Critique my proof of: $A \times (B \cap C) = (A \times B) \cap (A \times C)$

Solution 1:

I prefer to solve this kind of exercise by making use of the definitions and properties involved: \begin{align*} A\times(B\cap C) & = \{(x,y) \mid (x\in A)\wedge(y\in B\cap C)\}\\\\ & = \{(x,y) \mid (x\in A)\wedge(y\in B)\wedge(y\in C)\}\\\\ & = \{(x,y) \mid (x\in A)\wedge(x\in A)\wedge (y\in B)\wedge(y\in C)\}\\\\ & = \{(x,y) \mid ((x\in A)\wedge (y\in B))\wedge((x\in A)\wedge(y\in C))\}\\\\ & = \{(x,y) \mid (x\in A)\wedge(y\in B)\}\cap\{(x,y) \mid (x\in A)\wedge(y\in C)\}\\\\ & = (A\times B)\cap(A\times C) \end{align*}

Hopefully this helps !

Solution 2:

A first issue that seems minor, but causes all sorts of confusion: don't use commas as connectives; they often mean 'and' but sometimes mean 'if, then' and sometimes even mean 'or', and you'll get in the habit of reading them one way and not even notice when someone else uses (or you use!) them differently. Thus, say not "$y \in B$, $y \in C$" but "$y \in B$ and $y \in C$", and similarly elsewhere.

I think you ought not to start with "Suppose $p = (x, y)$ is an arbitrary element"; arbitrary elements (of what?) are not ordered pairs. Indeed, unless you work with an explicit universal set, then you ought not to fix $p$ outside of the case distinction at all. In case ($\subseteq$) you fix $p$, observe that by the definition of $A \times (B \cap C)$ we have that $p$ is of the form $(x, y)$ for some $x \in A$ and $y \in B \cap C$, and then proceed as before. Similarly for the other case (except that you have to wait a bit longer).

Otherwise your argument looks fine to me, depending on the level of detail expected. For example, as an instructor, I might ask you to explain why $p \in A \times B$ and $p \in A \times C$ by appealing to the definitions there.