Are there any sets A for which $\displaystyle (\mathcal{P}(A), \subseteq)$ is totally ordered? Prove your answer.

To be courteous, I will include the definitions for partial ordering and total ordering.

A relation is a partial order if the relation is reflexive, antisymmetric, and transitive. (in this case, the notation $\displaystyle (\mathcal{P}(A), \subseteq)$ just denotes that $\displaystyle \mathcal{P}(A) $ under the relation $\displaystyle \subseteq$ is a partially ordered set.

A partially ordered set A with partial order $\displaystyle \leq $ is said to be totally ordered if given any two elements a and b in A, either $\displaystyle a \leq b $ or $\displaystyle b \leq a $.

So, to attempt this problem. I tried making up examples first. The only set A that I could come up with for which $\displaystyle (\mathcal{P}(A),\subseteq)$ is totally ordered is a set with one element. Just to see this, let A = {1}. In this case, $\displaystyle \mathcal{P}(A) = ${$\displaystyle \emptyset$ , {1}}. So, if I pick any two elements, a and b. $\displaystyle a \leq b $ or $\displaystyle b \leq a $. For example, if I'd pick $\displaystyle \emptyset $ and {1}. Then, $\displaystyle \emptyset \subseteq ${1}. So I think it works. I don't know if there are any other sets, A, where this works or if I'm even thinking about this correctly. (once I figure out all the sets, I'll attempt to put a proof up). Thanks in advance.