Question

Given a function \(f: A \rightarrow B\) and subsets \(W, X \subseteq A,\) prove \(f(W \cap X) \subseteq f(W) \cap f(X)\).

Short answer

For any function \(f: A \rightarrow B\) and subsets \(W, X \subseteq A\), \(f(W \cap X) \subseteq f(W) \cap f(X)\) can be proved by showing that any element in the image of the intersection of \(W\) and \(X\) under \(f\) is also an element in the intersection of the image of \(W\) and the image of \(X\) under \(f\).

Step-by-step solution

State what needs to be proved

To prove that \(f(W \cap X) \subseteq f(W) \cap f(X)\), it is necessary to show that for any element \(y\) in \(f(W \cap X)\), that element will also be in \(f(W) \cap f(X)\). In other words, if \(y\) is in the image of the intersection of \(W\) and \(X\) under \(f\), then \(y\) will also be in the intersection of the images of \(W\) and \(X\) under \(f\).

Assume an element in f(W \cap X)

Assume \(y \in f(W \cap X)\). This means that there is some element \(x\) in \(W \cap X\) such that \(f(x) = y\). Because \(x\) is in \(W \cap X\), it follows that \(x\) is in \(W\) and \(x\) is in \(X\). Therefore, \(y \in f(W)\) and \(y \in f(X)\), as the same element \(x\) in \(W\) and \(X\) that resulted in \(y\) in \(f(W \cap X)\) would also result in \(y\) in \(f(W)\) and \(f(X)\).

Conclude the proof

The intersection of two sets is the common elements between these sets. Since \(y\) is in both \(f(W)\) and \(f(X)\), it must also be in their intersection. Hence, \(y \in f(W) \cap f(X)\). So, for the given \(y\) in \(f(W \cap X)\), we have showed that \(y\) is also in \(f(W) \cap f(X)\). Therefore, \(f(W \cap X) \subseteq f(W) \cap f(X)\).