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Chapter 1: Problem 9

Let \(\mathfrak{X}\) be a vector space over a field \(\mathbb{F} .\) A basis for \(\mathfrak{X}\) is a subset \(\mathfrak{B}=\left\\{e_{j} \mid j \in J\right\\}\) of linearly independent vectors from \(\mathfrak{X}\), such that every \(x\) in \(\mathfrak{X}\) has a (necessarily unique) decomposition as a finite linear combination of vectors from \(\mathfrak{B}\). Show that every vector space has a basis. Hint: A basis is a maximal element in the system of linearly independent subsets of \(\mathfrak{X}\).

Short Answer

Expert verified
By using Zorn's Lemma, we can conclude that every vector space has a basis.

Step by step solution

01

Define a Basis

A basis for a vector space \(\mathfrak{X}\) is a set \(\mathfrak{B}\) of linearly independent vectors such that every vector in \(\mathfrak{X}\) can be written uniquely as a linear combination of the vectors in \(\mathfrak{B}\). This means there are no linear dependencies in \(\mathfrak{B}\), and \(\mathfrak{X}\) cannot span any vector outside \(\mathfrak{B}\) without losing linear independence.
02

Use Zorn's Lemma

To prove that every vector space has a basis, we will use Zorn's Lemma. Zorn's Lemma states that if every chain (a totally ordered subset) of a partially ordered set has an upper bound, then the whole set has a maximal element.
03

Consider the Partially Ordered Set

Consider the set of all linearly independent subsets of \(\mathfrak{X}\) ordered by inclusion. This set is partially ordered because for any two linearly independent sets, one set is either contained in or distinct from the other.
04

Check the Chain Condition

If \(\mathfrak{C}\) is a chain (totally ordered set) of linearly independent subsets in \(\mathfrak{X}\), consider the union \(\bigcup \mathfrak{C}\). This union is also linearly independent because any dependency would imply a dependency in one of the smaller sets, contradicting our assumption of linear independence.
05

Apply Zorn's Lemma

By Zorn's Lemma, the set of linearly independent subsets of \(\mathfrak{X}\) has a maximal element. This maximal element, by definition, cannot be added to without introducing a linear dependency.
06

Conclude Existence of a Basis

Since the maximal linearly independent set cannot have more elements added to it without losing independence, it must span the entire space \(\mathfrak{X}\). Hence, this set is a basis for the vector space.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Independence
Linear independence is a fundamental concept in studying vector spaces. It refers to a set of vectors where no vector can be formed as a linear combination of the others. In simple terms, each vector provides unique information, so removing any vector would reduce the 'freedom' or dimensions in the space.

For example, consider the vectors
  • If vectors \( v_1, v_2, \ldots, v_n \) in a vector space are linearly independent, then the only solution to the equation \[ c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0 \]is with each coefficient \( c_i = 0 \).
  • This property ensures that none of the vectors is redundant. If you could express a vector as a combination of others, it wouldn't be needed for the basis.
Linear independence is crucial because it prevents redundancy in spanning a vector space, ensuring each vector contributes to the space's complexity.
Zorn's Lemma
Zorn's Lemma is a powerful tool in set theory often used to prove the existence of certain objects in mathematics, particularly in contexts like vector spaces. Here's how it applies to vector spaces:
  • **Understanding Zorn's Lemma:** It states that if every chain (or totally ordered subset) of a partially ordered set has an upper bound, the set itself must have a maximal element.
  • **Applying It to Vector Spaces:** In proving that every vector space has a basis, Zorn's Lemma helps us find a maximal linearly independent set. By considering all linearly independent sets ordered by inclusion, Zorn鈥檚 lemma asserts that there鈥檚 a linearly independent set that cannot be extended.
Essentially, Zorn's Lemma is an elegance of theoretical mathematics. It allows us to conclude the existence of a full basis in situations where constructing such a basis explicitly is complex or impractical.
Vector Space
A vector space is a foundational concept in linear algebra, comprising vectors and defined over a field of scalars. It is a collection of vectors, where you can perform vector addition and scalar multiplication, and these operations satisfy several axioms like associativity, commutivity, and distributivity.

Considerations in vector spaces include:
  • **Scalars and Vectors:** Scalars typically come from a field, such as real numbers \( \mathbb{R} \) or complex numbers \( \mathbb{C} \), and vectors are elements that can be added and multiplied by scalars.
  • **Axioms:** For any vectors \( u, v, w \) in the vector space and any scalars \( c, d \), the operations satisfy several axioms:- Additive identity: There exists a zero vector \(0\) such that \( v + 0 = v \).- Additive inverses: For every vector \(v\), there exists a vector \(-v\) such that \(v + (-v) = 0 \).
  • **Spanning the Space:** The idea of spanning involves expressing every element in a space as a combination of a set of vectors, known as a spanning set. A basis goes beyond this by ensuring these vectors are also linearly independent.
Understanding vector spaces is crucial as they provide the framework for analyzing vectors in any dimension, paving the way for complex applications in various fields such as physics, engineering, and computer science.
Linear Combinations
Linear combinations involve creating new vectors within a vector space by combining other vectors with scalars. This process is integral to understanding how vector spaces function and how dimensions are represented.
  • **Forming Linear Combinations:** Given vectors \( v_1, v_2, \ldots, v_n \) and scalars \( a_1, a_2, \ldots, a_n \), a linear combination is expressed as \[ a_1 v_1 + a_2 v_2 + \cdots + a_n v_n. \]
  • **Purpose of Combinations:** Linear combinations reflect ways to reach or construct any vector within the span of a set. It's about mixing available resources (vectors) to create needed results (other vectors in the space).
  • **Relation to Basis:** Every vector in a vector space can be uniquely represented by a linear combination of the basis vectors, which underscores the usefulness of linear combinations in exploring the dimensions and properties of a vector space.
Mastering linear combinations enables smoother navigation through vector spaces and enhances problem-solving capabilities in mathematical computations.

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Most popular questions from this chapter

A set is called countable (or countably infinite) if it has the same cardinality (cf. E 1.1.6) as the set No natural numbers. Show that there is a well-ordered set \((X, \leq)\), which is itself uncountable, but which has the property that each segment min \(\\{x\\}\) is countable if \(x \in X\). Hint: Choose a well-ordered set \((Y, \leq)\) that is uncountable. The subset \(Z\) of elements \(z\) in \(Y\) such that the segment \(\min \\{z\\}\) is uncountable is either empty (and we are done) or else has a first element \(\Omega\) Set \(X=\min \\{\Omega\\}\). The ordinal number (corresponding to) \(\Omega\) is called the first uncountable ordinal.

(Topology according to Kuratowski.) Let \(\mathscr{F}(X)\) denote the system of subsets of a set \(X\), and consider a function \(Y \rightarrow \operatorname{cl}(Y)\) of \(\mathscr{F}(X)\) into itself that satisfies the four closure axioms: (i) \(\operatorname{cl}(\emptyset)=\emptyset\) (ii) \(Y \subset \operatorname{cl}(Y)\) for every \(Y\) in \(\mathscr{F}(X)\). (iii) \(\operatorname{cl}(\mathrm{cl}(Y))=\operatorname{cl}(Y)\) for every \(Y\) in \(\mathscr{S}(X)\). (iv) \(\operatorname{cl}(Y \cup Z)=\operatorname{cl}(Y) \cup \operatorname{cl}(Z)\) for all \(Y\) and \(Z\) in \(\mathscr{S}(X)\). Show that the system of sets \(F\) such that \(\mathrm{cl}(F)=F\) form the closed sets in a topology on \(X\), and that \(Y^{-}=\operatorname{cl}(Y), Y \in \mathscr{S}(X)\).

Let \(f: X \rightarrow \mathbb{R}\) be a continuous function on a compact space \(X\). Show that \(f\) is bounded. Show that \(f\) attains its maximum and its minimum on \(X .\) Assume only that \(f\) is semicontinuous on \(X\) and prove "half" of the two previous assertions.

Show that a paracompact Hausdorff space is normal. Hint: If \(E\) and \(F\) are disjoint, closed subsets of \(X\), use regularity to cover \(E\) with a family \(\left\\{A_{j} \mid j \in J\right\\}\) of open sets such that \(A_{j}^{-} \cap F=\emptyset\). Use paracompactness to conclude that the covering may be taken to be locally finite. Set $$ A=\bigcup A_{j}, \quad B=X \backslash \bigcup A_{j}^{-}, \quad B^{o}=X \backslash\left(\bigcup A_{j}^{-}\right)^{-} $$ Show that \(E \subset A, F \subset B\) and \(A \cap B=\emptyset\). Use the local finiteness to conclude that \(B=B^{\circ}\).

We define an equivalence relation on the class of sets by setting \(X \sim Y\) if there exists a bijective map \(h: X \rightarrow Y\). Each equivalence class is called a cardinal number. Show that the natural numbers are the cardinal numbers for finite sets. Discuss the "cardinality" of some infinite sets, e.g. \(\mathbb{N}, \mathbb{Z}, \mathbb{R}\), and \(\mathbb{R}^{2}\).
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