Prove that every Non-Empty set of Positive Integers contains a "Least Element".

(In reply to re: {P, r, o, o, f} by friedlinguini)

I can find a counterexample. {2,2,3} doesn't have a least element, it has two equally least elements.

If you allow equally least elements, I can use the law of indirect reasoning.

(Unless all elements are the same, in which all of them are equally least) Assume there wasn't a least element (or least elements). This means that each number has a number in that set that it is greater than. Eventually you will create a "circle of greater thans", and this can be proven as a contradiction by the transitive law of inequality. So this means there must be a least element (or least elements)

Is that a good solution?

Posted by Gamer on 2003-05-08 12:09:07