Problem 13 - Advanced Calculus

Question

Let \(f: X \rightarrow Y\) be a function, and \(S\) and \(T\) arbitrary sets. Show that: (a) \(f f^{-1}(S) \subset S\) (b) \(T \subset f^{-1} f(T)\)

Step-by-Step Solution

Answer

Based on the definitions of inverse function and preimage, it has been shown that if \(f: X \rightarrow Y\) is a function, and \(S\) and \(T\) are arbitrary sets, then \(f f^{-1}(S) \subset S\) and \(T \subset f^{-1} f(T)\). The results are derived from the basic set operations and properties of inverse functions.

Step 1: Definition of inverse function

The first step is to recall the definition of an inverse function. An inverse function \(f^{-1}(S)\) is a function which will reverse the result done by the function \(f\). Therefore, when we execute \(f f^{-1}(S)\), if for every element \(x\) in \(X\) such that \(f(x)\) is in \(S\), then the result will be the element \(x\). This implies that, if \(x\) is not in \(S\), then \(f f^{-1}(S)\) will not include \(x\). Thereby, confirming that \(f f^{-1}(S)\) is a subset of \(S\).

Step 2: Demonstrating that \(f f^{-1}(S) \subset S\)

Given a set \(S\), let's take an arbitrary element \(y = f(x)\) from \(f f^{-1}(S)\). According to the definition of inverse function this implies that there existed an \(x\) in \(X\), such that \(f(x) = y\). Therefore, \(y \in S\) and \(x \in S\) which proves that every element of \(f f^{-1}(S)\) belongs to \(S\), i.e. \(f f^{-1}(S) \subset S\).

Step 3: Preimage

The preimage of a set, denoted as \(f^{-1}(T)\), is a set of elements in the function's domain \(X\) that are mapped into \(T\). Therefore, giving \(T = f(x)\) which means there is an \(x\) in \(X\) that \(f(x)\) that belongs to \(T\). Hence, \(T \subset f^{-1} f(T)\).

Step 4: Demonstrating that \(T \subset f^{-1} f(T)\)

Given a set \(T\), let's take an arbitrary element \(t\) from \(T\). The function of \(t\) would be \(f(t)\), and the preimage of \(f(t)\), denoted \(f^{-1}f(t)\), would include the element \(t\), confirming that every element of \(T\) belongs to \(f^{-1} f(T)\), i.e., \(T \subset f^{-1} f(T)\).

Key Concepts

Set Theory

Set theory is a fundamental concept that deals with collections of objects or elements. These collections are called sets. In mathematics, we often work with sets which can be thought of as well-defined collections of objects.

Understanding sets is crucial for exploring concepts like functions and their inverses. We denote a set with curly braces, such as \({1, 2, 3}\). Some important points to remember about sets include:

• Each element in a set is unique.
• Order of elements within a set is not important.
• Sets can be finite or infinite.

In the context of inverse functions, our focus is on how elements from these sets are related through mappings. Set theory provides the foundational language and structure necessary for describing these mappings.

Function Mapping

A function mapping is the core idea used to describe how each element from one set is associated with an element of another set. When we talk about a function \(f\) from set \(X\) to set \(Y\), denoted as \(f: X \rightarrow Y\), we're essentially defining a rule that assigns each element from the domain \(X\) to exactly one element in the codomain \(Y\).

Here's a breakdown:

• The set \(X\) is called the domain.
• The set \(Y\) is the codomain.
• Function \(f\) maps elements from \(X\) to elements of \(Y\).

Understanding mappings makes it easier to comprehend how inverse functions work because an inverse function essentially 'reverses' this mapping direction. Functions are essential tools for problem-solving and exploration in mathematics.

Preimage

The preimage is a key concept when dealing with functions and their inverses. When a function \(f\) maps an element from set \(X\) to set \(Y\), the preimage of a set \(T\) in \(Y\) is a set of all elements in \(X\) that \(f\) maps into \(T\).

To better understand preimages:

• If \(f(x) = y\), then \(x\) is the preimage of \(y\).
• The notation \(f^{-1}(S)\) signifies the preimage of \(S\).
• Preimages help us determine what inputs (from the domain) lead to particular outputs (in the codomain).

In problems involving inverses and subsets, understanding preimages helps reveal how elements originally in the set \(T\) maintain a relationship within the preimage set \(f^{-1}(T)\), linking back to the concept of inverse functions.