Analogy between prime numbers and singleton sets? While trying -- in vain -- to write an alternative answer for another question (If $\cup \mathcal{F}=A$ then $A \in \mathcal{F}$. Prove that $A$ has exactly one element.), I discovered the following property for sets: $$A \textrm{ is a singleton set} \;\equiv\; \langle \forall B : B \subseteq A : B = \emptyset \;\not\equiv\; B = A \rangle$$ and I noticed the similarity to the following definition for positive whole numbers: $$n \textrm{ is prime} \;\equiv\; \langle \forall d : d \textrm{ divides } n : d = 1 \;\not\equiv\; d = n\rangle$$  So it looks like singleton sets act a bit like prime numbers.  Which is not strange, come to think of it, given that both are 'indivisible atoms' of some sort.
So what common theory underlies both concepts of indivisibility?
 A: We can unify the two definitions by passing to their common generalisation in order theory. To begin, note that the set $\mathscr{P} (X)$ of all subsets of a fixed set $X$ is partially ordered by inclusion, and the set $P$ of all positive integers can be partially ordered by divisibility, and that both $\mathscr{P} (X)$ and $P$ has a bottom element ($\emptyset$ and $1$, respectively).
Moreover, both $\mathscr{P} (X)$ and $P$ have the property that, for any two elements $a$ and $b$, there exists a unique element $a \vee b$ such that, for all $c$ with $a \le c$ and $b \le c$, $a \vee b \le c$. In the case of $\mathscr{P} (X)$, this is the operation of taking the union of two subsets, and in the case of $P$, this is the operation of taking the l.c.m. of two numbers. A partially ordered set with a bottom element $\bot$ and this binary operation $\vee$ is said to be a join semilattice.
Definition. An indecomposable element of a join semilattice $L$ is an element $c$ such that, if $c = a \vee b$, then either $a = \bot$ or $a = c$ (but not both!).
As you have observed, the indecomposable elements of $\mathscr{P} (X)$ are precisely the singleton subsets, and the indecomposable elements of $P$ are precisely the prime numbers.
A: Let $P$ be the set of all prime numbers, and let $N$ be the collection of all multisets whose elements are in $P$.  Notice the fundamental theorem of arithmetic gives a bijection $N\leftrightarrow \mathbb{N}$, where $1$ corresponds to the empty multiset.
Under this bijection, the notion of divisibility of natural numbers corresponds to the notion of containment of sets, and the prime numbers correspond to singletons in $N$.  Also, notice that $\varnothing$ is contained in all elements of $N$, just as $1$ divides all natural numbers.
