the inverse image commutes with set operations


Theorem. Let f be a mapping from X to Y. If {Bi}i∈I is a (possibly uncountable) collectionMathworldPlanetmath of subsets in Y, then the following relationsMathworldPlanetmathPlanetmath hold for the inverse imagePlanetmathPlanetmath:

  1. (1)

    f-1⁢(⋃i∈IBi)=⋃i∈If-1⁢(Bi)

  2. (2)

    f-1⁢(⋂i∈IBi)=⋂i∈If-1⁢(Bi)

If A and B are subsets in Y, then we also have:

  1. (3)

    For the set complementPlanetmathPlanetmath,

    (f-1⁢(A))∁=f-1⁢(A∁).
  2. (4)

    For the set differenceMathworldPlanetmath,

    f-1⁢(A∖B)=f-1⁢(A)∖f-1⁢(B).
  3. (5)

    For the symmetric differenceMathworldPlanetmathPlanetmath,

    f-1⁢(A△B)=f-1⁢(A)△f-1⁢(B).

Proof. For part (1), we have

f-1⁢(⋃i∈IBi) = {x∈X∣f⁢(x)∈⋃i∈IBi}
= {x∈X∣f⁢(x)∈Bi⁢for some⁢i∈I}
= ⋃i∈I{x∈X∣f⁢(x)∈Bi}
= ⋃i∈If-1⁢(Bi).

Similarly, for part (2), we have

f-1⁢(⋂i∈IBi) = {x∈X∣f⁢(x)∈⋂i∈IBi}
= {x∈X∣f⁢(x)∈Bi⁢for all⁢i∈I}
= ⋂i∈I{x∈X∣f⁢(x)∈Bi}
= ⋂i∈If-1⁢(Bi).

For the set complement, suppose x∉f-1⁢(A). This is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to f⁢(x)∉A, or f⁢(x)∈A∁, which is equivalent to x∈f-1⁢(A∁). Since the set difference A∖B can be written as A∩Bc, part (4) follows from parts (2) and (3). Similarly, since A△B=(A∖B)∪(B∖A), part (5) follows from parts (1) and (4). □

Title the inverse image commutes with set operations
Canonical name TheInverseImageCommutesWithSetOperations
Date of creation 2013-03-22 13:35:24
Last modified on 2013-03-22 13:35:24
Owner matte (1858)
Last modified by matte (1858)
Numerical id 11
Author matte (1858)
Entry type Proof
Classification msc 03E20
Related topic SetDifference