Proving $A \subset B \implies A \cup B = B$

Sorry, but it's incorrect.

You want to show that “$A\subset B$” if and only if $A\cup B=B$”.

In order to do this, you have to prove

1. If $A\subset B$, then $A\cup B=B$
2. If $A\cup B=B$, then $A\subset B$

Proof of 1

Suppose $A\subset B$.

• Take $x\in A\cup B$; then $x\in A$ or $x\in B$; if $x\in A$, then, by assumption, $x\in B$. In either case, $x\in B$. Hence $A\cup B\subset B$
• Take $x\in B$; then $x\in A\cup B$. Hence $B\subset A\cup B$.

The two points above show that $A\cup B=B$.

Proof of 2

Suppose $A\cup B=B$. Take $x\in A$; then $x\in A\cup B$, so $x\in B$. Therefore $A\subset B$.

What you should have done for the proofs:

Proof of 1 Suppose $A\subset B.$

Take $x\in A\cup B.$ Then we have two possibilities: $x\in A$ or $x\in B.$ If $x\in A,$ $x\in B$ by assumption, and the case when $x\in B$ immediately holds. So $A\cup B\subset B.$ If $x\in B,$ then $x\in A\cup B,$ and so $B \subset A\cup B.$ Thus $A\cup B=B.$

Proof of 2 Suppose $A\cup B=B.$ Let $x\in A$. Then $x\in A\cup B.$ By hypothesis, $x\in B.$ Thus $A\subset B.$

($\implies$) Suppose that $ A\subseteq B $.

Take any $ x\in A \cup B $, so $ x\in A $ or $ x\in B $. Since $ A\subseteq B $ and $ x\in A $, so $ x\in B $. Then, we have $ A\cup B \subseteq B $.

Take any $ x\in B $. Since $ A\subseteq B $ and $x\in B$, so $x\in A$ or $x\in B$. Then, we have $B\subseteq A\cup B$.

Since $ A\cup B \subseteq B $ and $B\subseteq A\cup B$, so $A\cup B=B$.

($\impliedby$) Suppose that $A\cup B= B$.

Take any $ x\in A\cup B $, so $x\in A$ or $x\in B$, in particular, $x\in A$. Since $A\cup B=B$ and $x\in A\cup B$, then $x\in B$.

Since for any $x\in A \implies x\in B$, then we have $A\subseteq B$