Prove that $A \cup B = A$ if and only if $B$ is a subset of $A$ If $A \cup B = A$ then $A$ is a subset of $A$ and $B$ is a subset of $A$. Thus $A \cup B = A$.
If $B$ is a subset of $A$ then it follows that $A \cup B$ is a subset of $A$.
My solution. It seems very straightforward and simple. But is it valid?
 A: Its a good start! The key to a good proof is a sequence of if..then, therefore...statements that lead from the premise to the conclusion. And remember to use the definitions.
Also keep in mind that we have to prove both of these statements:


*

*if $A \cup B = A$ then $B \subseteq A$ 

*if $B \subseteq A$ then $A \cup B = A$ 


For the first one, try something like:
If $A \cup B = A$ then by the definition of union, $ A = \{x : x \in A\text{ or } x \in B\}$. So for each $x \in B$, $x \in A$. In other words, $B \subseteq A$.
Can come up with something similar for the second one?
A: First assume $A \cup B=A$
That means $x \in A \cup B \iff x \in A$ so:
$$x\in B \implies  x \in A \cup B \implies x \in A$$
The later statement is equivalent to $A \subset B$
Second assume $B \subset A$
That means $x \in B \implies x \in A$.
First what is obvious:
$$x \in A \implies x \in A \cup B$$
The later statement means $A \subset A \cup B$
Second for proof $A \cup B \subset A $
First if $x \in A \cup B$ then $x \in A$ or $x \in B$:
$x \in A \implies x \in A$ 
$x \in B \implies x \in A$ (by hipothesis)
So:
$x \in A \cup B \implies x \in A \vee x \in B \implies x\in A$
A: If $A \cup B = A$, then there is no element $e \in B$ such that $e \notin A$. Thus $B$ contains only elements that are also in $A$ (if at all: $B$ may be empty), which means that $B \subseteq A$.
If $B \subseteq A$; $A \cup B $ contains all elements of $A$ and $B$. But since $B \subseteq A$, there is no element $e \in B$ such that $e \notin A$, which means that $A \cup B = A$ since $B$ does not add any new elements to this set.
