Show that $3 \!\uparrow\uparrow\! 2$ is equal to the number of sets of rank at most $2$ with a $3$-valued notion of set membership

Question

This question is with regard to the answer of an older post. Please see the original post's answer for context.

The following answer was given to a question about what the tetration $k \!\uparrow\uparrow\! n$ is actually counting, where

$$‎k \!\uparrow\uparrow\! n := \underbrace{k^{k^{k^{.^{.^{.}}}}}}_{n - times}.$$

$k \!\uparrow\uparrow\! n$ is equal to the number of sets of rank at most $n$ when we adopt a $k$-valued notion of set membership.

For example, this view still gives us one set of rank $0$, namely the empty set $\{\}$, but now we have $k$ sets of rank $0$ or $1$, namely the sets containing only $\{\}$, with value between $0$ and $k-1$ (of course, the set containing $\{\}$ with value $0$ is identified with $\{\}$, so that we only get $k-1$ new sets of rank $1$, for a total of $k$ with rank $0$ or $1$). It is easy to check that we get $k^k$ "sets" of rank at most $2$, $k^{k^k}$ of rank at most $3$, and $k \!\uparrow\uparrow\! n$ sets of rank at most $n$.

My problem:

It is easy to write out the sets for $3 \!\uparrow\uparrow\! 0$ and $3 \!\uparrow\uparrow\! 1$, as they are simply

\begin{align} 3 \!\uparrow\uparrow\! 0 \ \ &: \ \ \{ \} \\ 3 \!\uparrow\uparrow\! 1 \ \ &: \ \ \{ \}, \big\{ \{ \}_{1}\big\}, \big\{ \{ \}_{2}\big\} \end{align} where each subscript indicates the intrinsic value of the member; however, I struggle to follow the described pattern for the 27 members that should be associated with $‎3 \!\uparrow\uparrow\! 2$.

Points will be awarded to the first answer that can correctly list all 27 members of $3 \!\uparrow\uparrow\! 2$, as described in the OP.

Answer 1

A set has rank at most $n$ iff each of its members has rank at most $n-1$. Therefore, each set of rank at most $n$ corresponds to a choice of one of the $k$ membership values for each set of rank at most $n-1$. This means our counting function $f$ satisfies $f(n, k)=k^{f(n-1,k)}$, from which we can conclude that $f(n,k)=k\!\uparrow\uparrow\!n$.

If your three sets of rank at most $1$ are denoted $A$, $B$, and $C$, then the $3^3$ sets of rank at most $2$ are:

$\{A_0,B_0,C_0\},\{A_0,B_0,C_1\},\{A_0,B_0,C_2\},$

$\{A_0,B_1,C_0\},\{A_0,B_1,C_1\},\{A_0,B_1,C_2\},$

$\{A_0,B_2,C_0\},\{A_0,B_2,C_1\},\{A_0,B_2,C_2\},$

$\{A_1,B_0,C_0\},\{A_1,B_0,C_1\},\{A_1,B_0,C_2\},$

...

$\{A_2,B_2,C_0\},\{A_2,B_2,C_1\},\{A_2,B_2,C_2\}$

where subscript $0$ denotes non-membership, so e.g. $\{A_0,B_0,C_0\}=\{\}$.