ABSTRACT
Solution to Exercise 1.1. We argue in the same notation as in the main text: f − 1 ( ∩ i ∈ I A i ) = { x ∈ X | f ( x ∈ ∩ i ∈ I A i ) } = { x ∈ X | for all i ∈ I : f ( x ) ∈ A i } = { x ∈ X | for all i ∈ I : f ( x ) ∈ A i } = ∩ i ∈ I f − 1 ( A i ) , f − 1 ( ∩ i ∈ I A i ) = { x ∈ X | f ( x ∈ ∩ i ∈ I A i ) } = { x ∈ X | exists i ∈ I : f ( x ) ∈ A i } = { x ∈ X | exists i ∈ I : x ∈ f − 1 A i } = ∪ i ∈ I f − 1 ( A i ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315195865/a5534f1d-1364-43dc-bd30-b3924321f544/content/eq4613.tif"/>
