Question

Let \(A, B,\) and \(C\) be sets, and let \(F: A \rightarrow C\) and \(G: B \rightarrow C\) be functions. (a) What condition must \(F\) and \(G\) satisfy for \(F \cup G\) to be a function from \(A \cup B\) to \(C ?\) (b) Give conditions on \(A\) and \(B\) such that \(F \cup G\) is a function for every \(F: A \rightarrow C\) and \(G: B \rightarrow C\)

Short answer

(a) \( F(x) = G(x) \) for all \( x \in A \cap B \). (b) \( A \) and \( B \) must be disjoint: \( A \cap B = \emptyset \).

Step-by-step solution

Understanding Function Union

To determine when \( F \cup G \) is a function, remember that a function must have exactly one output for every input in its domain. \( F \cup G \) takes the union of the graphs of \( F \) and \( G \). Therefore, each element in \( A \cup B \) should map to a unique element in \( C \).

Condition for Common Inputs

If \( x \) is an element in both \( A \) and \( B \), then both \( F(x) \) and \( G(x) \) must exist. For \( F \cup G \) to be a function, \( F(x) \) must equal \( G(x) \) for all \( x \in A \cap B \). This ensures that \( F \cup G \) assigns a unique output for each input.

Condition on Sets A and B

To ensure that \( F \cup G \) is a function irrespective of \( F: A \rightarrow C \) and \( G: B \rightarrow C \), \( A \) and \( B \) should be disjoint sets. That is, \( A \cap B = \emptyset \). With this condition, there are no common inputs between \( A \) and \( B \), eliminating the possibility of \( F(x) eq G(x) \).

Key concepts

Set Theory

In set theory, we study collections of objects, referred to as sets. These objects can be anything, such as numbers, symbols, or even other sets. Set theory is fundamental to understanding modern mathematics, as it lays the groundwork for topics like functions.

Sets are usually denoted by uppercase letters such as \( A \), \( B \), and \( C \). One important operation in set theory is the union, denoted by \( A \cup B \), which combines all elements from sets \( A \) and \( B \) into a single set. In other words, it includes every element from both sets, without repeating elements.

The concept of set union is vital when considering functions like \( F: A \rightarrow C \) and \( G: B \rightarrow C \), as we explore how these functions can be combined or overlapped. It allows us to understand how inputs can be mapped to outputs across larger domains.

Function Properties

Functions are special relationships between two sets, where each input from the first set (domain) is linked to exactly one output in the second set (range). This unique pairing is a crucial property of functions.

For example, if we have a function \( F: A \rightarrow C \), it assigns a specific output from \( C \) for each input from \( A \). Similarly, a function \( G: B \rightarrow C \) maps inputs from \( B \) to outputs in \( C \). The crucial point here is that there is always a unique corresponding output for every input.

When discussing the union of two functions \( F \) and \( G \), for \( F \cup G \) to be a function, this unique pairing must remain intact. This means that any input that is common to both functions should map to the same output in the union. Otherwise, the union would not satisfy the property that defines it as a function.

Domain and Range

The domain and range are essential components of any function. The domain refers to all possible input values, while the range is the set of possible outputs.

For a function \( F: A \rightarrow C \), \( A \) is the domain, and \( C \) is the range. Similarly, \( G: B \rightarrow C \) has \( B \) as its domain, with the same range \( C \). When considering the union of functions \( F \cup G \), understanding the domains' overlap helps decide whether this union results in a valid function.

To ensure \( F \cup G \) is a true function, if there's overlap in the domains (\( A \cap B ot= \emptyset \)), the functions must produce the same output for every common input. If \( F(x) \) and \( G(x) \) differ, then the union cannot define a unique output, violating the function property.

Disjoint Sets

Two sets are called disjoint if they have no elements in common. In the context of functions \( F: A \rightarrow C \) and \( G: B \rightarrow C \), this means that for each element \( x \) in \(\ A \), there should be no corresponding element \( y \) in \(\ B \) that is the same.

When \( A \) and \( B \) are disjoint \(A \cap B = \emptyset\), it simplifies the conditions under which \( F \cup G \) becomes a function. Because disjoint sets do not share any inputs, there is no risk of mapping the same input to different outputs in \( C \).

In other words, if \( A \) and \( B \) have no common elements, then \( F \cup G \) automatically respects the requirement of functions having one unique output for each input, making the union a well-defined function from \( A \cup B \) to \( C \). This is why having disjoint sets is a sufficient condition for ensuring that \( F \cup G \) remains a function.