# Math for Non-Geeks/Supremum and infimum
## Introduction

Supremum (from Latin „supremum“ = "the highest/supreme“) sounds, as if it were "the maximum“ (that is, the largest element of the set). In the course of the article, however, we will see that the supremum generalzes the maximum. Let's start by remembering the following:
While the maximum has to be an element of a considered set, this need not apply to the supremum. Therefore we should aptly translate "supremum“ as "the number immediately restricting from the top“. It is "restricting from the top“, because it is like the maximum greater than or equal to any number of the set. And it is "immediate" because it is the smallest of all "upward limiting numbers".
Similarly, the infimum is a generalization of the minimum. It is the "number that immediately restricts downwards", i.e. the largest of all the "numbers that restrict downwards" of a set. We will get to know concrete examples in the coming sections.
For us the concept of the supremum is important, because with it the completeness of the real numbers can be described alternatively. In addition, the supremum is a useful tool in proofs or in defining new terms.

## Explanation of the supremum

To explain the Supremum, we will examine how to arrive at its precise definition. For this we will determine how the supremum can be generalized from the maximum. Remember: the maximum of a set is its largest element. The maximum $m$ of a quantity $M$ has the following properties:
* $m$ is an element of $M$.
* For every $y\in M$ is $y\leq m$.
In the second property there is therefore a smaller-equal and no smaller sign, because in the statement could also be $y$ equal to $m$. For finite quantities, the maximum is always defined, but this is not necessarily the case for infinite quantities.
First of all, we may encounter the problem that the set under consideration is unlimited upwards. Take for example the set $\mathbb {R} ^{+}=\{x\in \mathbb {R} :x>0\}$. This set cannot have a maximum or the like, since there is a larger number of $\mathbb {R} ^{+}$ for each real number. This set cannot have a largest element. There is also no element that could be "directly the largest" element. Therefore, a question about this with this set simply does not make sense.
For the transfer of the term "maximum" to infinite sets, the set must therefore be limited upwards. So there must be a number $b$, which is greater than or equal to each element of the set. As a result $b$ does not necessarily have to be an element of the set.
But even then problems can still arise. Take for example the set $M=\{x\in \mathbb {R} :x<1\}$. This set is limited to the top, because for $b$ any number greater than $1$ can be selected.
Does the quantity $M$ have a maximum? Unfortunately not. For each $x\in M$ ${\tfrac {x+1}{2}}{2}$ is another number from $M$ with property $x<{\tfrac {x+1}{2}}$ (the number ${\tfrac {x+1}{2}}$ is in the middle between $x$ and $1$). However, $M$ cannot have a maximum element, because for each number from $M$ there is at least one larger number from $M$.
Thus, when looking at infinite quantities, the maximum loses one property. Namely, that it is element of the set:
* m is an element of M.
* For every $y\in M$ is $y\leq m$.
The only property that remains is that the number you are looking for is greater than any element in the set. Such a number is called the "upper limit" of the set:
Definition (upper bound)
Let $M$ be a subset of $\mathbb {R}$. Then a number $u$, which is greater than or equal to each element of $M$, is called an upper bound. So it is $x\leq u$ for all $x\in M$.
Similarly, a lower bound is a number that limits a quantity downwards:
Definition (Lower bound)
Let $M$ be a subset of $\mathbb {R}$. Then a number ${\tilde {u}}$, which is less than or equal to any element of $M$, is called a lower bound. So it is $x\geq {\tilde {u}}$ for all $x\in M$.
When we look at our new definition, we see two things. First: Upper and lower limits do not have to be elements of the considered set, because this is not required by the definition. And secondly: the definition says nothing about a possible uniqueness of the bounds.
For example, consider the set $M=\{x\in \mathbb {R} :x<1\}$. Here we certainly first think of $1$ as the upper bound. However, $17$ is also an upper bound and meets the requirements of the definition. Apart from the fact that $17$ is far above our example set, both numbers are not elements of the set. This example shows that there can be more than one upper bound. But it becomes even more disturbing: A limited subset of the real numbers always has infinitely many upper bounds. If $u$ is an upper bound of $M$, any larger number, i.e. $u+a$ for all $a>0$, is also an upper bound.
On closer inspection, the terms upper and lower bound are not very appropriate. They provide much less than a maximum term. The maximum is always unique: there can only be one of them. That is not the case with the upper bound. Let us therefore try to improve the concept.
On closer inspection, the terms upper and lower bound are not very accurate. They provide much less than a maximum term. The maximum is always unique: there can only be one of them. That is not the case with the upper barrier. Let us therefore try to improve the concept.
Consider as an example again the set $M=\{x\in \mathbb {R} :x<1\}$. Which number could be used to generalize the maximum for $M$? Intuitively the number $1$ occurs to us. But why choose this number?
We want a general term that works even when the set is no longer so clearly described. Therefore, all upper limits of $M$, i.e. all numbers greater than or equal to $1$, are possible. Now our number should be optimal in the sense that it is as small as possible. So we get to the number $1$. It is not only an upper bound, it is also the smallest upper bound of $M$. We have already seen that for each $x<1$ there is another number $y<1$ with $x<y$ (namely $y={\tfrac {x+1}{2}}$). Thus no number smaller than $1$ can be an upper bound of $M$. $1$ is what we consider to be the "immediately above" number of $\{x\in \mathbb {R} :x<1\}$.
Question: What might a set look like where it is not intuitively "clear" what number the Supremum could be?
Let's briefly have a look at this beautiful looking set of numbers: The Mandelbrot set. They are obtained by inserting all points in a two-dimensional coordinate system into a certain function $f$. It takes a coordinate $(x,y)$ and turns it into another coordinate $(x',y')$. This result is put back into this function and then again and again and again and again.... The coordinates you get with every step become very large for some starting points very fast, for others they remain small. Once the coordinates have moved far enough away from their starting point (have exceeded a limit $g$), they never come back and "run for it". If a point for the start value $(x,y)$ always remains below $g$, the point $(x,y)$ belongs to the set and is colored black. If it exceeds $g$, it gets a certain color, depending on when it exceeded $g$. What we see on the right is the resulting image.
The Mandelbrot set is now in the plane, its points have $x$- and $y$ coordinates, therefore it is not suitable for our supremum concept at first. But we can simply "look at the set of all $y$ coordinates of the Mandelbrot set" and try to find its supremum. To put it more clearly: We would like to know how far up the black dots in the picture reach and are looking for the smallest upper bound. Which value it has exactly, however, is completely unclear at the first (and also at the second) look.
Die kleinste obere Schranke $s$ wird durch folgende zwei Eigenschaften charakterisiert:
* $s$ ist obere Schranke von $M$: Für jedes $y\in M$ ist $y\leq s$.
* Jede obere Schranke $u$ von $M$ ist mindestens so groß wie $s$: Gilt $y\leq u$ für alle $y\in M$, so gilt auch $s\leq u$. Anders formuliert: Für jedes $u<s$ gibt es mindestens eine Zahl $y\in M$ mit $u<y$.
Das können wir als Definition des Supremums verwenden, da es offenbar die kleinste obere Schranke charakterisiert. Das Infimum wird analog als die größte untere Schranke definiert. Eine weitere Möglichkeit der Charakterisierung von Supremum und Infimum werden wir im Abschnitt „Suprema und Infima in Halbordnungen“ kennenlernen.}}

## Definition of the Supremum and Infimum

Die Definition des Supremums und des Infimums lautet:
Definition (Supremum)
Let $M$ be a subset of $\mathbb {R}$. The supremum $s$ of the set $M$ is the smallest upper bound of $M$. The supremum is characterized by the following two properties:
* For every $y\in M$ it holds $y\leq s$.
* There is no number $x$ less than $s$ that is an upper bound of $M$: For all $x<s$ there exists at least one number $y\in M$ with $x<y$.
Definition (Infimum)
Let $M$ be a subset of $\mathbb {R}$. The infimum ${\tilde {s}}$ of the set $M$ is the largest lower bound of $M$. The infimum is characterized by the following two properties:
* For every $y\in M$ it holds $y\geq {\tilde {s}}$.
* No number $x$ larger than ${\tilde {s}}$ is a lower bound of $M$: For all $x>{\tilde {s}}$ there exists at least one number $y\in M$ with $x>y$.

### The Epsilon Definition

In the second property in the definition of the supremum $s$ of the set $M$, which is an element of the set $M$, it says:
In mathematical literature and textbooks, authors often set $x=s-\epsilon$ with $\epsilon >0$. This is a way to write the second propery of the supremum as a formal mathematical claim. Namely, we could replace the second property given above with the equivalent statement:
Since both statements are equivalent and just differently worded, it is up to our discretion which variant we choose to use in proofs.
Question: What is the epsilon definition of the infimum?
${\tilde {s}}$ is an infimum of $M$ if ${\tilde {s}}$ is a lower bound of $M$ and if for any $\epsilon >0$ there exists some $y\in M$ such that ${\tilde {s}}+\epsilon >y$ holds.

### Maximum and Minimum

For the maximum and minimum we have the following well-known definitions:
Definition (Maximum)
The maximum $m$ of a set $M$ is a number with the two following properties:
* $m\in M$.
* For all $y\in M$ it holds $y\leq m$.
Definition (Minimum)
The minimum ${\tilde {m}}$ of a set $M$ is a number with the following two properties:
* ${\tilde {m}}\in M$.
* For all $y\in M$ it holds $y\geq {\tilde {m}}$.
From these definitions it follows immediately that the maximum of a set is also the supremum of the set. I.e. let $m$ be the maximum of the set $M$. For one, $m$ is by definition the upper bound of $M$. Furthermore, for every $x$ with $x<m$ there exists a $y\in M$ with $x<y$, namely $y=m$. On the other hand, not every supremum is a maximum, like we saw above using the set $\{x\in \mathbb {R} :x<1\}$. The number $1$ is the supremum of this set, but not the maximum! Similar statements are true for the minimum and infimum.

### Notation

| Notation | Meaning |
| $\sup M$ | Supremum of $M$ |
| $\sup _{x\in D}f(x)$ | Supremum of $\{f(x):x\in D\}$ |
| $\inf M$ | Infimum of $M$ |
| $\inf _{x\in D}f(x)$ | Infimum of $\{f(x):x\in D\}$ |
| $\max M$ | Maximum of $M$ |
| $\min M$ | Minimum of $M$ |

### The Duality Principle

We've already seen in the above definitions and explanations that the terms supremum and infimum can be considered and used similarly. This is a result of the fact that, by switching around the ordering of the real numbers, i.e. replacing $\leq$ with $\geq$, the supremum becomes the infimum and the infimum becomes the supremum. This means we can introduce a new ordering $\leq _{\text{neu}}$ such that $x\leq _{\text{neu}}y$ holds if and only if $x\geq y$ (this is the same as us reflecting the real numbers around zero). With this new ordering, the supremum acts as the infimum and vice versa. Both orderings $\leq _{\text{neu}}$ and $\leq$ have the same mathematical ordering properties. This means they are isomorphic to one another. Therefore, the properties of the infimum and supremum must be the same for this reversed ordering. This means that any statements we make in the future for suprema will also apply to infima and vice versa. The same applies to the interchangeability of statements regarding the maximum and minimum.
Example (Duality Principle)
For all $x\in M$ it is true that $x\leq \sup M$. Similarly for all $x\in M$ we have the inequality $x\geq \inf M$.

### Existence and Uniqueness

Up until this point we have exclusively been speaking of the supremum. This sounds as if the supremum always exists and is always unique. This hints at answers to some important mathematical questions, namely: why did we bother defining the supremum in the first place? If the maximum of a set does not always exist, defining a supremum does not actually solve this existence problem. What is the advantage of introducing the concept of a supremum? Intuitively we know that of all upper bounds of a set, there should be exactly one smallest upper bound, i.e. intuitively this smallest upper bound should always exist and be unique. However, we haven't yet proven this mathematically. Indeed, the existence and uniqueness is a true property of the supremum, which we will now show formally.
In the following theorem we will prove the uniqueness of the supremum (and infimum), i.e. that a set can have at most one supremum and one infimum.
Theorem (Uniqueness of the Supremum and Infimum)
A set can have at most one supremum and one infimum.
Proof (Uniqueness of the Supremum and Infimum)
We can use the standard proof method for showing uniqueness: first we assume there exists some set $M$ with two suprema $s_{1}$ and $s_{2}$. Then we will show $s_{1}=s_{2}$. Both of the suprema have the following properties:
* $s_{1}$ and $s_{2}$ are upper bounds of $M$.
* No number less than $s_{1}$ and $s_{2}$ is an upper bound of $M$.
By the second property and since $s_{2}$ is an upper bound o $M$, $s_{2}$ can't be smaller than $s_{1}$ and must therefore be greater than or equal to $s_{1}$. Similarly it must hold $s_{1}\geq s_{2}$. Since $s_{2}\geq s_{1}$ and $s_{1}\geq s_{2}$ we can conclude $s_{1}=s_{2}$. The proof for the uniqueness of the infimum is similar.
Using the completeness axiom we can also prove the existence of the supremum of a non-empty subset of the real numbers that is bounded above. However, we will not deal with this in this chapter. We can also prove a similar statement about the exitence of the infimum of a non-empty subset of the real numbers that is bounded below. It is indeed the case that the supremum and infimum of a non-empty subset of the real numbers exist when this set is bounded above and below and are always unique.

## Exegesis: Suprema and Infima in Partial Orderings

We introduced the above definitions for suprema and infima for sets of real numbers. This is sufficient for an Introduction to Real Analysis class, since such classes often deal with subsets of $\mathbb {R}$. In higher-level theoretical math classes, the concept of the partial ordering, which satisfies the reflexitivity, anti-symmetry, and transitivity properties, but does not satisfy the totality property, i.e. there may be elements $x,y$ for which neither $x\leq y$ nor $y\leq$ holds. In the case of partial orderings, the definitions we provided above are not sufficient to construct a sensible supremum or infimum. The main issue with these definitions in the case of partial orderings is that we lose the uniqueness of the supremum and infimum. In order to ensure the uniqueness of the supremum, we instead introduce the following definition:
Definition (Supremum in Partial Orderings)
In the partially ordered set $(A,\leq )$, an element $s\in A$ is the supremum of the set $M\subseteq A$ when it holds:
* $s$ is an upper bound of $M$: for every $y\in M$ it holds $y\leq s$.
* For every other upper bound $t$ of $M$ it holds: $s\leq t$
In order to show that these definitions is a sensible generalization of the supremum with respect to a partial ordering, we must show that both definitions coincide on a subset of the real numbers:
Theorem (Equivalent Definition of the Supremum)
Let $M\subseteq \mathbb {R}$ be arbitrary. Our definition of the supremum $s$ is:
* For every $y\in M$ it holds $y\leq s$.
* Every number $x$ less than $s$ is not an upper bound of $M$: for all $x<s$ there exists at least one number $y\in M$ with $x<y$.
This definition is equivalent to the definition of the supremum with respect to partial orderings:
* $s$ is an upper bound of $M$: for every $y\in M$ it holds $y\leq s$.
* For every other upper bound $t$ of $M$ it holds: $s\leq t$
Proof (Equivalent Definition of the Supremum)
Let $M\subseteq \mathbb {R}$ be arbitrary. Since the first property of each pairs of properties are identical, we only have to show that the second properties coincide. I.e. we have to show the equivalence of:
* Every number $x$ less than $s$ is not an upper bound of $M$: for al $x<s$ there exists at least one number $y\in M$ with $x<y$.
* For every other upper bound $t$ of $M$ it holds $s\leq t$.
We can formalize the two claims in the following:
* $t<s\Rightarrow \exists y\in M\colon t<y$
* $(\forall y\in M\colon y\leq t)\Rightarrow s\leq t$
We can show the equivalence of these two claims in the following way:
${\begin{array}{l}t<s\Rightarrow \exists y\in M\colon t<y\\[0.5em]\quad {\color {Gray}\left\Updownarrow \ {\text{In a totally ordered set like }}\mathbb {R} {\text{ it holds }}a<b\Leftrightarrow b\not \leq a\right.}\\[0.5em]s\not \leq t\Rightarrow \exists y\in M\colon y\not \leq t\\[0.5em]\quad {\color {Gray}\left\Updownarrow \ {\text{Contraposition: }}(A\Rightarrow B)\iff (\neg B\Rightarrow \neg A)\right.}\\[0.5em]\neg (\exists y\in M\colon y\not \leq t)\Rightarrow \neg (s\not \leq t)\\[0.5em]\quad {\color {Gray}\left\Updownarrow \ {\text{Negate the claims}}\right.}\\[0.5em](\forall y\in M\colon y\leq t)\Rightarrow s\leq t\end{array}}$
* ↑ The property that the maximum is greater than any element of the set is too characteristic of the term supremum to be deleted.
* ↑ See also http://math.stackexchange.com/questions/936462/supremum-of-all-y-coordinates-of-the-mandelbrot-set