An exercise given in Halmos' Naive Set Theory: required to prove that $$(A\cup B) \times X = (A \times X) \cup (B \times X).$$ My approach to this was as follows. Assume $z$ belongs to $(A\cup B) \times X$, then, since $z$ is a member of a cartesian product, it must be of the form $z = (x,y)$ with $x$ in $A\cup B$ and $y$ in $X$. Then $x$ is either in $A$, or in $B$. If $x$ is in $A$, then since $y$ is also in $X$, $(x,y)$ is, by definition, in $A \times X$. However, if $x$ is in $B$, then, again, by definition, $(x,y)$ must be in $B \times X$. So then $(x,y)$ is in $A \times X$, or it is in $B \times X$. From this it follows that $z$ is in $(A \times X) \cup (B \times X)$. We have just proved the "only if" part of the initial statement we set out to prove. Now, to prove the "if" part:
Assume $z$ is in $(A \times X) \cup (B \times X)$, then $(x,y)$ is either in $A\times X$ or $B \times X$. If the former case is true, then $x$ is in $A$, and $y$ is in $X$, however, if the latter case is true, then $x$ is in $B$, and $y$ is in $X$. Since in both cases $y$ is in $X$, it follows from disjunction elimination that $y$ is in $X$. Then $x$ is either in $A$ or $B$, and $y$ is in $X$; hence, $x$ is in $A\cup B$ and $y$ is in $X$. Then, by definition, $(x,y)$ is in $(A\cup B) \times X$.
I have never done a proof using cartesian products before, so I must ask for guidance from some more professional. Is there any fault in my reasoning, moreover, is it ok for a proof to be formated like this? I am still fairly new to proving set-theoretic facts so I require constant affirmation from the more professional. Thank you in advance.
The reasoning is correct. However, while working with sets, a notation that comes handy is working with the symbols "$\exists, \lor, \land, \forall$". These mean "exists", "or", "and", and "for all", respectively. With these, the proof can be reduced as follows:
$\begin{align*} (x,y) \in (A\cup B)\times X &\iff (x \in A\cup B) \land (y \in X) \\ &\iff (x \in A \cup B) \land (y \in X)\\ &\iff (x \in A \lor x \in B )\land (y \in X) \\ &\iff ((x \in A \land y \in X) \lor (x \in B \land y \in X) )\\ &\iff (x,y) \in A \times X \lor (x,y) \in B \times X) \\ &\iff (x,y) \in (A \times X) \cup (B \times X) \end{align*}$
Hence, as two sets $C,D \subset Y$ are equal iff $\forall x \in Y( x \in C \iff x \in D)$, we have proved the result.
If you notice, your reasoning of the "only if" and "if" parts are sort of the same, but one is in opposite order of the other. This was shortened by using the "$\iff$" symbol which is useful when deductions are, in some way, "invertible".