I have a question regarding the following proof(Thm 1.8 in Rudin's Real and Complex Analysis). I can see why R is the cartesian product of two open segments I1 and I2, but I don't understand how this translates into the next line. Why is it an intersection? Can someone explain?
For any map it holds that $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$. We can use this for $R$ since it can be written as an intersection of two sets: $R = I_1 \times I_2 = (I_1\times \Bbb{R}) \cap (\Bbb{R}\times I_2)$. Hence \begin{align*} f^{-1}(R) &= f^{-1}((I_1\times \Bbb{R}) \cap (\Bbb{R}\times I_2)) \\ &= f^{-1}(I_1\times \Bbb{R}) \cap f^{-1}(\Bbb{R}\times I_2)\\ &= u^{-1}(I_1) \cap v^{-1}(I_2). \end{align*}