## Totally Ordered Relations

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.