Let $A$ and $B$ be finite sets.Show that $A\bigcup B$ and $A\bigcap B$ are also finite sets,and that $\#A+\#B=\#(A\bigcup B)+\#(A\bigcap B)$.($\#A$ represents the cardinality of the set $A$)
Proof:First I'd like to do a related problem.
Let $A$ and $B$ be finite sets,then $A\bigcup B$ is finite and $\#(A\bigcup B)\leq \#A+\#B$.Further more,if $A\bigcap B=\varnothing$,then $\#(A\bigcup B)=\#A+\#B$.
Proof:Prove it by induction.When $\#B=0$,$B=\varnothing$
($\#B=0\Leftrightarrow B=\varnothing$.$\Leftarrow$ Very easy($\emptyset=\{1\leq i\leq 0|i\in\mathbf{N}\}$).
$\Rightarrow$ When $\#B=0$,if $B\neq \varnothing$,which means $B$ has at least one element.We know that $\#\varnothing=0$.This means that there is a bijection between $B$ and $\varnothing$,it is impossible.)
$A\bigcup B=A$.So $A\bigcup B$ is a finite set,$A\bigcap B=\varnothing$,and $\#A+\#B=\#A+0=\#A=\#(A\bigcup B)$.Now suppose that the proposition is already proven for some $n$,which means that when $\#B=n$,$\#(A\bigcup B)\leq \#A+\#B$,and if $A\bigcap B=\varnothing$,then $\#(A\bigcup B)=\#A+\#B$.We now prove it for $n+1$.In case of $n+1$,let $x\in B$,$\#(A\bigcup B)=\#(A\bigcup (B\backslash \{x\})\bigcup \{x\})$.If $x\in A$,then $\#(A\bigcup(B\backslash \{x\})\bigcup \{x\})=\#(A\bigcup (B\backslash \{x\}))\leq \#A+(\#B-1)< \#A+\#B$.If $x\not\in A$,and $(B\backslash \{x\})\bigcap A=\varnothing$,then $A\bigcap B=\varnothing$,and $\#(A\bigcup B)=\#(A\bigcup (B\backslash \{x\})\bigcup \{x\})=\#A+\#(B\backslash \{x\})+\#\{x\}=\#A+\#B.$(There is a lemma which can be easily verified:Let B be a finite set.$x\not\in B$,then $\#(B\bigcup\{x\})=\#B+1$.)According to induction,for any finite set B,the proposition holds. $\Box$
Now lets turn to the target problem:
From the above part,we can easily verify that $A\bigcup B$ is a finite set.And it is also easy for us to verify that $A\bigcap B$ is finite.(A subset of a finite set is a finite set.Suppose there is a finite set $A$ whose cardinality is $n$,which means that there is a bijection $f$ between $A$ and $\{i\in \mathbb{N}|1\leq i\leq n\}$.Suppose there is a subset of $A$,denoted by $X$.There is a bijection between $X$ and $f(X)$.)
Now we use induction to prove that $\#A+\#B=\#(A\bigcup B)+\#(A\bigcap B)$.When $\#(A\bigcap B)=0$(It means that $A\bigcap B=\varnothing$),we have proved that $\#A+\#B=\#(A\bigcup B).$ Suppose we have proved that when $\#(A\bigcap B)=n$,the proposition holds.In case when $\#(A\bigcap B)=n+1$,let $x\in A\bigcap B$,we know that $\#(A\backslash \{x\})+\#(B\backslash \{x\})=\#[(A\bigcup B)\backslash \{x\}]+\#[(A\bigcap B)\backslash \{x\}]$.(The cardinality of $(A\bigcap B)\backslash \{x\}$ is $n$,so we can make use of the assumption.)
Now add $x$ back to $A\backslash \{x\}$ and $B\backslash \{x\}$:
$\#A=\#(A\backslash \{x\})+1$,$\#B=\#(B\backslash \{x\})+1$,$\#(A\bigcup B)=\#[(A\bigcup B)\backslash \{x\}]+1$,$\#(A\bigcap B)=\#[(A\bigcap B)\backslash \{x\}]+1$.We still get $\#A+\#B=\#(A\bigcup B)+\#(A\bigcap B)$.
From the induction,we know the proposition holds .$\Box$