Problem 57 - Discrete Mathematics with Applications

Question

Let \(f: X \rightarrow Y\) be bijective. Let \(S\) and \(T\) be subsets of \(Y .\) Prove each. $$f^{-1}(S \cup T)=f^{-1}(S) \cup f^{-1}(T)$$

Step-by-Step Solution

Answer

To prove \(f^{-1}(S \cup T) = f^{-1}(S) \cup f^{-1}(T)\), we show that every element in \(f^{-1}(S \cup T)\) belongs to \(f^{-1}(S) \cup f^{-1}(T)\) and vice versa. For any \(x \in f^{-1}(S \cup T)\), either \(f(x) \in S\) or \(f(x) \in T\), implying \(x \in f^{-1}(S) \cup f^{-1}(T)\). Similarly, for any \(x \in f^{-1}(S) \cup f^{-1}(T)\), \(f(x) \in S \cup T\), implying \(x \in f^{-1}(S \cup T)\). Thus, \(f^{-1}(S \cup T) = f^{-1}(S) \cup f^{-1}(T)\).

Step 1: Show every element in \(f^{-1}(S \cup T)\) belongs to \(f^{-1}(S) \cup f^{-1}(T)\)

Let \(x \in f^{-1}(S \cup T)\). This means that \(f(x) \in S \cup T\). Now, \(f(x)\) must either belong to S or to T. - If \(f(x) \in S\), then \(x \in f^{-1}(S)\). Hence, \(x \in f^{-1}(S) \cup f^{-1}(T)\). - If \(f(x) \in T\), then \(x \in f^{-1}(T)\). Hence, \(x \in f^{-1}(S) \cup f^{-1}(T)\). Therefore, in both cases, \(x \in f^{-1}(S) \cup f^{-1}(T)\).

Step 2: Show every element in \(f^{-1}(S) \cup f^{-1}(T)\) belongs to \(f^{-1}(S \cup T)\)

Let \(x \in f^{-1}(S) \cup f^{-1}(T)\). We can have two cases: - If \(x \in f^{-1}(S)\), then \(f(x) \in S\) and \(f(x) \in S \cup T\). Thus, \(x \in f^{-1}(S \cup T)\). - If \(x \in f^{-1}(T)\), then \(f(x) \in T\) and \(f(x) \in S \cup T\). Thus, \(x \in f^{-1}(S \cup T)\). In both cases, \(x \in f^{-1}(S \cup T)\).

Step 3: Conclude that both sets are equal

Since we have shown that every element in \(f^{-1}(S \cup T)\) belongs to \(f^{-1}(S) \cup f^{-1}(T)\), and every element in \(f^{-1}(S) \cup f^{-1}(T)\) belongs to \(f^{-1}(S \cup T)\), we can conclude that these sets are equal: $$f^{-1}(S \cup T) = f^{-1}(S) \cup f^{-1}(T)$$

Key Concepts

Inverse Functions

An inverse function is a function that reverses the effect of the original function. If you have a function \( f : X \rightarrow Y \), finding the inverse means you want to find a function \( f^{-1} : Y \rightarrow X \) such that when you apply \( f \) and then \( f^{-1} \), you end up back where you started. Essentially, for each element \( y \) in the set \( Y \), \( f^{-1} \) gives you the corresponding element \( x \) in the set \( X \), so that \( f(x) = y \).

• The inverse function only exists if the original function is bijective (i.e., both injective and surjective).
• A function being bijective means every element in the target set \( Y \) is mapped by exactly one element from the domain set \( X \). This ensures each output has one unique input.
• Using inverse functions allows us to "undo" the function's effect, which is crucial in solving equations and finding pre-images of elements in terms of set theory.

Understanding inverse functions helps comprehend how sets relate through these types of mappings, making problems involving set operations more intuitive.

Set Theory

Set theory is a branch of mathematical logic that studies sets, which are collections of objects. Sets can be anything: numbers, points in space, or even functions. When dealing with functions, especially bijective ones, set theory provides a useful framework for understanding how elements from different sets relate to each other. Here are some fundamental aspects:

• The "domain" of a function is the set of all possible inputs, while the "codomain" is the set of all possible outputs.
• Subsets are smaller collections formed from larger collections and can be defined within the context of sets \( X \) and \( Y \).
• In logical expressions and proofs, we often deal with intersections, unions, and complements of sets.
• In the context of functions, the pre-image of a set \( S \) under a function \( f \) is the set of all original elements from the domain that map into \( S \).

Set theory provides the language and structures that underpin the behavior of functions and their inverses, which are essential for solving complex mathematical problems involving various set operations.

Union of Sets

The union of sets is an operation that combines all elements from two or more sets into a single set. If you have sets \( S \) and \( T \), the union \( S \cup T \) consists of all elements that are in \( S \), in \( T \), or in both.

• The union operation is commutative: \( A \cup B = B \cup A \).
• It is also associative: \((A \cup B) \cup C = A \cup (B \cup C)\).
• In the case of inverse functions, knowing how to handle unions helps show equalities involving pre-images, like proving \( f^{-1}(S \cup T) = f^{-1}(S) \cup f^{-1}(T) \).
• This property is intuitive: if an element maps into either \( S \) or \( T \), it will map into their union.

Understanding unions of sets is crucial in working with set expressions in mathematics and for solving problems that involve combining or comparing multiple groups of elements. It allows you to manipulate how different elements share the same attributes across various contexts.