Proposition 3.6.1.4: Let $X$ and $Y$ be finite sets. Then $X \cup Y$ is finite and $\#(X \cup Y ) \le \#(X) + \#(Y)$
Lemma 1: Let $n\in \mathbb{N}$. For any $n$ finite sets $A_1,...,A_n$ with $\#A_i < 2$ for any $i \in \{1,...,n\}$, $\#(\bigcup_{i\in\{1,...,n\}} A_i)\le n$,
We use induction on $n$. If $n=0$, then the statement is vacuously true since the union of nothing is undefined. Suppose inductively, we have shown that for some $n \in \mathbb{N}$ that for any $n$ finite sets $A_1,...,A_n$ with $\#A_i < 2$ for any $i \in \{1,...,n\}$, $\#(\bigcup_{i\in\{1,...,n\}} A_i)\le n$, We will now show that for for any $n+1$ sets $A_1,...,A_{n+1}$ with $\#A_i < 2$ for any $i \in \{1,...,n+1\}$, $\#(\bigcup_{i\in\{1,...,n+1\}} A_i)\le n$. Let there be $n+1$ sets $A_1,...,A_{n+1}$ with $\#A_i < 2$ for any $i \in \{1,...,n+1\}$, then by our inductive hypothesis $\#(\bigcup_{i\in\{1,...,n\}} A_i)\le n$. Then, by proposition 3.6.14, $\#((\bigcup_{i\in\{1,...,n\}} A_i) \cup A_{n+1})\le \#(\bigcup_{i\in\{1,...,n\}} A_i)+\#(A_{n+1})<n+2$. Equivalently, this also means that $\#((\bigcup_{i\in\{1,...,n\}} A_i) \cup A_{n+1})\le n+1$ as desired. This closes the induction.
Exercise 3.6.10: Let $A_1,...,A_n$ be finite sets such that $\#(\bigcup_{i\in\{1,...,n\}} A_i)>n$, then there exists $i \in \{1,..., n\}$ such that $\#(Ai) \ge 2$
We use contraposition. If there does not exist $i \in \{1,...,n\}$ st $\#(A_i)\ge 2$, then this will mean that for all $i \in \{1,...,n\}$, $\#(A_i) < 2$ . Then by lemma 1, this will mean that $\#(\bigcup_{i\in\{1,...,n\}} A_i)\le n$. $\square$
Could anyone help me check if i am using contraposition correctly? Also, is there any cleaner way to organise or write the proof for Lemma 1, been feeling that my proof-writing is needlessly long and messy but im not sure where to clean-up stuff.