# Consider the set S = {1/n : n ∈ N}. Does this set have a minimum Consider the set S = { 1 /n : n ∈ minimum?

N

} .

Does this set have a

Definition Given a subset S ⊂

R

, and element x ∈ S is said to be the minimum of S if for all y ∈ S , x ≤ y .

For a contradiction, assume that 1 /N ∈ S is the minimum. Then choose M > N (we need to know 1 > 0 for this). Then 1 /M ∈ S , but 1 /M < 1 /N (we need to prove this).

The set S is an example of a set that does not have a minimum.

It wants to have 0 as a “minimum”, but 0 is not in the set. We are going to define a concept of “greatest lower bound” that captures this number 0 in this case.

1

Consider the set { 1 /n : n ∈

N

} .

Lower bounds: 0, − 2, any negative number.

Upper bounds: 1, 1729

Maximum: 1 (because it is an element of the set and it is an upper bound)

Minimum: there isn’t one

Least upper bound: 1

Greatest lower bound: 0

How would you prove any of these statements?

2

Same questions for

N

.

Lower bound: 1

Upper bound: there isn’t one.

But how do we prove this?

It is tempting to do something like this: Suppose x is an upper bound. Then x ≥ n for all n ∈

N

. So take the integer part of x and add 1. The integer part is the greatest integer less than or equal to x . But there is a problem with this, because we have assumed x ≥ n for all n ∈

N

.

In fact, one cannot prove that the natural numbers are unbounded with the axioms we have so far.

3

Definition The least upper bound of a set S ⊂

R is a number x which satisfies two conditions

1. First, x is a upper bound for the set. That is, for all y ∈ S , we have x ≥ y .

2. If x

0 is any other upper bound, then x ≤ x

0

.

Completeness axiom

If S ⊂

R which is nonempty and has an upper bound, then S has a least upper bound.

4

Proof that

N is unbounded

Now suppose that

N is bounded above. Then

N has a least upper bound, m . Assume that m is not a natural number. Then one can find a smaller upper bound for

N by subtracting some small number from m (this needs some work to prove). Therefore m is a natural number. But then m + 1 is also a natural number.