# What is the infimum/supremum of a set in the extended real line?

May be a simple question, but I still don't know what the definition of the supremum or infimum of a set in the extended real line is. For example when we define the Lebesgue measure, we define the Lebesgue outer measure as the infimum of a set in the extended real line. My questions are:

1. What is the definition of the supremum/infimum of a set in the extended real line?

2. FORMALLY, when do we say that the infimum of a set in the extended real line is equal to positive infinity?

• Infimum of empty set is positive infinity and supremum is negative infinity. – jdods Jul 15 '16 at 18:06
• @jdods how is this relevant? – Topological cat Jul 15 '16 at 18:09
• Is it not an example of #2? – jdods Jul 15 '16 at 18:11
• @jdods in 2. I ask: "when do we say that the infimum of a set in the extended real is positive infinity?". You gave me an example of a set having infimum positive infinity, this isn't exactly helpful – Topological cat Jul 15 '16 at 18:14
• Wouldn't the infimum be $+\infty$ when the set is either empty or $\{\infty\}$? – jdods Jul 15 '16 at 18:19

What jdods wrote in the comments essentially covdrs it, but here I make things very explicit.

The supremum of a set is the smallest upper bound and the infimum is the largest lower bound. This is the formal definition that holds for every ordering.

Let's decode this. There is a usual ordering $\leq$ on $\mathbb{R}$, that one extends to the extended real line by letting $-\infty\leq r\leq$ for each real number $r$. If $A$ is a set of extended real numbers (in particular, a set of real numbers), then $u$ is an upper bound of $A$ if $r\leq u$ for all $r\in A$ and $l$ is a lower bound of $A$ if $l\leq r$ for all $r\in A$. Note that $-\infty$ is a lower bound of every set and $\infty$ is an upper bound of every set. Moreover, every extended real number is both an upper and lower bound of the empty set, since the condition "for all $r\in\emptyset$" holds vacuosly.

A smallest element in a set of extended real numbers is a lower bound of the set that is also an element of the set. If a set has a smallest element, it is unique. The set $(0,1]$ has no smallest element. Similarly, a largest element in a set is an upper bound of the set that is actually in the set.

Now let $A$ be a set of extended real numbers with infimum $\infty$. Then $\infty$ is a lower bound of $A$ and is the largest such lower bound. If $r\in A$, then we must have $\infty\leq r$, since $\infty$ is a lower bound of $A,$ and this is only possible if $r=\infty$. So the only element $A$ can have is $\infty$. There are then two possibilities, either $A$ is nonempty, then we must have $A=\{\infty\}$, in which case $\infty$ is indeed a lower bound of $A$, and, as the only lower bound, is also the largest lower bound. The other possibility is that $A=\emptyset$. Now every extended real number is a lower bound of the empty set, so the largest lower bound is the largest extended real number, namely $\infty$. So these are all possibilities for $\infty$ to be the infimum of a set of extended real numbers.

• +1. I don't think my answer adds too much to this, but I'll leave it there in case it helps someone. – 6005 Jul 15 '16 at 19:41

You are probably overthinking it a little. Infimum and supremum can be defined for any totally ordered set; the definition of $\inf$ is the "least upper bound", and the definition of $\sup$ is the "greatest lower bound" (explicit definitions given below). Both infimum and supremum are unique if they exist; on the other hand, not all totally ordered sets have the existence of supremum and infimum for every subset. However, crucially, the extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty,\infty\}$ does have a supremum and infimum for every subset.

### Formal details

We start out by ignoring addition, multiplication, and anything else about $\overline{\mathbb{R}}$ in particular, and focusing only on the structure $\overline{\mathbb{R}}$ under the order $\le$. Here are the definitions, which may be familiar:

Definition 1. Given a totally ordered set $T$ (under the order $\le$), and a subset $S \subseteq T$, an upper bound of $S$ is an element $a \in T$, such that $s \le a$ for all $s \in S$ (lower bound likewise);

Definition 2. Given a totally ordered set $T$ (under the order $\le$), and a subset $S \subseteq T$, a supremum of $S$ is a least upper bound of $S$, i.e. an upper bound that is less than or equal to all other upper bounds (infimum likewise).

As I already mentioned, the sup and inf are unique when they exist. In the totally ordered set $\mathbb{R}$, every bounded subset has a least upper bound. In $\overline{\mathbb{R}}$, every subset has a least upper bound (and you should prove this). This justifies using the notation $\sup S$ for any subset $S$ of the extended real line.

I believe that answers (1). To answer (2) and some related questions you may have, then, here are some specifics for $\overline{\mathbb{R}}$:

• For $S \subseteq \overline{\mathbb{R}}$, $\inf S = \infty$ exactly when $S \subseteq \{\infty\}$. That is, $\inf \varnothing = \infty$, $\inf \{\infty\} = \infty$, and there are no other examples.
• Likewise $\sup S = -\infty$ occurs exactly when $S = \varnothing$ or $S = \{-\infty\}$.

• For $S \subseteq \overline{\mathbb{R}}$, $\sup S = \infty$ exactly when $S$ is not bounded above by a finite real number; in particular this occurs if $\infty \in S$.

• Similarly $\inf S = -\infty$ exactly when $S$ is not bounded below by a finite real number.

• If $S$ is bounded both above and below by finite real numbers, then $\inf S$ and $\sup S$ agree with the supremum and infimum in $\mathbb{R}$.