Prove that any finite set has a maximum and a minimum.
I know that I have to prove this using mathematical induction...
I mean a finite set of real numbers.
I can use the field axioms, order axioms, absolute value, the completeness axiom, archimedean property, and density of the rationals to prove this.
I know I need to know that such a finite set of real numbers must be bounded above. I know I need to use the completeness axiom so that the set has a least upper bound. I think that I need to show that the LUB belongs to the set??? I just cant pull it all together...
Here is a completely different approach.
Lemma: If $\displaystyle \lambda = \sup (A)\;\& \;\lambda \notin A$ then $\displaystyle A$ is infinite.
Proof: $\displaystyle \lambda - 1$ is not an upper bound for $\displaystyle A$.
So $\displaystyle \left( {\exists a_1 \in A} \right)\left[ {\lambda - 1 < a_1 < \lambda } \right]\;\& \;\left( {\exists a_2 \in A} \right)\left[ {a_1 < a_2 < \lambda } \right]$.
Moreover, $\displaystyle n \geqslant 3\; \Rightarrow \;\left( {\exists a_n \in A} \right)\left[ {a_{n - 1} < a_n < \lambda } \right]$.
Now, in your proof you know there is least upper bound.
What would happen if it did not belong to the set?
HOW do you know there is a lub? Wouldn't you have to prove that a finite set has an upper bound first? I don't see that you need to use completeness on a finite set. Use induction on the size of the set:
Given two numbers, a, b, it is simple to determine which is larger and which is smaller. That is min(a,b) and max(a,b) always exists.
If S is contains a single number then that number IS both the max and min.
Assume that every set with k members has both a max and min and let S be a set with k+1 points. Let a be any member of S. Then S\{a} contains n-1 points and so has max $\displaystyle \alpha$ and $\displaystyle \beta$. Then the max of S is max{$\displaystyle \alpha$, a} and the min of S is min$\displaystyle \beta$, b}.