Well ordering of the subsets of a given set

Question

For a given set, does there always exists a well-ordering of the set of all its subsets which is stronger than the usual ordering (that is set-theoretic inclusion) of the sets of the subsets of the given set?

Again (now with symbols instead of words): Let $U$ be a set. Does there necessarily exists a well ordering of $\mathscr{P} U$ which is stronger than $\subseteq$ order?

I expect that there is a counter-example but haven't found one yet.

We can assume axiom of choice.

Related question: http://math.stackexchange.com/q/271003/4876 — porton, Jul 24 '16 at 20:22
@GitGud I am about finding a well-ordering $\leq$ such that $A\subseteq B \Rightarrow A\leq B$ — porton, Jul 24 '16 at 20:41
I would say that makes $\leq$ weaker. But OK, now that's clarified. — Git Gud, Jul 24 '16 at 20:44
@GitGud: No, it’s stronger, in much the same sense that a finer topology is stronger than a coarser topology: it has all of the same ordered pairs and more besides. — Brian M. Scott, Jul 24 '16 at 20:52
Not only is there some disagreement about what "stronger" and "weaker" mean in the case of orderings, as Git Gud and Brian Scott have just shown, but I've also encountered disagreement on what those words mean for topologies. As a result, I try to use those words only in the context of statements (or theories), where, as far as I know, it is always the case that the stronger implies the weaker. — Andreas Blass, Jul 24 '16 at 21:33

score 2 · Answer 1 · answered Jul 24 '16 at 20:51

2

Consider the infinite descending chain of intervals $(-1/n,1+1/n)$ of real numbers. If $A\subseteq B\implies A\le B$, as in the comment, then this is an infinite descending chain in the supposed well-ordering, but has no smallest element. (If we wish, we can restrict to the rationals in the given intervals.)

answered Jul 24 '16 at 20:51

André Nicolas

514,336

I ask about the set of all subsets of a given fixed set, not about an arbitrary set of sets! – porton Jul 24 '16 at 20:52
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In the example, $U$ is the reals (or the rationals). The intervals are elements of the power set of $U$. – André Nicolas Jul 24 '16 at 20:55

score 2 · Accepted Answer · answered Jul 24 '16 at 20:54

No. For $n\in\Bbb N$ let $T_n=\{k\in\Bbb N:k\ge n\}$. Then

$$T_0\supsetneqq T_1\supsetneqq T_2\supsetneqq T_3\ldots$$

is an infinite descending chain in $\langle\wp(\Bbb N),\subseteq\rangle$ and would necessarily remain so in any extension of $\subseteq$. This will happen with any infinite set.

Well ordering of the subsets of a given set

2 Answers2