Is there exactly one Y ∈ [X]R such that Y ∩ B = {}?

[Syrus]

Homework Statement

Suppose B ⊆ A and define a relation R on P(A) as follows:

R = {(X,Y) ∈ P(A) x P(A) | (X∆Y) ⊆ B}

a) Show that R is an equivalence relation on P(A).
b) Prove that for every X ∈ P(A) there is exactly one Y ∈ [X]_R such that Y ∩ B = { }.

*P(A) is the powerset of A

Homework Equations

The Attempt at a Solution

I have successfully completed part (a) of this exercise. I seem to be having a problem with part (b). My proof so far goes like this:

Let X ∈ P(A) and suppose Y = X\B ∈ [X]_R such that Y ∩ B = { }. Now let Z ∈ [X]_R such that Y ∩ B = { }. We must show that Y = Z, so let w ∈ Y. Then w ∈ X and w ∉ B...

 

[Tedjn]

First you must explicitly mention that Y = X\B ∈ [X]_R, which I imagine you might already have done.

Once this is done, consider this other Z ∈ [X]_R disjoint from B. We know that X\Z ∪ Z\X ⊆ B. What does this say about Z relative to X?