Let $\mathcal{P}(A)$ be the power set of $A$. Let $P_1\simeq P_2$ denote that posets $P_1, P_2$ are isomorphic. Let $D_{x}, x \in \mathbb{Z}$, be the set of divisors of $x$. I was given the following problem.
Decide whether $(D_{30}, \mid) \simeq (\mathcal{P}(\{a, b, c\}), \subseteq)$.
The Hasse diagrams of both posets show they are clearly isomorphic, which is enough to decide that the statement above is true. But I want to provide a proof of this fact, independent of the Hasse diagrams.
It seems to me I am to find a function $f : D_{30} \to \mathcal{P}(\{a, b, c\})$ such that $a | b \iff f(a) \subseteq f(b)$. This function is easy to construct without generality; this is, by providing the case-wise specification
$$ f(x) = \begin{cases} \{\} & x = 1 \\ \{a\} & x = 2 \\ \{b\} & x = 3 \\ \{c\} & x = 5 \\ \{a, b\} & x = 6 \\ ~~~~\vdots~~~~~~\vdots &~~~\vdots \end{cases} $$
Although such specification of $f$ is an isomorphism, I wonder whether there is a general function $f(a)$ that establishes the isomorphism by making use of some property inherent to both structures. At the end of the day, the fact that the posets are isomorphic means they share the same structure, so there must be an abstract, inherent similarity between the two that relates each $a \in D_{30}$ to each $S \in \mathcal{P}(\{a, b, c\})$.
Thanks in advance.