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Sergeu [11.5K]
3 years ago
9

Let A be a finite, non-empty subset of R. Prove that A has a maximum and a minimum. (Recall that a maximum of a set A is an uppe

r bound for the set that belongs to A, and a minimum of a set is a lower bound for the set that belongs to the set.) Hint: This result may seem so obvious that it isn't clear h One way is to use induction on the number of elements of the set.
Mathematics
1 answer:
julia-pushkina [17]3 years ago
5 0

Answer:

Let us use mathematical induction to prove the statement. So, we are going to start checking the statement for the first natural numbers.

n=1: Our set is \{x_1\}. So, obviously, x_1 is the maximum and minimum of our set. Then, the statement is true for n=1.

n=2: Our set is \{x_1,x_2\}. Necessarily, x_1 or x_1>x_2. In both cases, there is a minimum and a maximum.

Once we have our statement checked for the initial cases, we state our <em>induction hypothesis</em>:

For every finite set A of n elements there exists a maximum and a minimum.

Now, let us prove the that the above assertion is true for sets with n+1 elements.

Our set is A=\{x_1,x_2,\ldots,x_n,x_{n+1}\} and we want to find

\max\{x_1,x_2,\ldots,x_n,x_{n+1}\}.

Notice that this problem is equivalent to solve

\max\{\max\{x_1,x_2,\ldots,x_n\},x_{n+1}\},

i.e, to find the maximum among n+1 numbers, we can find first the miximum among n and then compare with the other one.

Now, using our induction hypothesis we know that there is a maximum in the set \{x_1,x_2,\ldots,x_n\}, because it has n elements. Let us write

x' =\max\{x_1,x_2,\ldots,x_n\}.

So, in order to find the maximum of A, we have to find the maximum of \A'={x',x_{n+1}\}. As we have checked at the beginning, there is a maximum in A', and it is the maximum of A.

Hence, we have completed the prove for the existence of the maximum of a set with n+1 elements. The prove for the existence of the minimum is analogue, we just need to change ‘‘maximum’’ for ‘‘minimum’’.

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