Exercises
[05S]Now consider instead the characteristic function defined as before, but considered as \({\mathbb 1}_ A: X→ℤ_ 2\) i.e. taking values in the group \(ℤ_ 2\) (more correctly referred to as \(ℤ/2ℤ\)).
In this case the above relations can be written as
\[ {\mathbb 1}_{A^ c} = {\mathbb 1}_ A +1 ~ ~ ,~ ~ {\mathbb 1}_{A∩ B} = {\mathbb 1}_ A {\mathbb 1}_ B ~ ~ ,~ ~ {\mathbb 1}_{A∪ B} = {\mathbb 1}_ A {\mathbb 1}_ B +{\mathbb 1}_ A +{\mathbb 1}_ B~ . \]Recall the definition of the symmetric difference \( AΔ B= (A⧵ B) ∪ (B⧵ A) \), and then
\[ {\mathbb 1}_{AΔ B} = {\mathbb 1}_ A + {\mathbb 1}_ B ~ ~ . \]With these rules we show that
\[ AΔ B = B Δ A ~ ~ ,~ ~ (AΔ B)^ c = A Δ (B^ c)=(A^ c) Δ B ~ ~ ,~ ~ AΔ B = C \iff A = B Δ C \]\[ (AΔ B)∩ C = (A∩ C) Δ (B∩ C) ~ ~ ,~ ~ A∪ (BΔ C) = (A∪ B)Δ (A^ c ∩ C) \]