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Math Help - Real Analysis Proof

  1. #1
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    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...
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    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...
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  3. #3
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    Quote Originally Posted by bearej50 View Post
    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?
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  4. #4
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    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}.
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