# Exercise 14, section 9 of Hammack's Book of Proof.

The exercise asks the reader to prove or disprove the following statement.

If $$A$$ and $$B$$ are sets, then $$\mathscr{P}(A) \cap \mathscr{P}(B) = \mathscr{P}(A \cap B)$$.

I know one way to prove such a statement is to simply produce a series of equalities that transform the LHS of the equation into the RHS. However, I am not sure if my translation of the powersets into set builder notation is valid. More specifically, the book never gives an example of a quantifier used in a set builder so I tried to translate from the book's definition of a subset written in English (as opposed to symbolic notation). Bellow is my proof.

Proof. Observe the following sequence of equalities. \begin{align} \mathscr{P}(A) \cap \mathscr{P}(B) &= \{ X: X \subseteq A \land X \subseteq B\} &(\text{def. of powerset}) \\ &= \{ X: \forall x \in X, x \in A \land x \in B \} &(\text{def. of subset}) \\ &= \{ X:\forall x \in X, x \in A \cap B \} &(\text{def. of intersection}) \\ &= \{ X: X \subseteq A \cap B \} &(\text{def. of subset}) \\ &= \mathscr{P}(A \cap B) &(\text{def. of powerset}) \end{align}

The proof is complete. $$\tag*{\blacksquare}$$

Any feedback on the validity of this proof or on ways it could be better written is greatly appreciated.

• The proof looks good to me, and the equalities are alright. I would just add the following caveat: $\{X|x\in A \textit { and } x\in B, \textit { then } x\in X\}$. Basically, make it a conditional with the "and" statement as the antecedent. It's just cosmetic, but it makes it more fluid. – Bertrand Wittgenstein's Ghost Feb 5 at 22:27
• @BertrandWittgenstein'sGhost: The OP’s version is correct; what you’re suggesting would not be, as it describes the collection of $X$ such that $A\cap B\subseteq X$. – Brian M. Scott Feb 5 at 22:30
• @BrianM.Scott that's weird, I don't read it like that. – Bertrand Wittgenstein's Ghost Feb 5 at 22:35
• @BertrandWittgenstein'sGhost: $x\in A\land x\in B\to x\in X$ clearly says that if $x$ belongs to both $A$ and $B$, then $x$ belongs to $X$. This certainly implies that $A\cap B\subseteq X$. And it says absolutely nothing about any $x$ that does not belong to both $A$ and $B$, so it is entirely possible that some of them also belong to $X$. Thus, we cannot conclude that $X=A\cap B$, and it is certainly the case that $X$ cannot be a proper subset of $A\cap B$. In short, it says that $X\supseteq A\cap B$. If you’re reading it differently, I’m afraid that you’re misreading it. – Brian M. Scott Feb 5 at 22:40
• @BertrandWittgenstein'sGhost: You’re very welcome! – Brian M. Scott Feb 5 at 22:50

$$X\in\wp(A)\cap\wp(B)$$ iff $$X\in\wp(A)$$ and $$X\in\wp(B)$$, which is the case iff $$X\subseteq A$$ and $$X\subseteq B$$. But that is true iff $$X\subseteq A\cap B$$, i.e., iff $$X\in\wp(A\cap B)$$, so $$\wp(A)\cap\wp(B)=\wp(A\cap B)$$.
• Your answer kind of leads to a latent question I have about proof writing. The reason I wanted to expand the statement into set builder notation is that I did not want to assume $X \subseteq A$ and $X \subseteq B \iff X \subseteq A \cap B$ although it makes intuitive sense. So, to what extend can we assume statements as these? Does it depend entirely on the context and the level of your readers? – madibam Feb 5 at 22:42