The Supremum Superman

To begin, let’s recall the definition of the supremum of a set.

Definition 1. Let $\mathbb K$ be an ordered field (for instance, $\mathbb Q$ or $\mathbb R$ , the latter we have technically not yet defined). For any $K \subseteq \mathbb K$ , we define the supremum $\sup K$ , if it exists, by the following conditions:

For any $x \in K$ , $x \leq \sup K$ .
For any upper bound $M$ of $K$ , $\sup K \leq M$ .

The defining property of $\mathbb R$ is that for any nonempty $K \subseteq \mathbb R$ that is bounded above, $\sup K$ exists and belongs to $\mathbb R$ . This we will finally construct in the next post.

The first of many interesting properties of the supremum is trivial to prove, since it’s a matter of bookkeeping, but is technically useful since it allows us to “squeeze” terms into inequalities.

Theorem 1. Let $K \subseteq \mathbb K$ such that $\sup K$ exists. Suppose there exists $M \in \mathbb K$ such that for any $x \in K$ , $x \leq M$ . Then $x \leq \sup K \leq M$ .

Proof. As an upper bound for $K$ , $x \leq \sup K$ . The number $M$ is an upper bound for $K$ . As the least upper bound, $\sup K \leq M$ . Therefore, $x \leq \sup K \leq M$ .

It should also make intuitive sense that the larger a set, the larger its supremum. We can formulate this more rigorously.

Theorem 2. Let $K, L \subseteq \mathbb K$ be defined such that $\sup K, \sup L$ exist. Then

$K \subseteq L \quad \Rightarrow \quad \sup K \leq \sup L.$

Consequently,

$\begin{aligned} \sup(K \cap L) &\leq \min\{\sup K,\sup L\} \\ &\leq \max\{\sup K,\sup L\} \\ &\leq \sup(K \cup L). \end{aligned}$

Proof. Since $\sup L$ is an upper bound for $L \supseteq K$ , $\sup L$ is an upper bound for $K$ . Therefore, $\sup K \leq \sup L$ .

If we consider arbitrary sums of elements, we should get corresponding supremum properties.

Theorem 3. Let $K, L \subseteq \mathbb K$ be defined such that $\sup K, \sup L$ exist. For $c \in \mathbb K$ with $c > 0$ , define

$\begin{aligned} K + L &:= \{x + y : x \in K, y \in L\} \subseteq \mathbb K,\\ cK &:= \{cx : x \in K\} \subseteq \mathbb K. \end{aligned}$

If $\sup(K+L)$ exists, then $\sup(K + L) =\sup K + \sup L$ .
If $\sup (cK)$ exists, then $\sup(cK) = c \cdot \sup K$ .

Proof. For the first claim, fix $x+y \in K + L$ . On the one hand,

$x + y \leq \sup K + \sup L \quad \Rightarrow \quad \sup(K+L) \leq \sup K + \sup L.$

On the other hand,

$\begin{aligned} x + y \leq \sup(K+L) \quad &\Rightarrow \quad x \leq \sup(K + L) - y \\ &\Rightarrow \quad \sup K \leq \sup(K+L) - y \\ &\Rightarrow \quad y \leq \sup(K+L) - \sup K \\ & \Rightarrow \quad \sup L \leq \sup(K+L) - \sup K \\ &\Rightarrow \quad \sup K + \sup L \leq \sup(K+L). \end{aligned}$

Combining the inequalities yields $\sup(K+L) = \sup K + \sup L$ , as required.

For the second claim we observe that for any $u \in K$ ,

$cu \leq c \cdot \sup(K) \quad \Rightarrow \quad \sup(cK) \leq c \cdot \sup K.$

Similarly

$\displaystyle u= \frac 1c \cdot cu \leq \frac 1c \cdot \sup(cK)\quad \Rightarrow \quad \sup K \leq \frac 1c \cdot \sup(cK)$

implies $c \cdot \sup K \leq \sup (cK)$ .

The supremum has a useful estimate property that is obvious by definition but whose result is useful for future proofs, as it gives us an “epsilon” of room to manipulate our inequalities.

Lemma 1. Let $a,b \in \mathbb K$ . Then $a \leq b$ if and only if for any $\epsilon > 0$ , $a \leq b + \epsilon$ .

Proof. The direction $(\Rightarrow)$ is obvious. For the direction $(\Leftarrow)$ , we will prove by contrapositive. Suppose $a > b$ . Define $\epsilon := (a-b)/2 > 0$ . Then

$\displaystyle b +\epsilon = b+ \frac{a-b}2 = \frac{a+b}2 < \frac{a+a}2 = a,$

as required.

Theorem 4. Let $K \subseteq \mathbb K$ have a supremum $\sup K$ . Then $M = \sup K$ if and only if $M$ is an upper bound of $K$ and for any $\epsilon > 0$ , there exists $x \in K$ such that $x > M -\epsilon$ .

Proof. The proof of $(\Rightarrow)$ is relatively straightforward. For $(\Leftarrow)$ , let $M'$ be another upper bound for $K$ . Then for any $\epsilon > 0$ , find $x \in K$ such that

$M < x +\epsilon \leq M' + \epsilon$

Since $\epsilon > 0$ is arbitary, by Lemma 1, we have $M \leq M'$ , so that $M$ is the least upper bound, i.e. $M = \sup K$ .

The supremum has a mirror twin, known as the infimum. It is defined almost identically to the supremum, but flipped. In fact, the two notions are intimately connected through a kind of “flipping” of the sets.

Definition 2. Let $\mathbb K$ be an ordered field. For any $K \subseteq \mathbb K$ , we say that $K$ is bounded below if there exists a lower bound $m \in \mathbb K$ such that for any $x \in \mathbb K$ ,

$x \in K \quad \Rightarrow \quad x \geq m.$

A greatest lower bound of $K$ is a lower bound $m$ of $K$ such that

For any lower bound $l$ of $K$ , $m \geq l$ .

If $K$ has a greatest lower bound, then it is unique, denoted $\inf K$ , called the infimum of $K$ .

Theorem 5. Let $K \subseteq \mathbb K$ . Suppose $K$ a supremum $\sup K$ . Define

$-K := \{-x : x \in K\}.$

Then $-K$ has an infimum $\inf(-K) = -\sup K$ .

Similarly, if $K$ has an infimum $\inf K$ , then $-K$ has a supremum

$\sup(-K) = -\inf K.$

Proof. Suppose firstly that $K$ has a supremum. For any $x\in K$ ,

$x \leq \sup K \quad \Rightarrow \quad -x \geq -\sup K.$

Thus, $-\sup K$ is a lower bound for $-K$ . Now, let $l$ be any lower bound for $-K$ . This means for any $x \in K$ ,

$-x \geq l \quad \Rightarrow \quad x \leq -l \quad \Rightarrow \quad \sup K \leq -l \quad \Rightarrow \quad -\sup K \geq l.$

Thus, $-\sup K$ is the greatest lower bound for $-K$ , i.e.

$\inf(-K) = -\sup K.$

The second property is left as an exercise.

Corollary 1. Let $K \subseteq \mathbb K$ have an infimum $\inf K$ . Then $m = \inf K$ if and only if $m$ is a lower bound of $K$ and for any $\epsilon > 0$ , there exists $x \in K$ such that $x < m+ \epsilon$ .

Proof. Exercise.

These slew of properties will be quintessential in establishing many real-analytic tools in future discussions. But they all predicate on one crucial question: do they exist? We will turn our attention to construct $\mathbb R$ so that we can have reasonably guaranteed access to these suprema quantities.

—Joel Kindiak, 17 Dec 24, 2348H

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The Supremum Superman

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