# Real Analysis Proof

• Feb 25th 2009, 04:10 PM
bearej50
Real Analysis Proof
Prove that any finite set has a maximum and a minimum.
I know that I have to prove this using mathematical induction...
• Feb 27th 2009, 05:28 AM
bearej50
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...
• Feb 27th 2009, 05:47 AM
Plato
Quote:

Originally Posted by bearej50
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 $\lambda = \sup (A)\;\& \;\lambda \notin A$ then $A$ is infinite.
Proof: $\lambda - 1$ is not an upper bound for $A$.
So $\left( {\exists a_1 \in A} \right)\left[ {\lambda - 1 < a_1 < \lambda } \right]\;\& \;\left( {\exists a_2 \in A} \right)\left[ {a_1 < a_2 < \lambda } \right]$.
Moreover, $n \geqslant 3\; \Rightarrow \;\left( {\exists a_n \in A} \right)\left[ {a_{n - 1} < a_n < \lambda } \right]$.

Now, in your proof you know there is least upper bound.
What would happen if it did not belong to the set?
• Feb 27th 2009, 06:47 AM
HallsofIvy
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 $\alpha$ and $\beta$. Then the max of S is max{ $\alpha$, a} and the min of S is min $\beta$, b}.