A Set Equality

Hello everyone. In this video, we're going to do another if and only if proof. This time we're going to use sets.

Question: Prove that $A\subseteq B$ if and only if $A\cap B=A$ .

Proof

So again with an if and only if proof, we're going to prove that the hypothesis implies the conclusion, and that the conclusion implies the hypothesis. So the forward direction and the reverse direction.

Forward Direction

For the forward direction, assume that $A\subseteq B$ .

Now we're trying to prove a set equality and remember to prove a set equality, you need to show that every element in the left-hand set is in the right-hand set, and every element in the right-hand set is indeed a member of the left-hand set.

So to show $A\cap B\subseteq A$ , we note that $A\cap B$ is the set of all elements contained in both $A$ and $B$ , and so all elements are contained in $A$ .

That's strictly by the definition. If you're in both $A$ and $B$ , then you're most certainly inside $A$ . So that's true, $A\cap B\subseteq A$ .

To show $A\subseteq A\cap B$ , we're going to do this by showing an arbitrary element of $A$ must be inside $A\cap B$ .

So I'm going to say let $x\in A$ . You could have done this for the other part as well, but the other part really does just follow by writing out the sentence of what the set is.

I'm going to take an arbitrary element in $A$ , and I'm going to show that it lives in this right-hand set, $A\cap B$ . So how do I do that? I can't do that on its own because on its own that's not true, but with the hypothesis I can get there. So the first statement we proved was true no matter what the hypothesis was, but this statement we need to use hypothesis.

Then since $A\subseteq B$ , we have that $x\in B$ . As $x\in A$ and $x\in B$ , we have that $x\in A\cap B$ , by definition.

Remember $A\cap B$ is the set of all elements that are in $A$ and $B$ . Well $x\in A$ and $x\in B$ , so we're done. So since $A\cap B\subseteq A$ and $A\subseteq A\cap B$ , we have that $A=A\cap B$ .

Alright, so the forward direction we're okay.

Reverse Direction

For the reverse direction, we're going to assume that $A\cap B=A$ . Alright, and so what are we trying to show? We're trying to show therefore that $A\subseteq B$ .

Let's look at this set right here $A\cap B$ . Notice that before we showed that $A\cap B\subseteq A$ , right, and we use nothing about the hypothesis. So in fact, this same argument is going to hold to show that $A\cap B\subseteq B$ .

So remember as mathematicians, we try to be as lazy as possible. Sort of a fun little joke, but we've already really done the same amount of work above. So really all I have to do is just copy the work except I change the $A$ to a $B$ .

So I'm going to say as before, we have that $A\cap B\subseteq B$ since $A\cap B$ is all elements contained in both $A$ and $B$ .

Great, so that's true. As $A=A\cap B$ , for any $x\in A$ , $x\in A\cap B$ and since $A\cap B\subseteq B$ , we have that $x\in B$ .

So we've shown that any arbitrary element of $A$ must live inside $B$ , that's precisely the statement that $A\subseteq B$ . So therefore, $A\subseteq B.\quad\blacksquare$

Always start sentences with words don't start sentences with symbols, it's bad form. That's it, that's a full if and only if proof using set notation. So hopefully this helps and all the best.