Inclusion-Exclusion proof for two sets

Question

I know this might sound silly but, it's easy to convince myself that $|A\cup B|=|A|+|B|-|A\cap B|$ but i'm not sure how i would go about proving it.

Suppose $A'=A\setminus\{a\}$ and $|A|=n+1$ for some positive integer $n$ then $\begin{align} |A\cup B|&=|A'\cup B|+|\{a\}|\\ & =|A'|+|B|-|A'\cap B|+ |\{a\}|\\ & =|A|+|B|-|A\cap B| \end{align}$

So is this correct??

Answer 1

If A and B are disjoint then it's straight forward that $|A\cup B|=|A|+|B|$. Now, $|A\cup B|=|A\cup(A\cap B)\cup(A^{c}\cap B)|=|A\cup (A^{c}\cap B)|=\\ =|A|+|A^{c}\cap B|=|A|+|B\setminus (A\cap B)|=|A|+|B|-|A\cap B|$

The third equality is due to the fact that $A$ and $A^{c}\cap B$ are disjoint sets.

Answer 2

I'm not convinced that induction here is the right method..

Hint: Rather prove that $|A\cap B|+|A\cup B|\ =\ |A|+|B|$ by counting the members on both sides of this equation. (Some members will be counted twice, but on both sides, which ones?)

Answer 3

Your calculation would be correct if and only if $a\in A$ but $a\not\in B$.