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

Consider the set S = {1/n : n ∈ N}.
minimum?
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?
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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.
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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 x0 is any other upper bound, then x ≤ x0.
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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