1
$\begingroup$

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 $=$."

$\endgroup$
5
  • $\begingroup$ For two sets $A,B$, $A=B\iff A\subseteq B\wedge B\subseteq A$. $\endgroup$ – poyea Jun 21 '18 at 8:32
  • 1
    $\begingroup$ You shouldn't see $\bigcup S$ as $S_1 \cup S_2 \cup \dots \cup S_n$ as the union can be infinite! $\endgroup$ – user370967 Jun 21 '18 at 9:49
  • $\begingroup$ @Math_QED Are you suggesting, that we can not put $=$ sign there in case of infinite sets? $\endgroup$ – Bakin Eugene Jun 21 '18 at 10:50
  • $\begingroup$ 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. $\endgroup$ – user370967 Jun 21 '18 at 11:13
  • $\begingroup$ I just ran into this in the same book, and I’m as confused as you are... Did you ever find an authoritative answer? $\endgroup$ – bellkev Feb 1 '20 at 7:46
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Here is a quote from book : "Show that in first formula we cannot replace $\subseteq$ with =" $\endgroup$ – Bakin Eugene Jun 21 '18 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.