Question

Let \(A\) and \(B\) be sets. Prove that \(A \subseteq B\) if and only if \(A \cap B=A\).

Short answer

It has been proven that \(A \subseteq B\) if and only if \(A \cap B = A\), by proving two directions: if \(A \subseteq B\), then \(A \cap B = A\); and if \(A \cap B = A\), then \(A \subseteq B\).

Step-by-step solution

Prove \(A \subseteq B\) implies \(A \cap B = A\)

Assume \(A \subseteq B\). Then by definition, for any element \(x\), if \(x\) is in \(A\), then \(x\) is also in \(B\). Now consider the intersection of \(A\) and \(B\), i.e., \(A \cap B\). By definition, \(x\) is in \(A \cap B\) if and only if \(x\) is in both \(A\) and \(B\). But since every element of \(A\) is also in \(B\), every element in \(A\) is therefore in \(A \cap B\). In other words, \(A \subseteq A \cap B\). Conversely, by definition of intersection, \(A \cap B \subseteq A\). Putting these together, we have \(A = A \cap B\). Hence if \(A \subseteq B\), then \(A = A \cap B\).

Prove \(A \cap B = A\) implies \(A \subseteq B\)

Now assume \(A \cap B = A\). By definition of intersection, each element \(x\) in \(A \cap B\) is also in \(B\). Hence if \(x\) is in \(A\), which is equal to \(A \cap B\), then \(x\) is in \(B\). By definition of subset, this means that \(A \subseteq B\). Hence if \(A = A \cap B\), then \(A \subseteq B\).

Summary

So we have proved the two directions individually: if \(A \subseteq B\), then \(A \cap B = A\); and if \(A \cap B = A\), then \(A \subseteq B\). Therefore, \(A \subseteq B\) if and only if \(A \cap B = A\). This completes the proof.