# Notation for Nested Set

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.

• What is the difference of your sequence and $\displaystyle{A^{2^n}}$? Commented Nov 27, 2022 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 $((A)^2)^2$. Commented Nov 27, 2022 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. Commented Nov 27, 2022 at 19:19
• 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). Commented Nov 27, 2022 at 19:29
• >"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. Commented Nov 27, 2022 at 19:36

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.