# Cartesian product of two unions of sets of sets $\subseteq$ union of products of sets

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 $=$."

• For two sets $A,B$, $A=B\iff A\subseteq B\wedge B\subseteq A$. – poyea Jun 21 '18 at 8:32
• You shouldn't see $\bigcup S$ as $S_1 \cup S_2 \cup \dots \cup S_n$ as the union can be infinite! – user370967 Jun 21 '18 at 9:49
• @Math_QED Are you suggesting, that we can not put $=$ sign there in case of infinite sets? – Bakin Eugene Jun 21 '18 at 10:50
• I'm suggesting that the union can run over an infinite index set, while you seem to think that you can index it over the set $\{1, \dots, n\}$ which is finite. – user370967 Jun 21 '18 at 11:13
• I just ran into this in the same book, and I’m as confused as you are... Did you ever find an authoritative answer? – bellkev Feb 1 '20 at 7:46

Then it should have an $=$ sign, not $⊆$. But exercise also asks to show, that you can not put equality here.
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.
• Here is a quote from book : "Show that in first formula we cannot replace $\subseteq$ with =" – Bakin Eugene Jun 21 '18 at 10:13