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In (1) your proof is fine but I'd reword it for clarity. You wrote, "I assumed [math]A = \{2,3,4,5]\}[/math] is the subset of the power set ..."

I would reword that as, "Suppose  [math]A = \{2, 3, 4, 5\}[/math]. Then [math]A \cap A = A \neq \emptyset[/math]. Or you could just say, For any nonempty set  [math]A[/math], we have [math]A \cap A = A \neq \emptyset[/math]. That would be sufficient for me. 

Also note that you meant that [math]A[/math] is an element of the power set. It's a subset of the integers, but an element of the power set.

Edited by wtf
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10 hours ago, wtf said:

Also note that you meant that A is an element of the power set. It's a subset of the integers, but an element of the power set.

That's a good point. This is an important distinction and one that technically makes the proof incorrect. A is an element of P(Z), not a subset. A is a subset of Z. The elements of P(Z) are the subsets of Z.

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