Problem 45 - Discrete Mathematics with Applications

Question

Let \(f: X \rightarrow Y\) and \(A, B \subseteq X^{\dagger} .\) Prove each. $$f(\mathrm{A} \cup \mathrm{B})=f(\mathrm{A}) \cup f(\mathrm{B})$$

Step-by-Step Solution

Answer

We proved that \(f(A \cup B) \subseteq f(A) \cup f(B)\) by showing that for any element y in f(A ∪ B), y must also belong to either f(A) or f(B). Then, we proved that \(f(A) \cup f(B) \subseteq f(A \cup B)\) by showing that any y in \(f(A) \cup f(B)\) must also belong to f(A ∪ B). Thus, we concluded that \(f(A \cup B) = f(A) \cup f(B)\).

Step 1: Prove \(f(A \cup B) \subseteq f(A) \cup f(B)\)

To begin, let's take an arbitrary element y from the set f(A ∪ B). By definition, this means there exists some element x in A ∪ B, such that \(f(x) = y\). Since x is in A ∪ B, then x must also belong to either A or B (or both). If x is in A, then \(y = f(x) \in f(A)\). If x is in B, then \(y = f(x) \in f(B)\). So, in either case, \(y \in f(A) \cup f(B)\). Therefore, \(f(A \cup B) \subseteq f(A) \cup f(B)\).

Step 2: Prove \(f(A) \cup f(B) \subseteq f(A \cup B)\)

Now, let's consider an arbitrary element y from the set \(f(A) \cup f(B)\). By definition, this means that y must belong to either f(A) or f(B) (or both). If \(y \in f(A)\), then there exists some element x in A, such that \(f(x) = y\). Since x is in A, it is also in A ∪ B, and therefore, \(y = f(x) \in f(A \cup B)\). If \(y \in f(B)\), then there exists some element x in B, such that \(f(x) = y\). Since x is in B, it is also in A ∪ B, and therefore, \(y = f(x) \in f(A \cup B)\). So, in either case, \(y \in f(A \cup B)\). Therefore, \(f(A) \cup f(B) \subseteq f(A \cup B)\). Since we have proved both subset relationships, we can conclude that: $$ f(A \cup B) = f(A) \cup f(B) $$

Key Concepts

Image of a Set

In set theory, the concept of the image of a set helps us understand the values that a function can take. When we have a function \( f: X \rightarrow Y \) and a subset \( A \) of \( X \), the image of \( A \) under \( f \), denoted as \( f(A) \), consists of all elements in \( Y \) that \( f \) maps some element of \( A \) to. In simple terms, imagine \( f \) as a machine that transforms inputs from a subset in its domain to outputs in its codomain.

• For any element \( y \) in \( f(A) \), there exists at least one element \( x \) in \( A \) such that \( f(x) = y \).
• If \( A \) is empty, then \( f(A) \) is also empty, since there’s nothing to map.

Understanding the image of a set is crucial in comprehending how functions transform or "map" entire collections of values from their domain into the codomain.

Union of Sets

The union of sets is a fundamental concept in set theory and deals with combining elements from multiple sets. If \( A \) and \( B \) are two sets, the union of these sets, written as \( A \cup B \), is a new set containing all distinct elements from \( A \) and \( B \). This includes all elements that are in either \( A \) or \( B \), or both.

• The union is inclusive, meaning it takes every unique element from the involved sets.
• If one set is a subset of another, the union will result in the larger set.

Understanding unions is important when working with simultaneous sets as it becomes the basis for other concepts, such as the distributive property when mapping with functions.

Function Mapping

Function mapping is how we define relationships between elements of two sets. When we say a function \( f: X \rightarrow Y \), we mean that every element \( x \) in set \( X \) is associated with exactly one element in set \( Y \). This relationship provides a structured way of associating pairs, forming a domain (input) to codomain (possible outputs) map.

• Functions can be visualized as arrows connecting inputs to outputs.
• Each input must have a unique output, but different inputs can share the same output.

In the context of set operations, function mapping helps us observe how set actions like unions affect the associated outputs.

Subset Relations

The concept of subsets is central to set theory. A set \( A \) is a subset of a set \( B \), denoted \( A \subseteq B \), if every element of \( A \) is also an element of \( B \). This means all values in \( A \) are "contained" within \( B \).

• A subset can be equal to its parent set, or it can contain fewer elements.
• The empty set is a subset of every set, providing a foundational starting point in set operations.

In proofs, such as the one provided, understanding subset relations is key to establishing equality between complex set mappings and operations.