Proof that $A \subseteq B \Leftrightarrow A \cup B = B$ I want to proof that:
$A \subseteq B  \Leftrightarrow  A \cup B = B$
At the moment I have no idea on how to start. Please give me a hint.
Thx in advance!
 A: A good place to start is always with definitions. Recall the definition of $A\subseteq{B}$:
$$x\in{A}\to{x}\in{B}$$
Now recll the definition of union:
$$A\cup{B}=\{x:x\in{A}\lor{x}\in{B}\}$$
And finally, rewrite the predicate:
$$A\cup{B}=B\leftrightarrow\{x:x\in{A}\lor{x}\in{B}\}=\{x:x\in{B}\}$$
A: Since you ask about how to start, here is how you can do that:
(1) Assume first that $A\subseteq B$. Then try to prove that $A\cup B = B$. That you can do by first proving that if $x\in A\cup B$ then $x\in B$. And after that, you prove that if $x\in B$ then $x\in A\cup B$ (this is obvious).
(2) Now assume that $A\cup B = B$. Then try to prove that $A\subseteq B$. This you can do by proving that if $x\in A$ then $x\in B$.
A: First, $A \cup B = B$ means $A \cup B \subseteq B$ and $A \cup B \supseteq B$. However, the latter is trivial, and so we will prove something stronger:
$$A \subseteq B \Longrightarrow A \cup B \subseteq B \tag{1}$$
$$A \subseteq B \Longleftarrow A \cup B \subseteq B \tag{2}$$
where $(1)$ follows from the monotonicity of $\cup$,
\begin{align}
A &\subseteq B \\
B &\subseteq B \\
A \cup B &\subseteq B \cup B
\end{align}
and $(2)$ is implied by transitivity of $\subseteq$
$$A \subseteq A \cup B \subseteq B.$$
I hope this helps $\ddot\smile$
A: As another answer says: start with the definitions.  The reason, in this case, is that that will bring you from the set level to the element and logic level, where you can use the (hopefully) familiar laws of logic.  Also, start at the most complex side and try to simplify.
So we calculate
\begin{align}
& A \cup B = B \\
\equiv & \;\;\;\;\;\text{"definition of $\;=\;$ on sets, i.e., set extensionality"} \\
& \langle \forall x :: x \in A \cup B \;\equiv\; x \in B \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\cup\;$"} \\
& \langle \forall x :: x \in A \lor x \in B \;\equiv\; x \in B \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify: $\;P \lor Q \equiv Q\;$ is one of several ways to write $\;P \Rightarrow Q\;$"} \\
& \langle \forall x :: x \in A \;\Rightarrow\; x \in B \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\subseteq\;$"} \\
& A \subseteq B \\
\end{align}
This completes the proof.
