Problem 10 - Algebra

Question

Begründen Sie: (a) Die Menge \(\mathbb{R}^{\mathbb{N}_{0}}\) aller reellen Folgen bildet mit der komponentenweisen Addition \(\left(a_{n}\right)_{n}+\left(b_{n}\right)_{n}:=\left(a_{n}+b_{n}\right)_{n}\) eine Gruppe. (b) Die Abbildungen $$ r:\left\\{\begin{array}{ccc} \mathbb{R}^{\mathbb{N}_{0}} & \rightarrow & \mathbb{R}^{\mathbb{N}_{0}}, \\ \left(a_{0}, a_{1}, \ldots\right) & \mapsto & \left(0, a_{0}, a_{1}, \ldots\right) \end{array} \text { bzw. } l:\left\\{\begin{array}{ccc} \mathbb{R}^{\mathbb{N}_{0}} & \rightarrow & \mathbb{R}^{\mathbb{N}_{0}} \\ \left(a_{0}, a_{1}, \ldots\right) & \mapsto & \left(a_{1}, a_{2}, \ldots\right) \end{array}\right.\right. $$ bei der die Folgenglieder um eine Stelle nach rechts verschoben bzw. nach links verschoben werden, sind Homomorphismen. (c) Die Abbildung \(r\) ist injektiv, aber nicht surjektiv, die Abbildung \(l\) ist surjektiv, aber nicht injektiv.

Step-by-Step Solution

Answer

(a) \( \mathbb{R}^{\mathbb{N}_{0}} \) is a group under addition. (b) Functions \( r \) and \( l \) are homomorphisms. (c) \( r \) is injective, not surjective; \( l \) is surjective, not injective.

Step 1: Define the Group Structure

To show that \( \mathbb{R}^{\mathbb{N}_{0}} \) forms a group under component-wise addition, we need to verify the group axioms: closure, associativity, identity element, and inverses.

Step 2: Show Closure

For any two sequences \( (a_n)_n, (b_n)_n \in \mathbb{R}^{\mathbb{N}_{0}} \), their sum \( (a_n + b_n)_n \) is also a sequence in \( \mathbb{R}^{\mathbb{N}_{0}} \). Thus, the operation is closed.

Step 3: Verify Associativity

For all sequences \( (a_n)_n, (b_n)_n, (c_n)_n \in \mathbb{R}^{\mathbb{N}_{0}} \), \( ((a_n + b_n) + c_n)_n = (a_n + (b_n + c_n))_n \). Therefore, addition is associative.

Step 4: Identify Identity Element

The sequence \( (0)_n \), where each component is zero, is the identity element because \( (a_n + 0)_n = (a_n)_n \) for any sequence \( (a_n)_n \).

Step 5: Determine Inverses

Each sequence \( (a_n)_n \) has an inverse \( (-a_n)_n \) such that \( (a_n + (-a_n))_n = (0)_n \). Thus, every sequence has an inverse.

Step 6: Define and Analyze Right Shift Function (Homomorphism)

The function \( r \) maps any sequence \( (a_0, a_1, \ldots) \) to \( (0, a_0, a_1, \ldots) \). It preserves addition: \( r((a_n)_n + (b_n)_n) = r((a_n + b_n)_n) = (0, a_0 + b_0, a_1 + b_1, \ldots) \), which equals \( r((a_n)_n) + r((b_n)_n) \). Thus, \( r \) is a homomorphism.

Step 7: Define and Analyze Left Shift Function (Homomorphism)

The function \( l \) maps any sequence \( (a_0, a_1, \ldots) \) to \( (a_1, a_2, \ldots) \). It preserves addition: \( l((a_n)_n + (b_n)_n) = l((a_n + b_n)_n) = (a_1 + b_1, a_2 + b_2, \ldots) \), which equals \( l((a_n)_n) + l((b_n)_n) \). Thus, \( l \) is a homomorphism.

Step 8: Show Injectivity of r

The function \( r \) is injective because if \( r((a_n)_n) = r((b_n)_n) \), then \( (0, a_0, a_1, \ldots) = (0, b_0, b_1, \ldots) \), implying \( (a_n)_n = (b_n)_n \).

Step 9: Show Non-Surjectivity of r

The function \( r \) is not surjective because no sequence of the form \( (a_0, a_1, \ldots) \) is mapped to a sequence beginning with a non-zero element.

Step 10: Show Surjectivity of l

The function \( l \) is surjective as for any sequence \( (b_0, b_1, \ldots) \), the sequence \( (a_0, a_1, \ldots) = (0, b_0, b_1, \ldots) \) maps to it under \( l \).

Step 11: Show Non-Injectivity of l

The function \( l \) is not injective because different sequences like \( (1, a_0, a_1, \ldots) \) and \( (0, a_0, a_1, \ldots) \) can map to the same sequence.

Key Concepts

Real Sequences

Real sequences are essentially ordered lists of real numbers. Think of them as an infinite list where each position corresponds to the next number in the sequence. Real sequences can take the form \(a_n\), where each element \(a_n\) is a real number and \(n\) represents the position in the sequence. This type of structure allows for operations like addition to be performed element by element, which is particularly useful when discussing algebraic structures.
Real sequences are foundational in mathematics for analyzing patterns and progressions over an infinite domain. They allow for systematic investigation into the behavior of series and functions and often appear in calculus and analysis.

Component-wise Addition

In the context of real sequences, component-wise addition means adding two sequences together by adding their corresponding elements. Given two sequences \( (a_n)_n \) and \( (b_n)_n \), their component-wise addition is the sequence \( (a_n + b_n)_n \). This operation is crucial because it adheres to the rules of combining sequences, ensuring operations can be performed consistently over the entire sequence domain.

• Closure: The sum of any two sequences is itself a sequence.
• Associativity: Grouping of operations does not affect the result.
• Identity Element: The sequence \( (0)_n \) acts as the neutral element.
• Inverses: Each sequence \( (a_n)_n \) has an inverse \( (-a_n)_n \).

Group Theory

Group theory provides a framework to study how different algebraic structures work with their own operations. An algebraic structure forms a group if it satisfies certain axioms: closure, associativity, identity, and invertibility.
In the context of real sequences and component-wise addition, they form a group because:

• Closure: Adding two sequences results in another sequence.
• Associativity: The order of addition doesn't affect the result.
• Identity: The zero sequence is the neutral element.
• Inverses: Every sequence has an additive inverse.

These properties ensure that real sequences with component-wise addition are robust under the group operation, meaning that operations can be performed predictably and consistently.

Homomorphisms

A homomorphism is a mapping between two algebraic structures that preserves the operations that define those structures. For real sequences, the shift operations defined are homomorphisms because they maintain the structure of addition for sequences.

• Right Shift (\(r\) Function): This function shifts all elements in a sequence to the right, inserting a zero at the beginning. It preserves the addition operation because the effect is uniform across sequences.
• Left Shift (\(l\) Function): This function removes the first element of a sequence, shifting all other elements left. It also preserves addition, keeping the sequence structure intact.

In terms of properties, the function \(r\) ensures injectivity but lacks surjectivity, meaning it assigns distinct elements uniquely but doesn’t cover the whole set. On the contrary, \(l\) is surjective, covering every possible output, but isn’t injective, so some sequences can map to the same result.