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.
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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.
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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.
Contradiction. So
N is unbounded.
There is a name for what we have proved:
Archimedean axiom Given any real number x , there exists a natural number n such that n > x .
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