Suppose that $S, T$ are sets such that $S$ is infinite and $T = \{t_i: i = 1, ..., n\}$ with $T$ having $n \geq 1$ distinct elements. If $S \cap T = \{\}$, show that $card(S) = card(S \cup T)$.
Hint: Construct a bijective function $\phi: S \rightarrow S \cup T$.
I tried defining various functions but cannot seem to construct one that satisfies the requirements above. I would appreciate if someone could provide such a bijective function.
On
$S$ is of infinite cardinal, i.e. $$ |S| \geq \aleph_0 $$ We define $\varphi : S \to S \sqcup T$ as it follows:
for $1\leq i \leq n$, there exits $s_i^0\in S$ s.t. $\varphi(s_i^0) = t_i$
recursively we define, for $1\leq i \leq n$, $s_i^{k+1} \in S\setminus \cup_{l=0}^k \{s_1^l, \cdots, s_n^l\}$, s.t. $\varphi(s_i^{k+1}) = s_i^{k}$
We put $S' = \cup_{k\geq 0}\{s_1^k,\cdots, s_n^k\}$, and for $s\notin S'$, we define $\varphi(s) = s$
If $S$ is infinite then there exists $C\subset S$ such that $C$ is countable infinite.
Therefore $C=\{c_1,c_2,...\}$. Now, consider $\phi(z)=z$ for $z\notin C$, $\phi(c_k)=t_k$ for $k\in\{1,2,...,n\}$ and finally $\phi(c_{n+k})=c_{k}$ for $k\geq 1$.
This function is surjective because $\phi(S\setminus C)=S\setminus C$, $\phi(\{c_1,c_2,...,c_n\})=\{t_1,t_2,...,t_n\}$ and finally $\phi(\{c_{n+1},c_{n+2},...\})=\{c_1,c_2,...\}=C$.
Let $a,b\in S$. If both are in $C$ then we have $\phi(a)=\phi(b)\Rightarrow a=b$. If $a\in C$ and $b\notin C$ then their image can not be equal because $\phi(a)\in C\cup T$ and $\phi(b)\in S\setminus C$ and $(C\cup T) \cap (S\setminus C)=\emptyset$.
We are left to prove that if $a,b\in C$ then $\phi(a)=\phi(b)\Rightarrow a=b$. Let $a=c_r$ and $b=c_s$ and consider $r<s$ without loss of generality.
If $s\leq n$ then $\phi(a)=\phi(b)\iff\phi(c_r)=\phi(c_s) \iff t_r=t_s$and so they are indeed different.
If $r\geq n+1$ then $\phi(a)=\phi(b)\iff\phi(c_r)=\phi(c_s) \iff c_{r-n}=c_{s-n}$ and so they are indeed different.
If $r\leq n< s$ then $\phi(a)=t_r$ and $\phi(b)=c_{s-n}$ which are different.
I hope this can help you.