Re: Constructibility of X -> X^2 bijection

On Tue, 31 Jul 2007 08:52:58 -0400, "Stephen J. Herschkorn"
<sjherschko@xxxxxxxxxxxx> wrote:

Herman Jurjus wrote:

Herman Rubin wrote:
[snip]

It is trivial to order the power set of a well-ordered set


Would you care to explain what you have in mind, here?


I am pretty sure that Herman R. refers to the following: Given a
well-ordered set X, order F = 2^X as follows. For f, g in F, let
x be least such that f(x) != g(x). Then f < g iff f(x) < g(x).

Without referring to 2^X, this becomes: Let A and B be subsets of
X. A < B iff and A contains the smallest element in the symmetric
difference of A and B. Exercise: Show this is a well-ordering of
P(X) without reference to 2^X.

Did you work out a solution to this "exercise" before posting it?


************************

David C. Ullrich
.