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 upper 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.

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

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

Once we have our statement checked for the initial cases, we state our induction hypothesis:

For every finite set A of [tex]n[/tex] elements there exists a maximum and a minimum.

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

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

[tex]\max\{x_1,x_2,\ldots,x_n,x_{n+1}\}[/tex].

Notice that this problem is equivalent to solve

[tex]\max\{\max\{x_1,x_2,\ldots,x_n\},x_{n+1}\}[/tex],

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

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

[tex]x' =\max\{x_1,x_2,\ldots,x_n\}[/tex].

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

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