Suppose that A is a totally ordered set. Use mathematical induction to prove that for any integer n ≥ 1 , every subset of A with n elements has both a least element and a greatest element.

To Prove:

For any integer n≥1 every subset of A with n elements has both a least element and a greatest element, using mathematical induction.

Given information:

A is totally order set.

Concept used:

Mathematical induction

Proof:

Let A be a set that is totally ordered with respect to a relation ≤. And let the property P(n) be the sentence “Every subset of A with n elements has both a least element and a greatest element”

Show that the property is true for n=1.

If A=ϕ then the property is true by default.

So assume that A has at least one element and suppose S={a1} is a subset of A with one element

Because ≤ is reflexive a1≤a1.

By the definition of least element and greatest element a1 ¡s both the least and greatest element of S.

Thus, the property is true for n=1.

Show that for all integers k≥1. If the property is true for n=k then it is true for n=k+1.

Let k be any integer with k≥1 and suppose that any subset of A with k elements has both a least element and a greatest element.

We must show that any subset of A with k+1 elements has both a least element and a greatest element. If A has fewer than k+1 elements, then the statement is true by default