Re: minimal element iff maximal element: how to prove?

On 27 Nov 2005 21:14:22 -0800, MoeBlee <jazzmobe@xxxxxxxxxxx> said:
> I'm stumped by an exercise in Suppes's 'Aximomatic Set Theory' (page
> 102, Dover). It's a crucial theorem for his treatment, so I am ardent
> to have a proof of it.
>
> Here's my formulation of it:
>
> Df: x is a minimal element of S iff x is a member of S and no member of
> S is a proper subset of x.
>
> Df: x is a maximal element of S iff x is a member of S and x is not a
> proper subset of any member of y.
>
> Theorem: Every non-empty set of subsets of B has a minimal element iff
> every non-empty set of subsets of B has a maximal element.
>
> I've tried everything I know, hoping to derive just the first direction
> (minimal then maximal).

Suppose every nonempty set of subsets of B has a minimal element. Let S
be a nonempty set of subsets of B. Consider the set S* = {B - s | s in
S}. By hypothesis, S* has a minimal element a. By definition of S*,
a = B - z, for some (unique) z in S. Claim: z is maximal in S.

.

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