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Hey so I searched quite a bit for this type of question and i couldn't find any help. - Prove: If B⊂A or A⊂B then P(A∪B) = P(A)∪P(B) and prove the opposite if P(A∪B) = P(A)∪P(B) then B⊂A

so I i think I got the first part correct- A⊂B→ x∈A→ x∈B→ {x}∈P(A)→ {x}∈P(B)→ x∈A∪B→ {x}∈P(A∪B) A⊂B→ x∈A→ x∈B→ {x}∈P(A)→ {x}∈P(B)→ {x}∈P(A)∪P(B) But for the second part it says I should do the negation of the problem but I don't exactly know where to start. Any help would be much appreciated! Also any good courses on set theory would also be appreciated!

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I think that for the second thing you want to prove, that you are also wanting to add an "or $A \subset B$". Please correct me if I'm wrong.

If we negate the problem, we get $P(A\cup B) = P(A)\cup P(B)$ and $B \not\subset A$ and $A \not\subset B$. We want to show this negation leads to a contradiction.

Since $B \not\subset A$, there exists $x\in B$ such that $x \notin A$. Similarly, since $A \not\subset B$, there exists $y\in A$ such that $y \notin B$.

Notice since $x \in B$, then $x \in A \cup B$ and since $y \in A$, then $y \in A \cup B$. Thus, $\{x,y\} \subset A \cup B$ and thus $\{x,y\} \in P(A \cup B)$. Since $P(A\cup B) = P(A)\cup P(B)$, we can then write $\{x,y\} \in P(A)\cup P(B)$. This means either $\{x,y\} \in P(A)$ or $\{x,y\} \in P(B)$. Consider the first case where $\{x,y\} \in P(A)$. This implies $\{x,y\} \subset A$ implying $x \in A$ and $y\in A$, a contradiction since we know $x \notin A$. Similarly, the case where $\{x,y\} \in P(B)$ leads to a similar contradiction.

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