Condition for Power Set to be Totally Ordered

Theorem
Let $\powerset S$ be the power set of a set $S$.

Let $\struct {\powerset S, \subseteq}$ be the set $\powerset S$ ordered by $\subseteq$.

Then $\struct {\powerset S, \subseteq}$ is totally ordered $S$ is either the empty set or a singleton.

Proof
From Subset Relation on Power Set is Partial Ordering we have that $\struct {\powerset S, \subseteq}$ is an ordered set.

We now need to show that $\struct {\powerset S, \subseteq}$ is a totally ordered set exactly when $S = \O$ or $S$ has exactly one element.

When $S = \O$ then $\powerset S = \set \O$ and it follows trivially that $\struct {\powerset S, \subseteq}$ is totally ordered.

Let $S = \set x$.

Then $\powerset S = \set {\O, \set x}$.

From Empty Set is Subset of All Sets we have $\O \subseteq \set x$.

Again it follows by definition that $\struct {\powerset S, \subseteq}$ is totally ordered.

Now suppose $S$ is neither the empty set nor a singleton.

Then:
 * $\exists x, y \in S: x \ne y$

and so:
 * $\exists \set x, \set y \in \powerset S: \set x \ne \set y$

But further, $\set x \nsubseteq \set y$ and $\set y \nsubseteq \set x$.

That is, $\set x$ and $\set y$ are non-comparable by $\subseteq$.

Thus, by definition, $\struct {\powerset S, \subseteq}$ is not totally ordered.

Hence the result.