# Question #00484

Jan 27, 2016

Let me offer you a suggestion of the proof.

As you would like to prove a "$\iff$" relation, you can prove it with proving first "$\Rightarrow$" and then "$\Leftarrow$".

1) Prove "$\Rightarrow$":

Let $A \cup B = A \cap B$.

We need to prove that $A = B$. To do that, we need to prove that for every $x \in A$, it follows $x \in B$ (i.e. $A \subseteq B$), and for every $x \in B$, $x \in A$ (i.e. $B \subseteq A$).

Let me also remind you of the formal definitions of $\cup$ and $\cap$:

$A \cup B = \left\{x | x \in A \vee x \in B\right\}$
$A \cap B = \left\{x | x \in A \wedge x \in B\right\}$

Let $x \in A$.

As $A \subseteq A \cup B$, so all elements from $A$ are included in $A \cup B$, we can safely assume that $x \in A \cup B$.

However, we know that $A \cup B = A \cap B$, so $\implies x \in A \cap B$.

But since $x \in A \cap B$ and all elements of $A \cap B$ are included in $B$, $x \in B$ must immediately hold.

Thus, $x \in A \implies x \in B$. $\implies A \subseteq B$

Let $x \in B$.

The reasoning is exactly the same, really. Let me formulate this with mathematical terms exclusively though.

$x \in B \stackrel{B \subseteq A \cup B}{\Rightarrow} x \in A \cup B$

$\stackrel{A \cup B = A \cap B}{\Rightarrow} x \in A \cap B$

$\stackrel{A \cap B \subseteq B}{\Rightarrow} x \in A$

Thus, $x \in B \implies x \in A$, thus $B \subseteq A$.

So we have proven that

$\left(A \cup B = A \cap B\right) \Rightarrow \left(A = B\right)$

2) Prove "$\Leftarrow$":

Let's prove the other direction then.

Let $A = B$. We need to prove that $A \cup B = A \cap B$ needs to be true now.

However since $A = B$, we see that

$A \cup B = A \cup A = A = A \cap B = A \cap B$.

In case of doubt, we can also check the formal definitions:

$x \in A \cup B \stackrel{A = B}{\iff} x \in A \cup A \iff x \in A \iff x \in A \cap A \stackrel{A = B}{\iff} x \in A \cap B$.

Thus, $A \cup B = A \cap B$.

Here, we have proven that

$\left(A \cup B = A \cap B\right) \Leftarrow \left(A = B\right)$

3)

Since both

$\left(A \cup B = A \cap B\right) \Rightarrow \left(A = B\right)$

and

$\left(A \cup B = A \cap B\right) \Leftarrow \left(A = B\right)$

are true, we know that

$\left(A \cup B = A \cap B\right) \iff \left(A = B\right)$

is true as well.

q.e.d.