Minima, maxima, infima and suprema

Infinite series: Introduction

Minima, maxima, infima and suprema

For sets $A\subseteq\mathbb R$ we define the maximum and minimum as follows:

The maximum of $A$ is an element $x\in A$ such that $y\leq x$ for all $y\in A$ .
The minimum of $A$ is an element $x\in A$ such that $y\geq x$ for all $y\in A$ .

You can show any $A$ has at most one maximum and one minimum. However, not every subset has a maximum or minimum.

The open interval

$(0,1) = \{x\in \mathbb R : 0 < x < 1\}$ has no maximum and no minimum.

The crux is that $0$ and $1$ aren't elements of $(0,1)$ and hence cannot be a minimum or maximum respecively.

For each $x\in (0,1)$ the number $\tfrac 12 \cdot x$ is smaller and still an element of $(0,1)$ . This means there does not exist a least number $x$ in this set; there always is a smaller element.

Similarly $\tfrac{1}{2}\cdot (1+x)$ a number in $(0,1)$ which is greater than $x$ . Therefore this set does not have a maximum either.

In the example we saw that $0$ is not the minimum of this set because $0$ is not an element of $(0,1)$ . Similarly $1$ not the maximum of $(0,1)$ . To express the special role of $0$ and $1$ we introduce the concepts of infimum and supremum.

A number $M$ is a lower bound for $A$ if $x \geq M$ for all $x\in A$ .
A number $M$ is an upper bound for $A$ if $x \leq M$ for all $x\in A$ .

The infimum $\inf(A)$ of a bounded set $A$ is the greatest lower bound of $A$ .
The supremum $\sup(A)$ of a bounded set of $A$ is the least upper bound of $A$ .

Returning to our example, we see that these values actually do exist.

We have $\inf (0,1) = 0$ and $\sup (0,1) = 1$ .

By definition $x > 0$ for all $x\in (0,1)$ , so the number $0$ is a lower bound for our interval. Suppose there exists a greater lower bound $L>0$ . Because $L > 0$ we know that $\tfrac 12 \cdot L \in (0,1)$ and because $\tfrac{1}{2} \cdot L < L$ this means that $L$ cannot be a lower bound for the entire interval. This is in contradiction with the assumption. We conclude that $0$ is the greatest lower bound.

In the same way you can prove that $1$ is the least upper bound.

In the definitions of the infimum and supremum we assume that $A$ is a limited collection. If $A$ is unbounded from above we write $\sup(A) = \infty$ and when $A$ contains arbitrarily small elements we write $\inf(A) = -\infty$ . These conventions make sure that the infimum and supremum of a subset of $\mathbb{R}$ always exist. One can think of the infimum as some kind of minimum that does not need to be inside the set itself. Similarly, one can think of the supremum as some kind of maximum is not necessarily an element of the set.

We conclude this section with a few examples.

For $A = [0,1]$ we have $\inf(A) = \min(A) = 0$ and $\sup(A) = \max(A) = 1$ .

For $B = (-\infty,1]$ it holds that $\inf(B) = -\infty$ and $\sup(B) = \max(B) = 1$ . There is no minimum.

For $C = \{x\in\mathbb Q : x > 0 \text{ and } x^2 < 2\}$ we have $\inf(C) = \min(C) = 0$ and $\sup(C) = \sqrt 2$ . Because $\sqrt 2\not\in\mathbb Q$ there is no maximum.