Given $A\cup B = A$, prove $B$ is the empty set Given the above statement, i've to prove that $B = \emptyset$
I started by using the theorem for set equality that says.

Given two sets $A$ and $B$ we have that:
$A = B \iff A \subset B \quad\land\quad B \subset A$

therefore i just plug things in and i got that
$A \cup B \implies x \in A \quad\lor\quad x \in B$
and
$A \cup B = A \iff A \subset A \cup B \quad\land\quad A \cup B \subset A$
After that i got a stuck, its my first time doing this kind of proofs.
Thanks in advance
 A: What you have given :

Prove :
$ (A\cup B = A) \implies B = \emptyset $

We can not Prove it , that Statement is not true.
Eg Consider :
$A=\{1,2,3\}$
$B=\{1\}$ or $B=\{2,3\}$ or $B=\{1,2,3\}$
In these Cases , $ (A\cup B = A)$ is true but $B = \emptyset $ is not true.
What you may have been given or what we can Prove :

Prove :
$ (A\cup B = A) \implies (B-A) = \emptyset $

This is true , & we can Prove it , & Intuitively check it.
When you merge 2 Sets , A & B , you get back Set A ,
... that means the Union with Set B did not add new elements to Set A ,
... which means Set B did not have elements which are not in A ,
... which means all elements of Set B are in Set A ,
... which means Set (B-A) , where all the common elements are removed , must have nothing left over ,
... which means Set (B-A) is the EMPTY SET.
Observation :
In Natural Numbers , $A+B=A \implies B=0$ , which is not Directly true with Sets.
A: Can't prove what isn't true.   Counter example.  $A= \mathbb R$ and $B = \mathbb Q$.  Then $A \cup B = \mathbb R \cup \mathbb Q = \mathbb R = A$.  But $B=\mathbb Q$ is certainly not empty.
So it's not true.
But what is true is $A\cup B =A \iff B \subset A$.
Pf: If $A\cup B = A$ then for any $b\in B$ we have $b \in A\cup B = A$ so $b \in A$.  So $B\subset A$.
So $A\cup B = A\implies B\subset A$.
If $B\subset A$ then.... If $x\in A \cup B$ then either $x \in A$ or $x \in B\subset A$.  So either way $x \in A$.  So $A\cup B \subset A$.  And if $a\in A$ then $a \in A\cup B$ so $A \subset A\cup B$.  So $A\cup B = A$.
So $B\subset A \implies A\cup B = A$
A: You simply forgot to write "Suppose A is disjoint from B". Then your statement is true and may be readily proven (Do you see how, assuming $A \cap B = \{\}$, that is, $A$ and $B$ are disjoint?)?
