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Let $A=\{a,b\}$. I'm looking for notation for $(A\times A)\;$,$\;(A\times A)\times (A\times A)$ , $((A\times A)\times (A\times A) )\times ((A\times A)\times (A\times A))$, and so on.

I don't think ${A^{2}}^{2}$ is a clear notation because this can be interpreted as being equal to $A^{4}=(A\times A\times A\times A\times A)$. By contrast, $(A^2)^2 \;$ is clearly equal to $\;(A\times A)\times (A\times A)$.

In general, I can write my set as $\overbrace{(((A^2)^2)^2...)}^{\text{n times}} \quad$, where I'm going to call $n$ the 'order' of the set.

I'm looking for a compact notation for this set. I am considering this notation: $A^{n:2} \equiv \overbrace{(((A^2)^2)...)}^{\text{n times}} \quad$. Do you have a suggestion?

Also, I would like to know what to call an object like $(((a,b),(a,a)),((b,b),(a,b)))$? I believe it is a kind of "tree". If I know what to call it, then I can hopefully go to the literature to see the work that has been done on these objects. So, is it a tree or a nest or a hierarchy, ect.? Thanks.

Asaf Karagila
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Chris
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  • What is the difference of your sequence and $\displaystyle{A^{2^n}}$? – Zerox Nov 27 '22 at 19:03
  • I might interpret $A^{2^2}$ as $A\times A \times A \times A$. But clearly $((A)^2)^2$ = $(A \times A) \times ( A \times A)$. In other words, $((a,b),(a,a)) $ belongs to $((A)^2)^2$ but it does not belong to $A^{2^2}$. And conversely, (a,b,a,a) belongs to $A^{2^2}$ but not $((A)^2)^2$. – Chris Nov 27 '22 at 19:15
  • But they are canonically isomorphic as sets by sending $((w,x),(y,z))$ to $(w,x,y,z)$. I don't think there is someplace in mathematics that needs particular notion to distinguish them. – Zerox Nov 27 '22 at 19:19
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    Hmmm... $A\times (A \times A) \times A$ is also canonically isomorphic to $A \times A \times A \times A$. But if I give you $(a,b,b,a)$, then by itself you don't know whether this should be mapped to $(a,(b,b),a)$ or to $((a,b),(b,a))$. In other words, I'm trying to make the argument that $(a,(b,b),a)$ contains more information than $(a,b,b,a)$. That is somehow my intuition (but I am only a physicist). – Chris Nov 27 '22 at 19:29
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    "I don't think there is a place in mathematics that needs to ... distinguish them." In my work, I believe I do need to make this distinction, or at least it makes the work more efficient. However, I will think more carefully about whether I really must make this distinction.

     – Chris Nov 27 '22 at 19:36
  • I think you can define the replacement operator $c$ recursively, too (see my answer). Clarify the meaning of "contains an $a$". In $( (((b,b),(b,b)), ((a,b),(b,b))) , \text{second element} )$ what do you replace the first element by? – Ethan Bolker Nov 28 '22 at 18:55

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Perhaps a straightforward recursive definition for a new operation (call it $!$ here) will work.

Define $$ A^{!1} = A $$ and for $n > 1$ $$ A^{!(n+1)} = A^{!n} \times A^{!n} $$

With this definition the order (in your sense) of $A^{!n}$ is simply $n$.

Replace $!$ by any symbol you like.

Ethan Bolker
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  • Thanks Ethan. I have a follow up question. I edited my original question and put a follow up question at the end. I was wondering if you could take a look? – Chris Dec 08 '22 at 17:59
  • @Chris I doubt that structure has a name. Invent one in your document. – Ethan Bolker Dec 08 '22 at 19:42