Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's say I take a set $S$, where $S$ can be well ordered. From what I understand, one can use that well ordering to totally order $\mathscr{P}(S)$.

How does a body actually use the well ordering of $S$ to construct a total ordering of $\mathscr{P}(S)$?

Perhaps construct is too strong a word. Is there a way to prove that a total ordering of $\mathscr{P}(S)$ just exists, no need to explicitly state what it is, since that might be very difficult.

share|improve this question

1 Answer 1

up vote 12 down vote accepted

If $\lt$ is a well-ordering of $S$, then we may define the lexical ordering on the power set $P(S)$, by which $A\lt_{lex} B$, if and only if the $\lt$-least element of the symmetric difference $A\triangle B$ is in $B$ and not in $A$. That is, we look to the first place where the sets differ, and then put the set without this element before the set with this element. This is just like the order in a dictionary, hence the name, since two words in a dictionary are put in order by comparing the first letter on which they differ.

To see that this is a total order, we check first that it is transitive. If $A\lt_{lex}B\lt_{lex} C$, then the first difference between $A$ and $C$ must be in $C$, since either this occurs before the first difference between $A$ and $B$, in which case it is in $C$ and not in $B$ and hence in $C$ and not in $A$, or else it occurs at or after the first difference between $A$ and $B$, in which case it occurs exactly at this difference, which is in $C$ and not in $A$. The order it linear since any two sets do indeed have a least difference, since $\lt$ is a well-order.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.