Why we need precise language for extremes
Once a set is ordered, the interesting questions usually happen near its extremes. Does the set have a largest element? If not, is there still a smallest number sitting above all of it? Is there a best lower barrier from the left?
The key chain of ideas is:
- maximum and minimum talk about elements that lie inside the set;
- upper and lower bounds may lie outside the set;
- supremum and infimum identify the best possible such bounds.
If you mix these ideas together, later arguments about completeness become hard to read. So this note separates them carefully.
Maximum and minimum versus upper and lower bounds
Definition
Maximum, minimum, upper bound, and lower bound
Let be a subset of an ordered set .
- A maximum of is an element such that for every .
- A minimum of is an element such that for every .
- An upper bound of is an element such that for every .
- A lower bound of is an element such that for every .
The difference is subtle but essential: a maximum or minimum must belong to the set, while an upper or lower bound only has to belong to the ambient ordered set.
Worked example
A finite set in Z
Let .
- The maximum is
3. - The minimum is
1. - Every integer is an upper bound.
- Every integer is a lower bound.
So the set has many upper and lower bounds, but only one maximum and one minimum.
Worked example
A set can have bounds without extremal elements
Consider as a subset of .
The number 0 is a lower bound of , but it is not an element of the
set. In fact, has no minimum, because for every positive rational
q, the smaller rational q/2 is still positive.
This example is a warning: lower bound and minimum are not interchangeable.
Supremum and infimum
Some upper bounds are better than others. The supremum is the smallest upper bound; the infimum is the largest lower bound.
Definition
Supremum and infimum
Let be a nonempty subset of an ordered set .
- A number is the supremum of , written , if
sis an upper bound of and every upper bounduof satisfies . - A number is the infimum of , written , if
tis a lower bound of and every lower bound of satisfies .
The ambient set matters. The same subset can have a supremum in but fail to have one in .
Worked example
The open interval (0,1)
Take .
- has no maximum, because
1is not in the set and every element is still smaller than some larger element of . - .
- has no minimum.
- .
So both extremal bounds exist even though neither belongs to the set.
Why supremum and infimum are unique
Theorem
A subset has at most one supremum and at most one infimum
If has a supremum, then that supremum is unique. The same holds for infimum.
The reason is short but important. Suppose s and s' are both suprema of
. Since s is an upper bound and s' is the least upper bound, we get
. Reversing the roles gives . By antisymmetry, .
Exactly the same argument works for infimum.
The approximation viewpoint
The definition of supremum can be turned into a more usable test.
if and only if
sis an upper bound of and for every there exists such that .
This says that a supremum is not just an upper wall; points of the set can come arbitrarily close to it from below.
Proof
Why the approximation property is equivalent
Common mistakes
Common mistake
Maximum is stronger than supremum
A maximum must belong to the set. A supremum only has to be the least upper
bound in the ambient ordered set. The open interval (0,1) has supremum 1
but no maximum.
Common mistake
The ambient ordered set matters
When you write , you are working inside some ordered set. A subset of can have a supremum in without having one in . Later, this is exactly how we see that is not complete.
Quick checks
Quick check
For Y=(0,1), does Y have a maximum? What are sup(Y) and inf(Y)?
Keep the inside/outside distinction clear.
Solution
Answer
Quick check
If a set A has a maximum m, what is sup(A)?
Compare the definitions directly.
Solution
Answer
Exercises
Quick check
Let A={1-1/n : n in N}. Find sup(A), inf(A), and decide whether A has a maximum.
List the first few terms and look at the trend.
Solution
Guided solution
Quick check
Show that if B is nonempty and bounded below, then inf(B) = -sup(-B).
Translate lower-bound statements for B into upper-bound statements for -B.
Solution
Guided solution
Related notes
Read this after 4.1 Total orders and ordered fields and continue to 4.3 Completeness and gaps in Q.