Suppose that $S$ is a well-ordered set; how can we prove that the following is a total order on the power set of $S$? $$A\prec B\Longleftrightarrow \min(A\triangle B)\in A.$$
powerset total order
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elementary-set-theory
1 Answers
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To prove that a relation $<$ is a total order you need to show:
- Given $A,B$ exactly one of the $A
- $A
The first one: Take two sets $A,B\in P(S)$. Then if $A\neq B$ their symmetric difference $A\triangle B\neq\varnothing$. Thus it has a least element (since $S$ is well ordered). If this element is a member of $A$ then $A
The second one: Let $A