Suppose $A$ and $B$ are sets. Prove that $A \subseteq B$ if and only if $A \cap B = A$.
Here's how I see it being proved.
If $A$ and $B$ are sets,and the intersection of $A$ and $B$ is equal to $A$, then the elements in $A$ are in both the set $A$ and $B$. Therefore, the set of $A$ is a subset of $B$ since all the elements are contained in the interesection of sets $A$ and $B$ are equal to $A$.
Can I prove it that way?
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$\begingroup$Your proof is almost perfect and let me rectify it a bit:
Let $A$ and $B$ be two sets. The intersection of $A$ and $B$ is equal to $A$, is equivalent to the elements in $A$ are in both the set $A$ and $B$ which's also equivalent to the set of $A$ is a subset of $B$ since all the elements of $A$ are contained in the intersection of sets $A$ and $B$ are equal to $A$.
$\endgroup$ $\begingroup$Your proof only really covers the $\Leftarrow$ direction of statement, as stated. I would suggest just using simple Boolean algebra to prove it:
$$(A \subseteq B) \iff ((A \cap B) = A)$$
$$(\forall x) (x \in A \Rightarrow x \in B) \iff (\forall y) (y \in A \land y \in B) \iff y \in A$$ $$(\forall x) (x \in A \Rightarrow x \in B) \iff (\forall y) (y \in A \Rightarrow y \in B)$$ $$\top$$
$\endgroup$ $\begingroup$Suppose A is subset of B. Let X belongs to A then by hypothesis, X will belong to B. Hence X belong to A and X belong to B implies that X belongs to A intersection B. Accordingly A is subset of A intersection B. But we know that A intersection B is always subset of A. Hence A intersection B is equal to A. On the other hand, suppose A intersection B is equal to A. Then in particular, A is subset of A intersection B. We know that, A intersection B is subset of B. Hence A is subset of B.
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