Well order of power set of an empty set

Question

For an empty set $A$, can we infer that $\left \langle \mathcal{P}(A),\subseteq \right \rangle$ is a well order? In the exact definition of well-order, it says to choose non-empty subsets and find whether that contains a minimal element or not. How about in this case?

Answer 1

If $A$ is empty, then $\mathcal{P}(A)=\{\varnothing\}$ has exactly one element. The order you have is the unique order on the one element set, $\{(\varnothing,\varnothing)\}$, which is certainly a well order: every nonempty subset (namely, $\{\varnothing\}$) has a least element (namely, $\varnothing$).

It is also true that the empty set, together with the empty order, is a well-ordered set, since it satisfies the condition by vacuity.

---

Note: if $(P,\leq)$ is a partially ordered set, then an element $x\in P$ is minimal if and only if for all $y\in P$, $y\leq x\implies y=x$; that is, there is no element of $P$ strictly smaller than $x$. An element $x\in P$ is a minimum (or a least element) if $x\leq y$ for all $y\in P$. Different concepts.

A well ordered set is a (partially) ordered set in which every nonempty subset has a least element, not “a minimal element”. If the definition were “a minimal element”, then every finite partially ordered set would be well-ordered, even if it wasn’t totally ordered.