Set Theory

[ofir]

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.

 

[Deleted member 4993]

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:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[ofir]

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?

 

[pka]

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]

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

 

[ofir]

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]

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]

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.

 

[Steven G]

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

 

[pka]

@pka
(AUB)\A = B\A
(B\A)UA = BUA
Now (B\A) = BUA if A = $$\emptyset$$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 $B\subset A$

 

[Steven G]

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 = $$\emptyset$$

 

[pka]

But I never assumed that B = $$\emptyset$$

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.
Think about it.
$(A\cup B)\setminus A\\ (A\cup B)\cap A^c\\(A\cap A^c)\cup (B\cap A^c)\\B\setminus A$ 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 $B\subset A$ $B\ne \emptyset$

 

[Steven G]

I too arrived in my previous post that that the problem is saying that B∖A=B∪A implies B⊂A (and B≠∅).

Now I see it, if A = ∅ and B⊂ A, the B= ∅