# Inferring a particular set is finite

I was given in my homework the following question:

Suppose that for some set A, the well ordered set P(A) (under inclusion) has the property that every non-empty subset of it has a minimal element. I was asked to show that A is finite.

I'd be happy to get clues/hints in this manner

• Saying that "$P(A)$ has the property that every non-empty subset of it has a minimal element" is the same as saying that $P(A)$ is well-ordered (as a partial order). So you could more simply say "Suppose that for some set $A$, the partially ordered set $P(A)$ (under inclusion) is well-ordered." – Alex Kruckman Nov 18 '15 at 22:44
• @AlexKruckman In the usual terminology, a well-ordered set must be totally ordered. If every nonempty subset of $P$ has a least element then $P$ is well-ordered. A partially ordered set in which every nonempty subset has a minimal element is called well-founded not well-ordered. If $A$ has more than one element, then the power set $P(A)$ is not well-ordered. – bof Nov 18 '15 at 22:49
• @bof Sure, that's why I wrote "well-ordered (as a partial order)". But you're right, well-founded is a better choice. In any case, the term "well-ordered" is out of place where it appears in the OP's question, which is the point I was trying to get across. – Alex Kruckman Nov 18 '15 at 22:52
• @AlexKruckman Oops, I didn't even see "well ordered" in the original post. – bof Nov 18 '15 at 23:16
• @AlexKruckman "well-founded" not well-ordered. – BrianO Nov 18 '15 at 23:22

If we assume for a contradiction that $A$ is infinite, then the set of all infinite subsets of $A$ is nonempty, and therefore has a minimal element. That is, there is an infinite set $B\subseteq A$ such that every proper subset of $B$ is finite. Can you get a contradiction from that?