# Set theory: Prove if $A \cup B = A \cap B$ then $A = B$

The question provides an answer, saying:

1. $A \subset A \cup B$, then, $A \cap B \subset B$, then $A \subset B$
2. $B \subset A \cup B$, then, $A \cap B \subset A$, then $B \subset A$

(1) and (2) $\Leftrightarrow A = B$

I know that when two sets are equal, they they're the subsets of each other. What I don't understand is where the OP has brought this answer from.

• Remember that two sets are equal if they have exactly the same elements. If one of them, say $A$, has an element that $B$ doesn't, then what does that say about $A \cup B$ compared to $A \cap B$? – user1390 Jun 12 '18 at 15:35
• I don't understand what you mean by OP here? Are you attempting to reference another question on this site? – Henning Makholm Jun 12 '18 at 15:36
• Do you have trouble understanding the argument for $A\subseteq B$ (and, with those two swapped, $B\subseteq A$)? Or do you have trouble understanding thy (1) and (2) imply that $A=B$? – Henning Makholm Jun 12 '18 at 15:37
• The obvious part is $A,B \subset A \cup B$ and $A \cap B \subset A,B$. – Mauro ALLEGRANZA Jun 12 '18 at 15:41
• But if $A \cup B = A \cap B$, then 1) becomes : $A \subset A \cap B \subset B$ and the subset relation is transitive. The same for 2). – Mauro ALLEGRANZA Jun 12 '18 at 15:41

Statement:

$A\cup B = A\cap B\rightarrow A=B$

Proof:

Assume $A\cup B = A\cap B$

This means $a\in A \lor a\in B \leftrightarrow a\in A\land a\in B$

Now assume $a\in A$. Thus $a\in A\lor a\in B$. As such $a\in A\land a\in B$. Therefore $a\in B$. As such$A\subseteq B$. (This line 1. of the quote).

And similar can be shown for $a\in B$ (line 2). Therefore $A=B$.

The wording of the phrases are weird. "then" doesn't really fit in the middle but if I edit them:

1) $A\subset A\cup B$, (and $A\cup B = A \cap B$) $\require{cancel} \bcancel{\text{then}}$ and $A\cap B\subset B$, $\bcancel{\text{then}}$ therefore $A\subset B$.

2) $B\subset A\cup B$, (and $A\cup B = A \cap B$) $\bcancel{\text{then}}$ and $A\cap B\subset A$, $\bcancel{\text{then}}$ therefore $B\subset A$.

(The results of) 1) and 2) $\iff A=B$.

Does it become more clear?

B $\subseteq$ A $\cup$ B = A $\cap$ B $\subseteq$ A.
Likewise, A $\subseteq$ B.

Contraposition: If A is not B, A U B is a proper superset of B is a proper superset of A intersection B. Thus the union is not the intersection.