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A question from my homework I'm having trouble understanding. We are given:

$A(1) = \{\varnothing\}$, $A(n+1) = A(n)\cup (A(n)\times A(n))$

$A=A(1)\cup A(2)\cup A(3)\cup \cdots \cup A(n)\cup A(n+1) \cup \cdots$ to infinity

The questions are:

1) show that $A\times A \subseteq A$

2) Is $A \times A = A$?

Thank you for your help.

I've tried writing $A(2)$ but it gets really complicated and I'm having trouble understanding what the sets are. Let alone solve the question.

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  • $\begingroup$ קבוצה = set, not group. $\endgroup$
    – Asaf Karagila
    Nov 5, 2013 at 21:22
  • $\begingroup$ $$A(2) = \emptyset \cup (\emptyset \times \emptyset) = \emptyset$$ Or do you have a different definition of the carthesian product? $\endgroup$
    – AlexR
    Nov 5, 2013 at 21:23
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    $\begingroup$ @Alex: $\{\varnothing\}\neq\varnothing$. $\endgroup$
    – Asaf Karagila
    Nov 5, 2013 at 21:23
  • $\begingroup$ What Asaf said. A(1) is the set that contains an element which is the empty set. the empty set itself, has no elements. and as such A(1) is not equal to the empty set $\endgroup$ Nov 5, 2013 at 21:24
  • $\begingroup$ Oh, I misread $A(1) = \emptyset$. nvm. $\endgroup$
    – AlexR
    Nov 6, 2013 at 12:53

2 Answers 2

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HINT: For (2), note that not all the elements of $A$ are ordered pairs.


Also, let's write $A(2)$, but to make it simpler let's call $A(1)=X$. Then $A(2)=X\cup(X\times X)=\{\varnothing\}\cup\{\langle\varnothing,\varnothing\rangle\}=\{\varnothing,\langle\varnothing,\varnothing\rangle\}$.

Not very difficult, $A(3)$ will have six elements.

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  • $\begingroup$ Ok Asaf, I understood what you mean by the ordered pairs. It's true not all the elements of A are ordered pairs, as opposed to $A\times A$ and so they can't be equal. $\endgroup$ Nov 5, 2013 at 21:44
  • $\begingroup$ Yup, that's the catch. If you assume $\sf ZF$ as your set theory, and $\langle x,y\rangle=\{\{x\},\{x,y\}\}$ (the Kuratowski definition for ordered pairs), then one can show that $A\times A=A$ implies $A=\varnothing$. $\endgroup$
    – Asaf Karagila
    Nov 5, 2013 at 21:48
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Hint: For (1), note that $A_n \times A_n \subseteq A_{n+1}$ is true and looks similar to what you are asked to prove.

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    $\begingroup$ You gave the very same hint I was going to write, but I got distracted by the new comment. :-) $\endgroup$
    – Asaf Karagila
    Nov 5, 2013 at 21:24
  • $\begingroup$ Are you suggesting mathematical induction? $\endgroup$ Nov 5, 2013 at 21:25
  • $\begingroup$ @Oria: Not really an induction. Just use the definition. $\endgroup$
    – Asaf Karagila
    Nov 5, 2013 at 21:27
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    $\begingroup$ @OriaGruber Asaf is right. Also, don't worry to much about the details of what's in your sets, just look at how they are defined in terms of each other. The argument for (1) won't really have anything to do with Cartesian products in particular. $\endgroup$ Nov 5, 2013 at 21:29

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