Are there any sets A for which (\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 (\mathcal{P}(A), \subseteq) just denotes that \mathcal{P}(A) under the relation \subseteq is a partially ordered set.

A partially ordered set A with partial order \leq is said to be totally ordered if given any two elements a and b in A, either a \leq b or 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 (\mathcal{P}(A),\subseteq) is totally ordered is a set with one element. Just to see this, let A = {1}. In this case, \mathcal{P}(A) = { \emptyset , {1}}. So, if I pick any two elements, a and b. a \leq b or b \leq a . For example, if I'd pick \emptyset and {1}. Then, \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.