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Subsets of $\Bbb N$ are well ordered. So two subsets of $\Bbb N$, A and B could be compared by comparing the least elements in A\B and B\A; whichever has the lesser is the lower. This relation appears also transitive.

It looks to me that this "ordering" will find a "least" in any set of subsets of $\Bbb N$.

But this would produce an effective well ordering of the continuum, which is impossible. So the proposed "order" is either non-transitive or it will fail to find the "least" in some set of subsets of $\Bbb N$.

Is there immediately obvious where the mistake (if any)?

PS The proposed relation is not even an ordering, as detailed below. Although the counter example from the other post will work here as well.

Dacian Bonta
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2 Answers2

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OK, so if I understand you correctly, you say that $A < B$ if the least element in $A-B$ is smaller than the least element in $B-A$

First, a question: what if $A \subseteq B$?

then $A -B = \emptyset$ ... What would be the 'least element' of that?

It might be appealing to say that $A<B$ whenever $A\subseteq B$, but then you have a problem with transitivity:

$A=\{ 1,2\}$

$B=\{ 0,1,2\}$

$C=\{0,3\}$

then

$A \subseteq B$ so $A<B$

$B-C=\{1,2\}$ => least element is 1

$C-B=\{3\}$=> least element is 3

So, $B<C$

$A-C=\{1,2\}$=> least element 1

$C-A=\{0,3\}$=> least element 0

So, $C <A$

In sum, there is a 'hole' in your definition, and the 'obvious' way to plug it does not work.

Bram28
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  • I'm thinking that if the larger set extends "lower" than the smaller/included set, transitivity could be saved by making the larger set the lower one. Maybe, have to think about it. It would fix your counterexample – Dacian Bonta Mar 30 '17 at 09:35
  • @DacianBonta OK, so you propose that if $A \subseteq B$ then say $B < A$. Notice that this would make $\mathbb{N}$ the smallest element of the set of all subsets of natural numbers ... which is ok ... but what would be the least element when you take out $\mathbb{N}$? With the earlier definition (and where if $A \subseteq B$ then $A < B$, we had that $\emptyset$ is the least element, and ${ 0 }$ the least elements of all other subsets ... but what is the 'next' least element if the least element is $\mathbb{N}$? So this new approach may fix the order issue, but it may not be well-ordered. – Bram28 Mar 30 '17 at 16:30
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This is not well ordered: let $X_i $ be all numbers $<i $. As $i $ goes up, $X_i $ goes down.

Another useful counterexample to such attempts, though it doesn't work here, is to replace "$< $" with "$> $" in the above example.

Noah Schweber
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