Consider the set $\mathcal{P}(\mathbb{R})$ of all subsets of $\mathbb{R}$, the set of real numbers. It has a natural partial order: $A \leq B$ iff $A \subseteq B$.


Can one extend this order to a total order?


(I was discussing this with a friend and we didn't know if this is possible. If we replace $\mathbb{R}$ by any finite set, this is possible. We were not sure even when we replace $\mathbb{R}$ by $\mathbb{N}$.)



Extensions of the partial order of the power set
Michael Greinecker's answer (in a comment) is correct in the context of the usual axioms of set theory, including the axiom of choice. If one works in set theory without the axiom of choice, then one cannot prove that $\mathcal P(\mathbb R)$ admits any total order at all (even without the requirement that it extend $\subseteq$). On the other hand, the $\subseteq$ ordering of $\mathcal P(\mathbb N)$ can be explicitly extended to a total ordering, given by lexicogaphically ordering the characteristic functions of subsets of $\mathbb N$.


