$$$$

7  $★$ Infinity

Definition 7.1 The symbol $\mathrm{\infty }$ is a formal symbol: that is, a symbol that we agree to write, but do not attach any specific value to.

By default, any expression involving the symbol $\mathrm{\infty }$ is considered undefined. We will use define certain contexts where the symbol $\mathrm{\infty }$ is meaningful below.

7.1 Order

Our first use of the symbol $\mathrm{\infty }$ is to expand interval notation of the real numbers. Right now, using the order $<$ we have rigorously defined intervals such as $(a,b)$, $[a,b)$ and $[a,b]$ for $a,b\in \mathbb{R}$.

Definition 7.2 For any real number $a$, we define the following intervals with $\pm \mathrm{\infty }$ as an endpoint:

$$(-\mathrm{\infty },a)=\{x\in \mathbb{R}\mid x<a\}$$ $$(-\mathrm{\infty },a]=\{x\in \mathbb{R}\mid x\le a\}$$ $$(a,\mathrm{\infty })=\{x\in \mathbb{R}\mid x>a\}$$ $$[a,\mathrm{\infty })=\{x\in \mathbb{R}\mid x\ge a\}$$

But we can take this farther, by actually adding the formal symbols $\pm \mathrm{\infty }$ to our number system, to create a set called the extended reals.

Definition 7.3 (The Extended Reals) The extended real number line is the set $$\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },\mathrm{\infty }\}.$$

Definition 7.4 (Ordering on $\bar{\mathbb{R}}$) The order $<$ on $\mathbb{R}$ can be extended to $\bar{\mathbb{R}}$ by the following two rules: $$\mathrm{\forall }x\in \mathbb{R},x<\mathrm{\infty }\mathrm{\forall }x\in \mathbb{R},-\mathrm{\infty }<x$$

This allows for interval notation on $\bar{\mathbb{R}}$ where, we may may write intervals such as $[-\mathrm{\infty }1]$ to mean the points $\{x\mid \bar{R}\mid x\le 1\}$ etc.

In $\bar{\mathbb{R}}$ then, $\mathrm{\infty }$ is an upper bound for every set, and $-\mathrm{\infty }$ is a lower bound for every set. On the real numbers alone, the completeness axiom tells us that the supremum of bounded nonempty sets exist, but unbounded sets do not have a supremum. In the extended reals, we see that $\pm \mathrm{\infty }$ naturally satisfy the definitions of

Proposition 7.1 (Unbounded Above means $sup=\mathrm{\infty }$) Let $A$ be a nonempty subset of $\mathbb{R}$ which is not bounded above. Then as a subset of of the extended reals, $supA=\mathrm{\infty }$.

Proof. By the definition of $\mathrm{\infty }$, we see that $\mathrm{\infty }$ is an upper bound for $A$ always, so we need only show it is the supremum. Let $x\in \bar{\mathbb{R}}$ be any element less than $\mathrm{\infty }$. Then $x$ must be an element of $\mathbb{R}$, and since $A$ is not bounded above in $\mathbb{R}$, there is some $a\in A$ with $a>x$. Thus $x$ is not an upper bound, and so every element less than $\mathrm{\infty }$ fails to be an upper bound: that is, $\mathrm{\infty }$ is the least upper bound as claimed.

Exercise 7.1 (Unbounded Below means $inf=-\mathrm{\infty }$)  

Corollary 7.1 (Sup and Inf in the Extended Reals) Every nonempty subset of the extended real line has both an infimum and a supremum.

Proof. Let $A$ be a nonempty subset of $\bar{\mathbb{R}}$. First, if $A$ contains $\mathrm{\infty }$, then $supA=\mathrm{\infty }$ as it is the maximum. So, we can consider the case that $\mathrm{\infty }\notin A$. If $A$ is bounded above by a real number, then $supA$ is also a real number by completeness, and if $A$ is not bounded above, then $supA=\mathrm{\infty }$ by Proposition 7.1.

The same logic applies to lower bounds: after taking care of the case where $infA=minA=-\mathrm{\infty }$, if $A$ is bounded below completeness furnishes a real infimum, and if it is not, Exercise 7.1 shows the infimum to be $-\mathrm{\infty }$.

In the extended reals, it is still common to take the infimum and supremum of the empty set to be undefined. But there is also another option: one can assign $inf\varnothing =\mathrm{\infty }$ and $sup\varnothing =-\mathrm{\infty }$: if we do this then every set in the extended reals has an infimum and supremum!

7.2 Arithmetic

We know from the previous chapter that complete ordered fields cannot contain infinite numbers, yet in the section above we added $\pm \mathrm{\infty }$ to $\mathbb{R}$ in a way that did not mess up the order, or completeness properties. So, the addition of this new symbol must cause trouble with the field axioms. And, indeed it does!

Its extremely important to remember that $\bar{\mathbb{R}}$ is not a field. We have not extended any of the operations to the formal symbol $\mathrm{\infty }$, so things like $\mathrm{\infty }+1$ or $\mathrm{\infty }-\mathrm{\infty }$ or $3\mathrm{\infty }+\mathrm{\infty }/2$ are currently undefined.