Problem:
Show there exists a set of sets $ \{ A_x : x \in \Bbb R \} $ s.t. :
$\bullet $ $ \forall x \neq y. A_x \cap A_y = \emptyset $
$ \bullet $ $ \bigcup_{x \in \Bbb R }A_x = \Bbb R $
$ \bullet $ $ \forall x \in \Bbb R. |A_x| = \Bbb \aleph $
Answer : Since we know $ |\Bbb R \times \Bbb R| = | \Bbb R | $ then there exists a bijection $ f: \Bbb R \times \Bbb R \to \Bbb R $. Given $ x \in \Bbb R $ we'll define the set $ A_x = f[ \{ x \} \times \Bbb R] = \{ f( \langle x,y \rangle) : y \in \Bbb R \} $ . [ Then the answer goes ahead to prove all three properties are satisfied]
Question: The set $ A_x $ given above is correct, but I wanted to know if you have an idea for another set, I tried really hard to sit on this problem but had no idea so I resorted to the answer above. I still think I couldn't have come up with an example straight ahead that satisfies all the three conditions ( nor come up with the example above ). Also, how would you approach the problem above?