# Set theory proof problem

1. Apr 9, 2007

### GregA

1. The problem statement, all variables and given/known data
Suppose B is a set and suppose $$\mathcal{F}$$ is a family of sets.
Prove that $$\cup$${$$A\setminus B|A \in \mathcal{F}$$}$$\subseteq \cup(\mathcal{F}\setminus \mathcal{P}(B))$$

For want of a better way I'm denoting powerset of B as $$\mathcal{P}(B)$$)

2. Relevant equations

3. The attempt at a solution

Whilst trying to interpret the above I figured that I should argue that since any elements of a powerset are themselves sets that I need only show that the LHS is a set containing all elements of sets in F minus those that are in B whilst RHS is a set containing all elements of all sets in F because no element in such sets would actually be sets anyway such that if $$\exists C\in \cup$$ {$$A\setminus B|A \in \mathcal{F}$$}$$(x \in C)$$ then it would definitely be true that $$\exists D\in \cup(\mathcal{F}\setminus \mathcal{P}(B))(x \in D)$$. But at this point asked myself why I have justification for saying this, I need the above statement to be true for all sets regardless of what's in them...in fact if I let each set in F be a set containg a set for example:

A1= {{1,2}}
A2= {{2,3}}
B = {1,2}

then $$\cup$${$$A\setminus B|A \in \mathcal{F}$$} = {{1,2}}$$\cup$${{2,3}} = {{1,2},{2,3}}
whilst $$\cup(\mathcal{F}\setminus \mathcal{P}(B))$$ = $$\emptyset$$$$\cup$${2,3} = {{2,3},$$\emptyset$$}
But now I can find some x in LHS that isn't in RHS such that LHS cannot be a subset.

Is my reasoning/interpretation wrong or should I move on to another querstion?

Last edited: Apr 9, 2007
2. Apr 9, 2007

### Dick

Translated into english, x is in the LHS if x is an element of some A in F and x is not in B. x is in the RHS if x is an element of some A in F where A is not a subset of B. It should be clear LHS->RHS, but not vice-versa.

3. Apr 10, 2007

### GregA

I agree with you here that it should be clear but the problem I have is that if all A's in F are themselves families of sets (this is not ruled out in the problem) whilst B is a set; then supposing F was comprised of 2 A's and their elements comprised as follows: A1 = {{1,2}}, A2 = {{1},{2}} Also if B was comprised as follows: {1,2} then:

the elements of$$\mathcal{P}(B) =$${1},{2},{1,2},$$\emptyset$$ whilst the elements of B are 1,2

Now for LHS, if x was in either A in F then it would definitely not be in B for all elements of A1 or A2 are sets. The union of LHS would be {{1},{2},{1,2}}. Considering RHS however, if x was in either A in F it would also be in $$\mathcal{P}(B)$$ and so the union of RHS would be empty. :grumpy:

Last edited: Apr 10, 2007
4. Apr 10, 2007

### Dick

Not true. The RHS is the same as the LHS. Neither A1 nor A2 is in P(B). They have ELEMENTS that are in P(B) but they themselves are not in P(B).

5. Apr 10, 2007

### GregA

Cheers Dick I've had to really look at what the RHS means (as well as your first post again) and notice that I interpreted it wrong