Question

Consider a function \(f: A \rightarrow B\) and a subset \(X \subseteq A\). We observed in Example 12.14 that \(f^{-1}(f(X)) \neq X\) in general. However \(X \subseteq f^{-1}(f(X))\) is always true. Prove this.

Short answer

The statement is proven by showing that every element of \(X\) belongs to \(f^{-1}(f(X))\), which means that \(X\) is a subset of \(f^{-1}(f(X))\).

Step-by-step solution

Understanding the inverse image of a set

The inverse image \(f^{-1}(Y)\) of a set \(Y \subseteq B\) under a function \(f: A \rightarrow B\) is defined as the set of all elements in \(A\) that are mapped to any element of \(Y\) by \(f\). In other words, \(f^{-1}(Y) = \{x \in A : f(x) \in Y\}\). So \(f^{-1}(f(X))\) is the set of all elements in \(A\) that are mapped to any element of \(f(X)\) by \(f\).

Proving the superset

To prove that \(X \subseteq f^{-1}(f(X))\), take any element \(x \in X\). By definition, \(f(x)\) belongs to \(f(X)\). This means that \(x\) belongs to the set of all elements in \(A\) that get mapped to \(f(X)\) under \(f\), i.e. \(x \in f^{-1}(f(X))\). Therefore, every element of \(X\) belongs to \(f^{-1}(f(X))\), which proves that \(X \subseteq f^{-1}(f(X))\).