Can someone help me with this proof please.
Let n be a positive integer. If A1, A2, ... , An are pairwise disjoint finite sets then | A1 U A2 U ... U An| = |A1| + |A2| + ... + |An|.
Yeah sure. So we proved about the base case when you adjoin 1 disjoint set to A that is how you count them.
So suppose for induction hypothesis that for disjoint sets $\displaystyle \{A,A_1,A_2,...,A_n\}$ we have $\displaystyle |A\cup A_1 \cup A_2 \cup ... \cup A_{n}|=|A|+ |A_1| + |A_2| +... + |A_n|$, we now show it must be true for adjoining n+1 disjoint sets. So $\displaystyle \{A,A_1,A_2,...,A_n,A_{n+1}\}$ are all disjoint. By induction hypothesis $\displaystyle |A\cup A_1 \cup A_2 \cup ... \cup A_{n}|=|A|+ |A_1| + |A_2| +... + |A_n|$. Now we notice $\displaystyle (A\cup A_1 \cup A_2 \cup ... \cup A_{n})\cap A_{n+1}=\emptyset$ so by our base case in the first post
$\displaystyle |(A\cup A_1 \cup A_2 \cup ... \cup A_{n}) \cup A_{n+1}|=|A\cup A_1 \cup A_2 \cup ... \cup A_{n}|+|A_{n+1}|$$\displaystyle =(|A|+ |A_1| + |A_2| +... + |A_n|)+|A_{n+1}|=|A|+ |A_1| + |A_2| +... + |A_n|+|A_{n+1}|$
Which completes the induction.
Assume the Nth case is true: $\displaystyle \left| {\bigcup\limits_{k = 1}^N {E_k } } \right| = \sum\limits_{k = 1}^N {\left| {E_k } \right|} $.
Then let $\displaystyle E = \bigcup\limits_{k = 1}^N {E_k } \;\& \,E \cap E_{N + 1} = \emptyset $.
By the base case we know that $\displaystyle \left| {E \cup E_{N + 1} } \right| = \left| E \right| + \left| {E_{N + 1} } \right|$.
Now finish.