordered. 2. Let {Ai }i∈I be a collection of non-empty sets. 3. Well-order each Ai , and select the least element from each set. 4. This defines a choice function, proving the Axiom of Choice. 1 Problem 5: Prove that every chain in a partially ordered set has a least upper bound implies Zorn’s Lemma. Solution: 1. Let S be a partially ordered set where every chain has an upper bound. 2. Assume there is no maximal element. 3. Construct a sequence of increasing elements in S that contradicts the assumption of upper bounds. 4. Use the assumption of least upper bounds to show a maximal element exists. 5. This proves Zorn’s Lemma. Problem 6: Using Zorn’s Lemma, prove that every ring with unity has a maximal ideal. Solution: 1. Let R be a ring with unity, and consider the set of all proper ideals of R. 2. Order these ideals by inclusion. 3. Any chain of proper ideals has an upper bound, which is their union. 4. By Zorn’s Lemma, there exists a maximal proper ideal. 5. This ideal cannot be further extended while remaining proper, proving the claim. Problem 7: Show that the Axiom of Choice is equivalent to the statement that every surjective function has a right inverse. Solution: 1. Assume the Axiom of Choice. 2. Let f : X → Y be surjective. For each y ∈ Y , choose an x ∈ X such that f (x) = y. 3. Define g : Y → X by g(y) = x. 4. This g is a right inverse of f . 5. Conversely, if every surjective function has a right inverse, use it to construct a choice function, proving the Axiom of Choice. 2