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

1. $A \cap B \subseteq A$ (i.e., every element of $A \cap B$ is also an element of $A$).
2. $A \subseteq A \cap B$ (i.e., every element of $A$ is also an element of $A \cap B$).

Proof:

1. Show that $A \cap B \subseteq A$:

By definition, the intersection of two sets, $A \cap B$, contains elements that are in both $A$ and $B$. So, if $x \in A \cap B$, then $x \in A$ and $x \in B$.

Thus, any element in $A \cap B$ is clearly an element of $A$, which means $A \cap B \subseteq A$.

2. Show that $A \subseteq A \cap B$:

We are given that $A \subseteq B$, which means that if $x \in A$, then $x \in B$. Therefore, if $x \in A$, it must also be in $B$, meaning $x \in A \cap B$.

Thus, any element in $A$ is also in $A \cap B$, which means $A \subseteq A \cap B$.

Conclusion:

Since we've shown both $A \cap B \subseteq A$ and $A \subseteq A \cap B$, it follows that $A \cap B = A$.

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Do you want further details or have any questions? Here are some related problems to explore:

1. Prove that if $A \subseteq B$, then $A \cup B = B$.
2. Prove that $A \cap A = A$.
3. Prove that $A \cap B \subseteq B$.
4. Prove that $A \subseteq B$ if and only if $A \cup B = B$.
5. Prove that $A \cap \emptyset = \emptyset$.

Tip: When proving set equalities, always break down the proof into subset relations to systematically establish both directions of the equality.