To prove that $A \subseteq B$ if and only if $A \cap B = A$, we need to show two things:

1. If $A \subseteq B$, then $A \cap B = A$:

Assume $A \subseteq B$. This means that every element of $A$ is also an element of $B$.

• Let $x \in A \cap B$. By definition of intersection, this means $x \in A$ and $x \in B$. Since $A \subseteq B$, we already know that $x \in B$ whenever $x \in A$. Thus, $A \cap B \subseteq A$.

• Now, let $x \in A$. Since $A \subseteq B$, $x \in B$ as well. Therefore, $x \in A \cap B$, which implies that $A \subseteq A \cap B$.

Hence, $A = A \cap B$.

2. If $A \cap B = A$, then $A \subseteq B$:

Assume $A \cap B = A$. We want to show that every element of $A$ is also an element of $B$.

• Let $x \in A$. From the assumption $A \cap B = A$, we know that $x \in A \cap B$, which means $x \in B$.

Therefore, every element of $A$ is also in $B$, so $A \subseteq B$.

Conclusion:

We have shown both directions:

• If $A \subseteq B$, then $A \cap B = A$.
• If $A \cap B = A$, then $A \subseteq B$.

Thus, $A \subseteq B$ if and only if $A \cap B = A$.

Would you like more details, or do you have any questions?

Here are 5 related questions:

1. How does the proof change if we deal with proper subsets?
2. Can you extend this proof to multiple sets?
3. How would the proof look using Venn diagrams?
4. What is the difference between this proof and the union of sets?
5. How does this concept apply to functions or mappings?

Tip: Always try to visualize set operations like intersection and union using Venn diagrams for clarity.