- Thread starter ofir
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since you did not show any work/thought about the problem, let us make sure we have understanding of common definitions.I need to find an example thatdisproveeach 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.

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

Please show us what you have tried and

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

Please share your work/thoughts about this problem.

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?

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Does the notation \(\bf{P}(A)\) stand for the powerset of \(A~?\)I need to find an example thatdisproveeach 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)

I am looking for an example to

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.

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That statement is true.P(A\B) = P(A) if and only if A and B are disjoint sets

|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 \(A\cap B=\emptyset\)

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The above is a logically true statement.2. If (AUB)\A = (B\A)UA then B ⊂ A. (B is not equal to the empty set)

That takes a bit of explaining.

The statement

The statement \((A\cup B)\setminus A=(B\setminus A)\cup A\) is false.

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Note that the problem states \(B\ne \emptyset\) and if \((A\cup B)\setminus A=(B\setminus A)\cup A\) then \(B\subset A\)@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.

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

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I understand that you did not but the problem does.But I never assumed that B = \(\displaystyle \emptyset\)

It says that

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 \(B\subset A\) \(B\ne \emptyset\)