# Set Theory

#### ofir

##### New member
I need to find an example that disprove each statement. A and B are sets. (It can be an infinity set)
1. P(A\B) = P(A) if and only if A and B are disjoint sets
2. If (AUB)\A = (B\A)UA then B ⊂ A. (B is not equal to the empty set)

Thank you.

#### Subhotosh Khan

##### Super Moderator
Staff member
I need to find an example that disprove each statement. A and B are sets. (It can be an infinity set)
1. P(A\B) = P(A) if and only if A and B are disjoint sets
2. If (AUB)\A = (B\A)UA then B ⊂ A. (B is not equal to the empty set)

Thank you.
since you did not show any work/thought about the problem, let us make sure we have understanding of common definitions.

Please tell us - What is the definition of disjointed set/s?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

#### ofir

##### New member
The definiation of disjointed sets are if x is a set and y is a set, they are disjointed, there intersection are empty.
I tried to do to define A (inifity set) as { ∅,{∅},{∅,{∅}}…}, and B is {∅}, but I don't know if it is ok? And if it is not, which example can be suitable?

• lookagain

#### pka

##### Elite Member
I need to find an example that disprove each statement. A and B are sets. (It can be an infinity set)
1. P(A\B) = P(A) if and only if A and B are disjoint sets
2. If (AUB)\A = (B\A)UA then B ⊂ A. (B is not equal to the empty set)
Does the notation $$\bf{P}(A)$$ stand for the powerset of $$A~?$$

#### ofir

##### New member
Yes. P(A) is the powerset of A.
I am trying to find an example that disprove this statement.

#### ofir

##### New member
P(A\B) = P(A) if and only if A and B are disjoint sets
I am looking for an example to disprove it.
I found that:
A = { n | n * n=1, n ∈ Z}
B = {1}

Is it correct? If not, can you please give me an example of sets A and B that disprove the statement.

Thank you.

#### pka

##### Elite Member
P(A\B) = P(A) if and only if A and B are disjoint sets
That statement is true.
|f $$A\cap B=\emptyset$$ then $$A\setminus B = A$$ therefore $$\mathscr{P}(A\setminus B)=\mathscr{P}(A)$$.
We prove the converse by contradiction.
Suppose that $$\mathscr{P}(A\setminus B)=\mathscr{P}(A)~\&~\exists t\in A\cap B$$.
But that means $$\{t\}\in \mathscr{P}(A)$$ but $$\{t\}\notin \mathscr{P}(A\setminus B)$$.
Thus if $$\mathscr{P}(A\setminus B)=\mathscr{P}(A)$$ then $$A\cap B=\emptyset$$

#### pka

##### Elite Member
2. If (AUB)\A = (B\A)UA then B ⊂ A. (B is not equal to the empty set)
The above is a logically true statement.
That takes a bit of explaining.
The statement If $$2+2=5$$ then $$(-1)^2<0$$ is true. Because a false implies any statement.
The statement $$(A\cup B)\setminus A=(B\setminus A)\cup A$$ is false.

#### Jomo

##### Elite Member
@pka
(AUB)\A = B\A
(B\A)UA = BUA
Now (B\A) = BUA if A = $$\displaystyle \emptyset$$
But pka claims (I think) that (AUB)\A = (B\A)UA is never true.

#### pka

##### Elite Member
@pka
(AUB)\A = B\A
(B\A)UA = BUA
Now (B\A) = BUA if A = $$\displaystyle \emptyset$$
But pka claims (I think) that (AUB)\A = (B\A)UA is never true.
Note that the problem states $$B\ne \emptyset$$ and if $$(A\cup B)\setminus A=(B\setminus A)\cup A$$ then $$B\subset A$$

#### Jomo

##### Elite Member
Note that the problem states $$B\ne \emptyset$$ and if $$(A\cup B)\setminus A=(B\setminus A)\cup A$$ then $$B\subset A$$
But I never assumed that B = $$\displaystyle \emptyset$$

#### pka

##### Elite Member
But I never assumed that B = $$\displaystyle \emptyset$$
I understand that you did not but the problem does.
It says that if $$(A\cup B)\setminus A=(B\setminus A)\cup A$$ then $$B\subset A$$ {B not empty.) If A is empty then B is also.
$$(A\cup B)\setminus A\\ (A\cup B)\cap A^c\\(A\cap A^c)\cup (B\cap A^c)\\B\setminus A$$ AND $$(B\setminus A)\cup A\\(B\cap A^c)\cup A\\(B\cup A)\cap (A^c\cup A)\\B\cup A$$ It is saying that $$B\setminus A=B\cup A$$ implies $$B\subset A$$ $$B\ne \emptyset$$