Let $A,B,C$ be sets, and $B \cap C=\emptyset$. Show $|A^{B \cup C}|=|A^B \times A^C|$
Let $A,B,C$ be sets, and $B \cap C=\emptyset$. Show $|A^{B \cup C}|=|A^B \times A^C|$ by defining a bijection $f:A^{B \cup C} \rightarrow A^B \times A^C$.
Any hints on this one?
Thank you!
Solution 1:
Hint: If $f$ is a function is a function from $B\cup C$ to $A$, let $f_B$ ($f$ restricted to $B$) be the function from $B$ to $A$ defined by $f_B(b)=f(b)$. Define $f_C$ analogously.
Now show that the mapping $\varphi$ which takes any $f$ in $A^{B\cup C}$ to the ordered pair $(f_B,f_C)$ is a bijection from $A^{B\cup C}$ to $A^B\times A^C$.