When it comes to proving that two sets are equal, say $A = B$, we're usually told that we have to prove that $A \subset B$ and $B \subset A$. However, I'm under the impression that this strategy isn't unique. Two sets are equal if they have the same elements, and this commonly used strategy is just one way to prove that.
A better way to prove that two sets are equal, at least in my opinion, is to use set notation. However, because I never see this being used, I'm unsure if this method is correct. Example:
Prove that $ (A \cup B) \times C = (A \times C) \cup (B \times C) $.
Proof. We have,
\begin{align}
(A \cup B) \times C &= \{x; x \in A \operatorname{or} B \} \times \{y; y \in C \} \\
&= \{(x, y); x \in A \operatorname{or} B, y \in C \} \\
&= \{(x, y); x \in A, y \in C \} \cup \{(x, y); x \in B, y \in C \} \\
&= (A \times C) \cup (B \times C).
\end{align}
So the question is, is this method correct? Thanks in advance.
Best Answer
This series of equalities is correct, and is kind of hiding the argument that one is a subset of the other. It covers both directions of containment. For example, to show that $(A \cup B) \times C \subseteq (A \times C) \cup (B \times C)$, replace all $=$ with $\subseteq$, and then for showing that $(A \cup B) \times C \supseteq (A \times C) \cup (B \times C)$, replace all $=$ with $\supseteq$. These replacements work because equality implies containment on either side. So the given method is correct and sort of basically hides the usual method.
When it comes to equality of sets, at least at the very basic level, you will be using this and only this method. Sometimes, you can switch things up : for example, if you want to prove that $A=B$ you can try proving instead that $A^c = B^c$. Or you could go for a mixture : if $A \subset B$ and $A^c \subset B^c$ then $A=B$ as well. You could also prove that $A \cap B^c = \emptyset$ and $A^c \cap B = \emptyset$.
The advantage of using complement/union/intersection modifications is that sometimes a set $A$ is better understood by its complement (e.g. $A = $ "non-multiples of $6$" then the complement of $A$ is "multiples of $6$" which are easier to list), and sometimes two sets are difficult to understand but it is easy to understand their intersection. But it all boils down to that kind of reasoning in the end.