Question

Let \(A\) and \(B\) be sets with \(A_{1}, A_{2} \subseteq A,\) and let \(F: A \rightarrow B\). Let \(F\left(A_{i}\right)\) denote \(\mid F(x)\) : \(\left.x \in A_{i}\right\\}\) for \(i=1,2\). Show that: (a) If \(A_{1} \subseteq A_{2},\) then \(F\left(A_{1}\right) \subseteq F\left(A_{2}\right)\). (b) \(F\left(A_{1} \cup A_{2}\right)=F\left(A_{1}\right) \cup F\left(A_{2}\right)\). (c) \(F\left(A_{1} \cap A_{2}\right) \subseteq F\left(A_{1}\right) \cap F\left(A_{2}\right)\). (d) \(F\left(A_{1}\right)-F\left(A_{2}\right) \subseteq F\left(A_{1}-A_{2}\right)\). (e) \(A_{1} \subseteq F^{-1}\left(F\left(A_{1}\right)\right)\). (f) Find an example in which \(A_{1} \subset A_{2}\) but \(F\left(A_{1}\right)=F\left(A_{2}\right)\). (g) Find an example in which \(A_{1} \neq F^{-1}\left(F\left(A_{1}\right)\right)\).

Short answer

Steps demonstrate various relationships between images and subsets under a function, with examples illustrating when equalities and inequalities hold or do not hold.

Step-by-step solution

Show that If A1 ⊆ A2, then F(A1) ⊆ F(A2)

Assume that \(x \in F(A_{1})\). By definition, there exists an element \(a \in A_{1}\) such that \(F(a) = x\). Since \(A_{1} \subseteq A_{2}\), \(a \in A_{2}\), and therefore \(x \in F(A_{2})\). Hence, \(F(A_{1}) \subseteq F(A_{2})\).

Prove F(A1 ∪ A2) = F(A1) ∪ F(A2)

Consider any element \(x \in F(A_{1} \cup A_{2})\). It means there is some \(a \in A_{1} \cup A_{2}\) such that \(F(a) = x\). Therefore, \(a \in A_{1}\) or \(a \in A_{2}\). So \(x \in F(A_{1})\) or \(x \in F(A_{2})\), implying \(x \in F(A_{1}) \cup F(A_{2})\). Conversely, any element in \(F(A_{1}) \cup F(A_{2})\) is clearly in \(F(A_{1} \cup A_{2})\). Thus, \(F(A_{1} \cup A_{2}) = F(A_{1}) \cup F(A_{2})\).

Prove F(A1 ∩ A2) ⊆ F(A1) ∩ F(A2)

Suppose \(x \in F(A_{1} \cap A_{2})\). There exists \(a \in A_{1} \cap A_{2}\) such that \(F(a) = x\). Since \(a \in A_{1}\) and \(a \in A_{2}\), \(x \in F(A_{1})\) and \(x \in F(A_{2})\). Therefore, \(x \in F(A_{1}) \cap F(A_{2})\). Hence, \(F(A_{1} \cap A_{2}) \subseteq F(A_{1}) \cap F(A_{2})\).

Show that F(A1) − F(A2) ⊆ F(A1 − A2)

Assume \(x \in F(A_{1}) - F(A_{2})\). Then, there exists \(a \in A_{1}\) such that \(F(a) = x\) and \(x otin F(A_{2})\). This implies \(a otin A_{2}\), hence \(a \in A_{1} - A_{2}\). Thus, \(x \in F(A_{1} - A_{2})\). Therefore, \(F(A_{1}) - F(A_{2}) \subseteq F(A_{1} - A_{2})\).

Show that A1 ⊆ F^{-1}(F(A1))

Consider \(a \in A_{1}\). We have that \(F(a) \in F(A_{1})\). Thus, \(a \in F^{-1}(F(A_{1}))\) because \(F^{-1}(F(A_{1}))\) contains all preimages of elements of \(F(A_{1})\). Hence, \(A_{1} \subseteq F^{-1}(F(A_{1}))\).

Example where A1 ⊂ A2 but F(A1) = F(A2)

Let \(A = \{1, 2\}\) and \(B = \{a\}\), with \(F(a) = a\) for any \(a \in A\). If \(A_{1} = \{1\}\) and \(A_{2} = \{1, 2\}\), then \(F(A_{1}) = \{a\}\) and \(F(A_{2}) = \{a\}\). Here, \(A_{1} \subset A_{2}\) but \(F(A_{1}) = F(A_{2})\).

Example where A1 ≠ F^{-1}(F(A1))

Let \(A = \{1, 2\}\), \(B = \{a, b\}\), \(F(1) = a\), and \(F(2) = a\). Set \(A_{1} = \{1\}\). Then \(F(A_{1}) = \{a\}\) and \(F^{-1}(F(A_{1})) = \{1, 2\}\). Therefore, \(A_{1} eq F^{-1}(F(A_{1}))\).

Key concepts

Set Theory

In the realm of discrete mathematics, **set theory** is a fundamental branch that studies how to collect and analyze different sets of objects. A set is essentially a well-defined collection of distinct objects, which can be numbers, letters, or even other sets. Sets are typically denoted using uppercase letters, such as \(A\), \(B\), or \(C\). If you wish to show that an item belongs to a set, you use the "element of" symbol \(\in\). If an object is not in a set, you use the "not an element of" symbol \(otin\).

Subsets are a vital concept, where a set \(A_1\) is a subset of another set \(A_2\) if every element of \(A_1\) is also an element of \(A_2\). This is expressed as \(A_1 \subseteq A_2\). When working with sets, operations like unions, intersections, and set differences allow you to combine or compare them in various ways. The **union** \(A_1 \cup A_2\) contains all elements from both sets, while the **intersection** \(A_1 \cap A_2\) comprises only the elements common to both sets. A **set difference** \(A_1 - A_2\) contains the elements in \(A_1\) that are not in \(A_2\). Recognizing and applying these operations is crucial in solving problems related to functions and their images or preimages.

Function Image

When studying functions in discrete mathematics, the **image of a function** represents the set of all output values that occur when elements from one set, usually referred to as the domain, are mapped to another set, the codomain, through the function. In simpler terms, if you consider a function \(F: A \rightarrow B\), the image of a subset \(A_i \subseteq A\) through \(F\) is the collection of all elements \(y \in B\) that can be written as \(y = F(x)\) for some \(x \in A_i\).

To illustrate, suppose \(A_1\) is a subset of \(A\). The function image \(F(A_1)\) contains every element \(F(x)\) where \(x\) belongs to \(A_1\). This concept is essential when exploring how different subsets of the domain relate to each other under transformation by function \(F\). Understanding function images assists in proving subset relationships and decomposing union, intersection, and differences of function outputs, as demonstrated in various proofs of set operations concerning function mapping.

Preimage of a Function

The **preimage of a function** pertains to the set of all potential inputs in the domain that are mapped to specific outputs in the codomain. For a function \(F: A \rightarrow B\) and a subset \(C\) of \(B\), the preimage \(F^{-1}(C)\) is defined as the set of all \(x \in A\) such that \(F(x) \in C\). This concept allows us to trace back or "reverse" the action of the function from its output to its input, making it a crucial tool in set-based function analysis.

Using preimages helps verify the existence of certain elements in the domain that correspond to given images. For instance, showing that a subset of the domain \(A_1\) is contained within the preimage \(F^{-1}(F(A_1))\) highlights that even beyond specific choices in the domain, every element maps back into the function's target set. This can lead to interesting findings, such as discovering scenarios where \(A_1\) does not equal \(F^{-1}(F(A_1))\), demonstrating the intricate nature of how functions operate over different domains and codomains.

Subset Relationships

**Subset relationships** are crucial in understanding the properties and behavior of functions within set structures. When working with functions, examining how subsets of the domain are transformed into the codomain can reveal important insights.

For example, if you have subsets \(A_1\) and \(A_2\) such that \(A_1 \subseteq A_2\), it can be shown that the image under a function \(F\) also maintains a subset relationship: \(F(A_1) \subseteq F(A_2)\). This means that every element that \(A_1\) maps to through \(F\) is also mapped from \(A_2\).

Additionally, examining the results of operations like unions and intersections with respect to function images yields other interesting subset relationships. For instance, while the image of a union \(F(A_1 \cup A_2)\) equals the union of images \(F(A_1) \cup F(A_2)\), the intersection tends to produce a more complex relationship, typically \(F(A_1 \cap A_2) \subseteq F(A_1) \cap F(A_2)\). These relationships provide deep insights into how functions manipulate data across different sets, underpinning much of the theoretical framework of discrete mathematics.