Atom of Lattice/Examples/Singleton in Lattice of Sets

Examples of Atoms of a Lattice
Let $\powerset S$ denote the power set of a set $S$.

Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the subset relation $\subseteq$.

From Power Set is Lattice, $\struct {\powerset S, \subseteq}$ is a lattice.

The atoms of $\struct {\powerset S, \subseteq}$ are the singleton subsets of $S$.

Proof
From Empty Set is Bottom of Lattice of Power Set, the bottom of $\struct {\powerset S, \subseteq}$ is the empty set $\O$.

Let $A$ be an atom of $\struct {\powerset S, \subseteq}$.

Recalling the definition of atom.
 * $\forall B \in S: B \subsetneqq A \implies B = \O$

We note that $\O$ itself is not an atom of $\struct {\powerset S, \subseteq}$ by definition.

Let $T \subseteq S$ be a subset of $S$ with strictly more than $1$ element.

Let $a, b \in T$.

Then
 * $\set a \subsetneqq T$

but:
 * $\set a \ne \O$

Hence for such a $T$, there exists a proper subset of $T$ which is not an atom of $\struct {\powerset S, \subseteq}$.

Hence a subset of $S$ with more than $1$ element is not an atom of $\struct {\powerset S, \subseteq}$.

Let $a \in S$ be arbitrary.

Then we have:


 * $\set a \in \powerset S$

that is:
 * $\set a \subseteq S$

Let $B \subseteq \set a$ such that $B \ne \set a$.

Let $b \in B$ such that $b \ne a$.

Then:
 * $b \notin \set a$

and so:
 * $B \not \subsetneqq A$

Hence if $B \subsetneqq A$ it follows that:
 * $B = \O$

Hence, by definition, $\set a$ is an atom of $\struct {\powerset S, \subseteq}$.