I'm reading the book "The Foundations of Mathematics” by Ian Stewart and David Tall, and I am puzzled about one of exercises:
Chapter 4. Exercise 2. Given $S$ and $T$ are sets of sets, prove that $$\left(\bigcup S\right) \times \left(\bigcup T\right) \subseteq \bigcup \{ X \times Y \mid X \in S, Y \in T \}$$
I can write the LHS as $$\left(\bigcup S\right) \times \left(\bigcup T\right) = (S_1 \cup S_2 \cup S_3 \cup \ldots \cup S_n ) \times (T_1 \cup T_2 \cup T_3 \cup \ldots \cup T_n) $$
where each $S_i\in S$ and $T_i\in T$. Also, the RHS is
$$ \bigcup \{ X \times Y \mid X \in S, Y \in T \} = (S_1 \times T_1) \cup (S_1 \times T_2) \cup \ldots \cup (S_1 \times T_n) \cup (S_2 \times T_1) \cup \ldots \cup (S_2 \times T_n) \cup \ldots \cup (S_n \times T_n) $$
If I understand correctly, it is a distributive property. Like in this answer. Then, in this case, it should have an $=$ sign, not $\subseteq$. But exercise also asks to show that there is no equality here.
Update: Here is a quote from book: "Show that in first formula we cannot replace $\subseteq$ by $=$."
No, $\subseteq$ is not the same as $\subset$ or more clearly $\subsetneq$ . Just as, $\leq$ is not the same as $\lt$ or more clearly $\lneq$ .
To prove $\sf A=B$ you must prove $\sf A\subseteq B$ and $\sf A\supseteq B$. All the exercise is asking is for you to prove the former and just not worry about whether or not the other is true.