Question

Prove that for any mapping \(f\) and any sets \(A, B, A_{i}(i \in I),\) (a) \(f^{-1}[A \cup B]=f^{-1}[A] \cup f^{-1}[B] ;\) (b) \(f^{-1}[A \cap B]=f^{-1}[A] \cap f^{-1}[B]\); (c) \(f^{-1}[A-B]=f^{-1}[A]-f^{-1}[B]\) (d) \(f^{-1}\left[\bigcup_{i} A_{i}\right]=\bigcup_{i} f^{-1}\left[A_{i}\right]\) (e) \(f^{-1}\left[\bigcap_{i} A_{i}\right]=\bigcap_{i} f^{-1}\left[A_{i}\right]\) Compare with Problem 3 . [Hint: First verify that \(x \in f^{-1}[A]\) iff \(x \in D_{f}\) and \(f(x) \in A\).]

Short answer

The properties of inverse images for union, intersection, and differences are identical to the direct image properties.

Step-by-step solution

Understanding the Inverse Image

We need to understand that for any set \( A \), the inverse image \( f^{-1}[A] \) refers to the set of all elements \( x \) in the domain of \( f \) such that \( f(x) \in A \). Thus, \( x \in f^{-1}[A] \) if and only if \( f(x) \in A \).

Proving (a) for Union

We want to show \( f^{-1}[A \cup B] = f^{-1}[A] \cup f^{-1}[B] \). Choose any element \( x \in f^{-1}[A \cup B] \), then \( f(x) \in A \cup B \). This means \( f(x) \in A \) or \( f(x) \in B \), implying \( x \in f^{-1}[A] \) or \( x \in f^{-1}[B] \), thus \( x \in f^{-1}[A] \cup f^{-1}[B] \). Conversely, if \( x \in f^{-1}[A] \cup f^{-1}[B] \), then \( f(x) \in A \) or \( f(x) \in B \), thus \( f(x) \in A \cup B \), and hence \( x \in f^{-1}[A \cup B] \).

Proving (b) for Intersection

To show \( f^{-1}[A \cap B] = f^{-1}[A] \cap f^{-1}[B] \), consider \( x \in f^{-1}[A \cap B] \), which implies \( f(x) \in A \cap B \). Therefore, \( f(x) \in A \) and \( f(x) \in B \), so \( x \in f^{-1}[A] \) and \( x \in f^{-1}[B] \), leading to \( x \in f^{-1}[A] \cap f^{-1}[B] \). Conversely, if \( x \in f^{-1}[A] \cap f^{-1}[B] \), then \( f(x) \in A \) and \( f(x) \in B \), meaning \( f(x) \in A \cap B \) and \( x \in f^{-1}[A \cap B] \).

Proving (c) for Set Difference

We need to prove \( f^{-1}[A - B] = f^{-1}[A] - f^{-1}[B] \). Consider \( x \in f^{-1}[A - B] \), meaning \( f(x) \in A \) and \( f(x) otin B \), so \( x \in f^{-1}[A] \) and \( x otin f^{-1}[B] \), thus \( x \in f^{-1}[A] - f^{-1}[B] \). Conversely, if \( x \in f^{-1}[A] - f^{-1}[B] \), then \( f(x) \in A \) and \( f(x) otin B \), hence \( f(x) \in A - B \), leading to \( x \in f^{-1}[A - B] \).

Proving (d) for Union of Index Sets

To prove \( f^{-1}\left[\bigcup_{i} A_{i}\right] = \bigcup_{i} f^{-1}[A_{i}] \), let \( x \in f^{-1}\left[\bigcup_{i} A_{i}\right] \). Then, \( f(x) \in \bigcup_{i} A_{i} \), so there exists an \( i \) such that \( f(x) \in A_{i} \), implying \( x \in f^{-1}[A_{i}] \) for some \( i \), thus \( x \in \bigcup_{i} f^{-1}[A_{i}] \). Conversely, if \( x \in \bigcup_{i} f^{-1}[A_{i}] \), there exists an \( i \) with \( x \in f^{-1}[A_{i}] \), meaning \( f(x) \in A_{i} \), hence \( f(x) \in \bigcup_{i} A_{i} \), and \( x \in f^{-1}\left[\bigcup_{i} A_{i}\right] \).

Proving (e) for Intersection of Index Sets

For \( f^{-1}\left[\bigcap_{i} A_{i}\right] = \bigcap_{i} f^{-1}[A_{i}] \), let \( x \in f^{-1}\left[\bigcap_{i} A_{i}\right] \). This implies \( f(x) \in \bigcap_{i} A_{i} \), meaning \( f(x) \in A_{i} \) for all \( i \), so \( x \in f^{-1}[A_{i}] \) for each \( i \), thus \( x \in \bigcap_{i} f^{-1}[A_{i}] \). Conversely, if \( x \in \bigcap_{i} f^{-1}[A_{i}] \), then \( x \in f^{-1}[A_{i}] \) for all \( i \), giving \( f(x) \in A_{i} \) for all \( i \), hence \( f(x) \in \bigcap_{i} A_{i} \), and \( x \in f^{-1}\left[\bigcap_{i} A_{i}\right] \).

Key concepts

Properties of Mappings

Mappings in set theory, also known as functions, are relationships between two sets. A mapping from set \( X \) to set \( Y \) is a rule that assigns each element of \( X \) exactly one element of \( Y \). The set \( X \) is called the domain while \( Y \) is called the codomain.

The notion of inverse images involves mappings too. If \( f: X \to Y \) is a mapping, and \( A \) is a subset of \( Y \), the inverse image of \( A \) under \( f \), denoted \( f^{-1}[A] \), is the set of all elements in \( X \) that are mapped to elements in \( A \).

Here are some key properties of mappings with respect to inverse images:

• If \( x \in f^{-1}[A] \), then \( f(x) \in A \).
• The inverse image of the union of sets is the union of the inverse images, i.e., \( f^{-1}[A \cup B] = f^{-1}[A] \cup f^{-1}[B] \).
• The inverse image of the intersection of sets is the intersection of the inverse images, i.e., \( f^{-1}[A \cap B] = f^{-1}[A] \cap f^{-1}[B] \).
• The inverse image of a set difference is the difference of the inverse images, i.e., \( f^{-1}[A - B] = f^{-1}[A] - f^{-1}[B] \).

These properties highlight the consistency and predictability when dealing with mappings and inverse images in set operations.

Set Operations

Set operations such as union, intersection, and difference are fundamental concepts in set theory. Understanding these operations helps to grasp how they affect mappings and their inverse images.

- **Union** \( (A \cup B) \): Combines all elements from both sets \( A \) and \( B \). An element is in \( A \cup B \) if it is in \( A \) or \( B \) or both.

- **Intersection** \( (A \cap B) \): Contains only elements that are in both sets \( A \) and \( B \). An element is in \( A \cap B \) if it is in both \( A \) and \( B \).

- **Difference** \( (A - B) \): Contains elements that are in \( A \) but not in \( B \). An element is in \( A - B \) if it is in \( A \) and not in \( B \).

When dealing with mappings, these operations on sets translate directly to operations on their inverse images, preserving the properties of set interactions.

Proof Techniques

Proving properties involving mappings and set operations requires foundational proof techniques. Here are some strategies commonly employed in such proofs:

- **Element Chase**: This involves logically following an arbitrary element through the sets, verifying each necessary condition row. For instance, to prove that \( f^{-1}[A \cap B] = f^{-1}[A] \cap f^{-1}[B] \), we consider any element \( x \) and show it belongs to both sides of the equation.

- **Biconditional Proof**: Prove both directions, "if and only if". For many properties, you'll need to show two things: if \( x \) is in one set, it must be in another, and vice versa. This method is important for properties like inverse images of unions or intersections.

- **Set Builder Notation**: Sometimes using set builder notation (e.g., \( \{ x \mid \text{property}(x) \} \)) can clarify conditions and make the logic of proofs more transparent.

Mastering these techniques is crucial for rigorously proving properties related to mappings and set theory, ensuring a clear and reliable argument.