How to prove $A \subset B \iff A \cup B = B$

Taking the "longer" road. Let us review the definitions:

1. $A\subseteq B$ if and only if for every $x\in A$, $x\in B$.
2. $x\in A\cup B$ if and only if $x\in A$ or $x\in B$.
3. $A=B$ if and only if $A\subseteq B$ and $B\subseteq A$.
4. $P\iff Q$ means that if we assume that $P$ holds, then $Q$ must hold; and vice versa.

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Now assume $A\subseteq B$. This means that for every $x\in A$ we have $x\in B$. We want to show that $A\cup B=B$.

• So we take $x\in B$, then $x\in A$ or $x\in B$, and therefore $x\in A\cup B$.

• Now take $x\in A\cup B$, we want to show that $x\in B$. By definition either $x\in A$ or $x\in B$.

  • If $x\in B$ we are done.
  • If $x\in A$ then by the assumption that $A\subseteq B$ we have that $x\in B$.

  Either way we have that $x\in B$.

We have shown that if $A\subseteq B$ then $A\cup B\subseteq B$ and $B\subseteq A\cup B$, which is by fact number $3$ to say $A\cup B=B$.

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Now we need to assume that $A\cup B=B$, and deduce that $A\subseteq B$. So we need to show that if $x\in A$ then $x\in B$.

Take $x\in A$ to be an arbitrary element. Because $x\in A$ we have that $x\in A$ or $x\in B$, and therefore $x\in A\cup B$. The assumption was, however, that $A\cup B=B$ and therefore we have that $x\in B$ as wanted.

Let $A\subset B$. Since $B\subset B$, we have $A\cup B\subset B$. Clearly, $B\subset A\cup B$. Hence $A\cup B=B$.

Let $A\not\subset B$. Then there is some $x\in A$ with $x\not\in B$. Clearly, $x\in A\cup B$. Hence $A\cup B\neq B$.

Here are the (very straightforward) first steps you should have thought of beginning with:

In one direction, suppose $A\subseteq B$: then $A\cup B\subseteq B\cup B\dots$

In the other direction, suppose $A\cup B=B$: then $A\subseteq A\cup B\subseteq\dots$