Properties of the pre-image of a function

From Maths
Jump to: navigation, search
Stub grade: A*
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Tidy up and then demote

Contents

 [hide] 

Statement

Let X and Y be sets and let f:X→Y be a function between them. Then:

  1. For {Aα}α∈I⊆P(Y)[f−1(⋃α∈IAα)=⋃α∈If−1(Aα)]
  2. For {Aα}α∈I⊆P(Y)[f−1(⋂α∈IAα)=⋂α∈If−1(Aα)]
  3. For A,B∈P(Y)[f−1(A−B)=f−1(A)−f−1(B)
  4. For A∈P(Y)[f−1(Y−A)=X−f−1(A)] - corollary to 3

Proof

Grade: C
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
These are easy and routine

This proof has been marked as an page requiring an easy proof

3

  1. f−1(A−B)⊆f−1(A)−f−1(B) (we use the implies-subset relation to see this is equivalent to ∀x∈f−1(A−B)[x∈f−1(A)−f−1(B)]
    • Let x∈f−1(A−B) be given, then f(x)∈A and f(x)∉B (as if f(x)∈B then f(x)∉A−B so x∉f−1(A−B))
      • so x∈f−1(A) and x∉f−1(B) (as if x∈f−1(B), then f(x)∈B, which we've established is not the case)
  2. f−1(A)−f−1(B)⊆f−1(A−B) (we use the implies-subset relation to see this is equivalent to ∀x∈f−1(A)−f−1(B)[x∈f−1(A−B)]
    • Let x∈f−1(A)−f−1(B) be given. Then x∈f−1(A) and x∉f−1(B) (by definition of relative complement)
      • Then f(x)∈A and f(x)∉B (as if f(x)∈B then x∈f−1(B) which we've established is not the case)
        • So f(x)∈A−B (by definition of relative complement)
          • thus x∈f−1(A−B)
  3. We combine f−1(A)−f−1(B)⊆f−1(A−B) and f−1(A−B)⊆f−1(A)−f−1(B) to see:
    • f−1(A)−f−1(B)=f−1(A−B)
Retrieved from "http://www.maths.kisogo.com/index.php?title=Properties_of_the_pre-image_of_a_function&oldid=3302"