Verify Proof that All Finite sets in order Field of Real numbers has a maximum and a minimum.

Question

I use the following facts:

1. The completeness axiom.

2. If supreme of a set an element the said set, then it is the maximum of the set.

Suppose S is finite non-empty set of real number. By the completeness axiom we are granted a supremum

exists for set. Call it is M. Assume as to acquire a contradiction that M is not in set S. Now take

any element s_1 in S. We must have s_1 < M, and hence there is s_2 in S such that s_1 < s_2

[note s_2 is not equal to M because of our assumption]

We may repeat the same argument over and over and generate infinitely many elements of S, which is

absurd because it is a finite set.

A similar argument should work for the case of a minimum.