# Power sets of the set difference and difference of the power sets

There are two statements:

$$(a)\space\mathcal P(A\setminus B)\subseteq \mathcal P(A)\setminus \mathcal P(B)$$

$$(b)\space \mathcal P(A\setminus B)\supseteq \mathcal P(A)\setminus \mathcal P(B)$$

Generally speaking: $$\emptyset\in(\mathcal P(A)\cap \mathcal P(B))$$

Is it correct to say power sets can be disjunctive because the empty set is disjunctive to itself? $$\emptyset\cap\emptyset=\emptyset, edited$$ My counter-example for the statement: $$(a)\space\mathcal P(A\setminus B)\subseteq \mathcal P(A)\setminus \mathcal P(B)$$ is a pair of two disjunctive sets $$A$$ and $$B$$ so my claim is equivalent to: $$\mathcal P(A)\nsubseteq\mathcal P(A)\setminus \{\emptyset\}$$ but is kind of a contradiction with the question above.

The counter-example for the statement:$$(b)\space \mathcal P(A\setminus B)\supseteq \mathcal P(A)\setminus \mathcal P(B)$$ is a pair of sets $$A$$ and $$B$$ such that: $$A\cap B\neq \{\emptyset\}$$ /this should be $$\emptyset$$ instead of $$\{\emptyset\}$$, as noticed in comments/ I also took into consideration the cardinality of the power sets, but it was insignificant when dealing with disjunctive sets. It works better for a Cartesian product.

• $\{\varnothing\}$ is not the empty set. Nov 3 '19 at 11:09
• What is a disjunctive set? Nov 3 '19 at 12:04
• The term you're looking for is "disjoint sets". Nov 3 '19 at 12:55
• Thank you very much. I had problems with translating from Croatian to English. Nov 3 '19 at 12:58

## 1 Answer

Counterexamples:

1. The empty set is in $$\mathcal P(A \setminus A)$$ while $$\mathcal P(A) \setminus P(A)$$ is empty.
2. The set $$\{0,\pi\}$$ is in $$\mathcal P(R) \setminus \mathcal P(Q)$$ but not in $$\mathcal P(R \setminus Q)$$.