# Proving general property of power set of set difference

I'm taking an introductory course in set theory. We were presented with the following property for the power set of the difference between sets:

$$\mathcal{P}(A\setminus B ) \subseteq (\mathcal{P}(A)\setminus\mathcal{P}(B)) \cup \ \{\varnothing\}$$

The prove was not given or attempted. I can formulate multiple examples that satisfy the propoerty but can't come up with a formal demonstration. I would like to know how it can be proved, as I don't know how to start with the demonstration.

• Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Sep 12 '21 at 21:16
• If you want to prove that for two sets $R$ and $S$ you have $R \subseteq S$ a formal proof usually loos like this: "Let $x$ be an element of $R$. Then (argument using $x$ here ...) so $x$ is also an element of $S$." Try that. edit the question to show us what you did if you get stuck. Sep 12 '21 at 21:26
• In general, when working on inclusion or equality problems in elementary set theory you should start by writing out what the sets are. for example $\mathcal{P}(A)\setminus \mathcal{P}(B)=\{ S\subseteq A : S\not\subseteq B\}$ Sep 12 '21 at 21:35

To prove that $$\mathcal{P}(A\setminus B ) \subseteq (\mathcal{P}(A)\setminus\mathcal{P}(B)) \cup \ \{\varnothing\}$$, it suffices to show that for any $$S \in \mathcal P(A \setminus B)$$ with $$S \neq \varnothing$$, $$S \in \mathcal P(A) \setminus \mathcal P(B)$$. In other words: show that if $$S \subseteq A \setminus B$$ with $$S \neq \varnothing$$, then it holds that $$S \subseteq A$$ and it is not the case that $$S \subseteq B$$. It is useful to note that if $$A \setminus B$$ is non-empty, then there exists at least one element $$x$$ such that $$x \in A$$ but $$x \notin B$$.