Typo?

In ZF, any finite set with n elements can be counted in n! ways.

Therefore it can be totally ordered in n! ways.

Not sure. But the real point here is that if the paper you're reviewing makes the statement but does not substantiate it with either a short explanation or a citation that serves then the paper needs to be returned with the note that these things are required.

Dear Prof.Koch,

There is no explanation. But, they provide a reference which is not available to me.

Thank you very much for your suggestions.

Best regards,

BKTripathy

Typo?

In ZF, any finite set with n elements can be counted in n! ways.

Therefore it can be totally ordered in n! ways.

In Zermelo-Fraenkel set theory + the axiom of choice (ZFC), which is equivalent to ZF + Zorn's Lemma, it can even be proved that *every* set can be well-ordered, not only the finite sets. See Bell's article "The axiom of choice" in the Stanford Encyclopedia of Philosophy, Article The axiom of choice

For finite sets, using the axiom of choice would be like shooting a mosquito with a gun. As Mani A describes above, for finite sets of n elements, one can simply enumerate all the n! different total orderings.

Dear Prof. Hartonas,

Thanks for your suggestions.

regards,

B. K. Tripathy

You don't need Zorn's lemma or axiom of choice for ordering a finite set. A finite set by definition has a bijection with a finite cardinal which is also a finite ordinal which is naturally ordered.

Thanks Prof. Colin.

regards,

B. K. Tripathy

Let S be a finite set, say |S| = n. Then S is in bijection with the set {1,2,...,n}, say via a function f. Then the same bijection f determines an order on the set S, meaning x < y in S iff f(x) < f(y) in {1,2,...,n} (with the usual order). This is a simple way to obtain an explicit order of any finite set. As it was noticed by other RG users before, the Axion of Choice (or, equivalently, the Zorn's Lemma) is not necessary to argue that any finite set can be endowed with a total order. I hope this helps.

Dear Prof. Felix,

Thanks for your point. Still the ordering is dependent upon 'f' . The way one defines f, the ordering changes. Isn't it !!! I mean the ordering is not unique.

With regards,

B. K. Tripathy

Dear Dr. Tripathy,

You are right; the total order of S depends on the choice of the bijection f. There are a total of n! bijections from S to {1,2,...,n}, and each of them yields a different total order on S. Therefore we have a total of n! different total orders on S, as already pointed out by Dr. Mani A.

Thanks Dr. Felix.

regards,

B. K. Tripathy