Using supremum and infimum

To make using suprema and infima even easier, we want to be able to always write sup A and inf A
without worrying about A being bounded and nonempty. We make the following natural definitions
Definition 1.2.6. Let A  R be a set.
(i) If A is empty, then sup A := 􀀀¥.
(ii) If A is not bounded above, then sup A := ¥.
(iii) If A is empty, then inf A := ¥.
(iv) If A is not bounded below, then inf A := 􀀀¥.
For convenience, we will sometimes treat ¥ and 􀀀¥ as if they were numbers, except we will
not allow arbitrary arithmetic with them. We can make R := R[f􀀀¥;¥g into an ordered set by
letting
􀀀¥ < ¥ and 􀀀¥ < x and x < ¥ for all x 2 R:
The set R is called the set of extended real numbers. It is possible to define some arithmetic on R,
but we will refrain from doing so as it leads to easy mistakes because R will not be a field.
Now we can take suprema and infima without fear. Let us say a little bit more about them. First
we want to make sure that suprema and infima are compatible with algebraic operations. For a set
A  R and a number x define
x+A := fx+y 2 R j y 2 Ag;
xA := fxy 2 R j y 2 Ag:
Proposition 1.2.7. Let A  R.
(i) If x 2 R, then sup(x+A) = x+sup A.
(ii) If x 2 R, then inf(x+A) = x+inf A.
(iii) If x > 0, then sup(xA) = x(sup A).
(iv) If x > 0, then inf(xA) = x(inf A).
(v) If x < 0, then sup(xA) = x(inf A).
(vi) If x < 0, then inf(xA) = x(sup A).
Do note that multiplying a set by a negative number switches supremum for an infimum and
vice-versa.
1.2. THE SET OF REAL NUMBERS 29
Proof. Let us only prove the first statement. The rest are left as exercises.
Suppose that b is a bound for A. That is, y < b for all y 2 A. Then x+y < x+b, and so x+b is
a bound for x+A. In particular, if b = sup A, then
sup(x+A)  x+b = x+sup A:
The other direction is similar. If b is a bound for x+A, then x+y < b for all y 2 A and so
y < b􀀀x. So b􀀀x is a bound for A. If b = sup(x+A), then
sup A  b􀀀x = sup(x+A)􀀀x:
And the result follows.
Sometimes we will need to apply supremum twice. Here is an example.
Proposition 1.2.8. Let A;B  R such that x  y whenever x 2 A and y 2 B. Then sup A  inf B.
Proof. First note that any x 2 A is a lower bound for B. Therefore x  inf B. Now inf B is an upper
bound for A and therefore sup A  inf B.
You have to be careful about strict inequalities and taking suprema and infima. Note that x < y
whenever x 2 A and y 2 B still only implies sup A  inf B, and not a strict inequality. This is an
important subtle point that comes up often.
For example take A := f0g and take B := f1=n j n 2 Ng. Then 0 < 1=n for all n 2 N however
sup A = 0 and inf B = 0 as we have seen.
1.2.4 Maxima and minima
By Exercise 1.1.2 we know that a finite set of numbers always has a supremum or an infimum that
is contained in the set itself. In this case we usually do not use the words supremum or infimum.
When we have a set A of real numbers bounded above, such that sup A 2 A, then we can use the
word maximum and notation maxA to denote the supremum. Similarly for infimum. When a set A
is bounded below and inf A 2 A, then we can use the word minimum and the notation minA. For
example,
maxf1;2:4;p;100g = 100;
minf1;2:4;p;100g = 1:
While writing sup and inf may be technically correct in this situation, max and min are generally
used to emphasize that the supremum or infimum is in the set itself.