Say $A$ is a a non-empty set, is it possible to have $A^n \times A^m \in A$? For any integers $n,m$, I'm not asking if this is true in general, I'm only asking if it can be the case for some particular set. If this is true, could you provide an example? If this is always false could you provide a proof? Thank you.

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    $\begingroup$ Should that be $A^n \times A^m \subset A$? $\endgroup$ – Ben Grossmann Jul 26 at 9:11
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    $\begingroup$ No no no, it can't be "depends on the answer". If you want your questions to be taken seriously, you need to take them seriously yourself. Clarify your question. $\endgroup$ – Asaf Karagila Jul 26 at 9:23
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    $\begingroup$ I do not know what A contains because in my mind it's a set that can be expanded unlmitedly by always taking cartesian product of its elements so I do not know whether every element of the cartesian products coincide with A itself. I leave $∈$ because my idea is that since A is always expanding the cartesian product can be only element of the set A. I'm super serious, you have no reason to be rude. $\endgroup$ – Kolmogorovwannabe Jul 26 at 9:31
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    $\begingroup$ I am sorry if I was rude. Nevertheless, when someone asks you to clarify the meaning of your words, you can't say "let's see what I mean". This is not a reality show. People spend their free time here helping you and others. Respect that. Respect them. $\endgroup$ – Asaf Karagila Jul 26 at 9:35
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    $\begingroup$ I understood that, but I don't know what is inside A, otherwise I would have not posted the question. Given that I do not know the content of A, how can I say if the right notation is $∈$ or $⊂$? My lack of explanation is due to ignorance, not laziness. $\endgroup$ – Kolmogorovwannabe Jul 26 at 9:40

The answer depends on the underlying set theory and the actual symbol under consideration, whether $\in$ or $\subseteq$.

In standard (ZF) set theory, the axiom of foundation prevents the existence of any set as specified. The reason is that sets have a rank, and the rank of any member of a set $A$ is strictly smaller than that of $A$. However, the rank of powers of $A$ and of Cartesian products is at least as large as that of $A$.

There are, however, set theories (such as ZFA, where the A stands for Aczel's anti-foundation axiom) where there are sets $\Omega$ such that $\Omega=\{\Omega\}$. For any such $\Omega$ and the standard definition of ordered pair, $$(\Omega,\Omega)=\{\{\Omega\},\{\Omega,\Omega\}\}=\{\Omega,\Omega\}=\Omega$$ and so all finite powers and all Cartesian products of finite powers of $\Omega$ just coincide with $\Omega$ itself, and for any $n,m$ positive integers, $\Omega^n\times\Omega^m=\Omega\in\Omega$. The anti-foundation axiom also implies the existence of sets $A$ with $A^n\times A^m\in A$ even if $n$ or $m$ is 0. It also implies the existence of such sets if powers are taken in the sense of functions rather than Cartesian products (i.e., for instance., if $A^2$ is understood as the set of functions $f:\{0,1\}\to A$ rather than as the product $A\times A$).

The existence of sets $A$ such that $A^n\times A^m\subset A$ is less problematic (the axiom of foundation plays no role here). It is easy to give examples: the empty set, for instance. Or $V_\omega=\bigcup_k\mathcal P^k(\emptyset)$. In general, given any set $A$, you can easily find a larger set $A'$ with $(A')^n\times(A')^m\subseteq A'$ for any $n,m$: just build $A'$ by a recursive process: $A'=\bigcup_k A_k$, where $A_0=A$, and $$A_{s+1}=A_s\cup\bigcup_{n,m}A_s^n\times A_s^m.$$ (The set $A'$ so constructed in the smallest set containing $A$ and satisfying the desired property.)

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  • $\begingroup$ This is a marvellous answer. I'm going to deepen my knowledge of all the theories that you quoted. The first question that I would like to ask is: the anti-foundation axiom implies the existence of sets $A$ with $A^n \times A^m \in A$. In this case, like for $\Omega$. we have that $A^n \times A^m = A$ or there can also be $A^n \times A^m \neq A$? Thank you very much. $\endgroup$ – Kolmogorovwannabe Jul 26 at 14:24
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    $\begingroup$ @Kolmogorovwannabe Yes, anti-foundation gives you sets $A$ satisfying your condition with $A\ne\{A\}$. $\endgroup$ – Andrés E. Caicedo Jul 26 at 14:28
  • $\begingroup$ Reading again your answer: what do you mean with "even if n or m is o"? I'm asking this because I'm trying to figure it out if this result holds only for finite number of Cartesian products or if it can be extented to countably infinite number of Cartesian products. Thank you again. $\endgroup$ – Kolmogorovwannabe Jul 26 at 15:16
  • $\begingroup$ @Kolmogorovwannabe $\Omega^0=\{\emptyset\}\ne\Omega$, so none of $\Omega^0$, $\Omega^0\times\Omega^n$, $\Omega^m\times\Omega^0$, etc., are members of $\Omega=\{\Omega\}$. But anti-foundation gives us other sets that satisfy the requirement also in that case (i.e., for all $m,n$ natural numbers, not just positive integers). Similarly, anti-foundation also gives you sets as needed for $n,m$ countably infinite. $\endgroup$ – Andrés E. Caicedo Jul 26 at 19:56

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