Prove $|A_1\cup\dotsb\cup A_n|\leq|A_1|+|A_2|+\dotsb+|A_n|$ using Induction I need help proving Prove  $|A_1\cup\dotsb\cup A_n|\leq|A_1|+|A_2|+\dotsb+|A_n|$ (probably using induction. I have already proven that  $|A_1\cup A_2|\leq|A_1|+|A_2|$
by 
$|A_1\cup A_2|= (|A_1|+|A_2|)-|A_1\cap A_2|$ & Def of union of sets
    $-|A_1\cap A_2|=|A_1\cup A_2|-(|A_1|+|A_2|)$  & Addition is well-defined
    $|A_1\cap A_2|\geq 0$ & Def of sets (cant have neg num of elements)
    $-|A_1\cap A_2|\leq 0$ & Multiplication is well defined
    $|A_1\cup A_2|-(|A_1|+|A_2|)=-|A_1\cap A_2|\leq 0$ & Def of $\leq$
    $|A_1\cup A_2|-(|A_1|+|A_2|)\leq 0$ & Def of $\leq$
    Thus $|A_1\cup A_2|\leq|A_1|+|A_2|$ is true & Addition is well-defined 
Please explain each step because I really want to understand what/why those steps are used.
Thanks

We have the result: 
$$\left\vert A_1 \cup A_2 \right\vert \leq \left\vert A_1 \right\vert + \left\vert A_2 \right\vert.$$
Now suppose that
$$\left\vert \bigcup_{i=1}^n A_i \right\vert \leq \sum_{i=1}^n \left\vert A_i \right\vert,$$
where $n$ is a natural number. 
Then we have 
$$
\begin{align}
\left\vert \bigcup_{i=1}^{n+1} A_i \right\vert &= \left\vert \left( \bigcup_{i=1}^n A_i \right) \cup A_{n+1} \right\vert \ \mbox{ using the associative property of the union of sets } \\
&\leq \left\vert  \bigcup_{i=1}^n A_i \right\vert + \left\vert A_{n+1} \right\vert \\
& \ \ \ \mbox{ using the result for union of two sets, where the first set is a union of $n$ sets} \\  
&\leq \sum_{i=1}^n \left\vert A_i \right\vert +  \left\vert A_{n+1} \right\vert \ \mbox{ using our induction hypothesis } \\
&= \sum_{i=1}^{n+1} \left\vert A_i \right\vert .
\end{align}
$$
