§5 The Symbols +∞ and −∞

The symbols $+\infty$ and $-\infty$ are extremely useful even though they are not real numbers. We often write $+\infty$ simply as $\infty$.

We adjoin $+\infty$ and $-\infty$ to $\mathbb{R}$ and agree that $-\infty \le a \le +\infty$ for all $a \in \mathbb{R} \cup \{-\infty, +\infty\}$. This ordering satisfies O1, O2 and O3 of §3. But we give $\mathbb{R} \cup \{-\infty, +\infty\}$ no algebraic structure. Never apply a theorem or exercise stated for real numbers to the symbols $\pm\infty$.

Why this matters

The symbols are bookkeeping, not new numbers. They let us state results without a list of special cases: “every nonempty set has a supremum” (this section) and “every monotone sequence has a limit” (Corollary 10.5) are only true once $\pm\infty$ are allowed as answers.

Unbounded intervals

For $a, b \in \mathbb{R}$:

$$\begin{aligned} [a, \infty) &= \{x \in \mathbb{R} : a \le x\}, & (a, \infty) &= \{x \in \mathbb{R} : a < x\}, \\ (-\infty, b] &= \{x \in \mathbb{R} : x \le b\}, & (-\infty, b) &= \{x \in \mathbb{R} : x < b\}. \end{aligned}$$

We sometimes write $(-\infty, \infty)$ for $\mathbb{R}$. $[a, \infty)$ and $(-\infty, b]$ are closed (unbounded) intervals, and $(a, \infty)$ and $(-\infty, b)$ are open (unbounded) intervals.

sup and inf of any nonempty set

If $S$ is bounded above, $\sup S$ is a real number by the Completeness Axiom 4.4. If $S$ is bounded below, $\inf S$ is a real number by Corollary 4.5. We now define

$$\sup S = +\infty \text{ if } S \text{ is not bounded above}, \qquad \inf S = -\infty \text{ if } S \text{ is not bounded below}.$$

Intuition

“$\sup S = +\infty$” is just shorthand for a statement about real numbers: for every real $M$ there is $s \in S$ with $s > M$. Whenever a proof involves $\sup S = +\infty$, unpack it into this form.

Summary.

Let $S$ be any nonempty subset of $\mathbb{R}$. The symbols $\sup S$ and $\inf S$ always make sense. If $S$ is bounded above, $\sup S$ is a real number; otherwise $\sup S = +\infty$. If $S$ is bounded below, $\inf S$ is a real number; otherwise $\inf S = -\infty$. Moreover, $\inf S \le \sup S$.

Example 1.

For nonempty bounded $A, B \subseteq \mathbb{R}$, Exercise 4.14 says

$$\sup(A + B) = \sup A + \sup B \quad\text{and}\quad \inf(A + B) = \inf A + \inf B. \tag{1}$$

We verify that the first equality holds even if $A$ or $B$ is unbounded. Exercise 5.7 asks you to do the same for the second.

Verification

If $x \in A + B$, then $x = a + b$ with $a \in A$, $b \in B$, so $x \le \sup A + \sup B$. Thus $\sup A + \sup B$ is an upper bound for $A + B$, and $\sup(A + B) \le \sup A + \sup B$.

For the reverse inequality, note that neither supremum is $-\infty$ (the sets are nonempty), so the undefined sum $-\infty + \infty$ cannot occur. If $\sup A + \sup B = +\infty$, then one of them, say $\sup B$, is $+\infty$. Pick $a_0 \in A$. Then $\sup(A + B) \ge \sup(a_0 + B) = a_0 + \sup B = +\infty$.

Otherwise $\sup A + \sup B$ is finite. Let $\varepsilon > 0$. There are $a \in A$ and $b \in B$ with $a > \sup A - \frac\varepsilon2$ and $b > \sup B - \frac\varepsilon2$. Then $a + b \in A + B$, so $\sup(A + B) \ge a + b > \sup A + \sup B - \varepsilon$. Since $\varepsilon > 0$ is arbitrary, $\sup(A + B) \ge \sup A + \sup B$.

∎

Common pitfallforms that have no value

Expressions such as $\infty - \infty$, $0 \cdot \infty$ and $\frac{\infty}{\infty}$ are left undefined on purpose, because no single value would be right. For instance $n \to \infty$ and $n^2 \to \infty$, yet $n - n \to 0$, $n^2 - n \to +\infty$ and $n - n^2 \to -\infty$. Each of these is “$\infty - \infty$”. The proof above is careful to rule out the sum $-\infty + \infty$ before it can arise.

Reading alongside Abbott

Abbott has no section corresponding to this one and treats $\pm\infty$ informally. Ross’s conventions here are what make later statements such as Theorem 11.8 true for every sequence, bounded or not.

The exercises below tie up loose ends, mostly by extending results of §4 to sets that need not be bounded.

Check your understanding

Take the quiz (3 questions)

What is $\sup\{n^2 : n \in \mathbb{N}\}$?

$+\infty$It does not exist.The largest perfect square

For a nonempty set $S \subseteq \mathbb{R}$, which is always true?

$\inf S \le \sup S$$\inf S < \sup S$$\sup S$ is a real number.

Let $-S = \{-s : s \in S\}$ for nonempty $S$. Which identity holds?

$\inf S = -\sup(-S)$$\inf S = \sup(-S)$$\inf S = \frac{1}{\sup S}$

Exercises

Exercise 5.1

Write the following sets in interval notation:

1. $\{x \in \mathbb{R} : x < 0\}$
2. $\{x \in \mathbb{R} : x^3 \le 8\}$
3. $\{x^2 : x \in \mathbb{R}\}$
4. $\{x \in \mathbb{R} : x^2 < 8\}$

Solution

1. $(-\infty, 0)$
2. $(-\infty, 2]$
3. $[0, \infty)$
4. $(-2\sqrt2, 2\sqrt2)$

Tags: Warm-upComputation

Used in Exercise 5.2

Exercise 5.2

Give the infimum and supremum of each set listed in Exercise 5.1.

Solution

(a) $-\infty$, $0$ (b) $-\infty$, $2$ (c) $0$, $+\infty$ (d) $-2\sqrt2$, $2\sqrt2$

Tags: Warm-upSup & infComputation

Exercise 5.3

Give the infimum and supremum of each unbounded set listed in Exercise 4.1.

Solution

(h) $\inf = 2$, $\sup = +\infty$ (k) $0$, $+\infty$ (l) $-\infty$, $2$ (o) $-\infty$, $0$ (t) $-\infty$, $2$ (u) $0$, $+\infty$

Tags: Warm-upSup & infComputation

Exercise 5.4

Let $S$ be a nonempty subset of $\mathbb{R}$, and let $-S = \{-s : s \in S\}$. Prove $\inf S = -\sup(-S)$.

Hint

For the case $-\infty < \inf S$, simply state that this was proved in Exercise 4.9. If $\inf S = -\infty$, then $S$ is not bounded below, so $-S$ is not bounded above.

Solution

We use the conventions $-(+\infty) = -\infty$ and $-(-\infty) = +\infty$.

If $S$ is bounded below, then $-S$ is nonempty and bounded above, $\inf S$ is a real number, and $\inf S = -\sup(-S)$ by Exercise 4.9.

If $S$ is not bounded below, then $\inf S = -\infty$. Then $-S$ is not bounded above (an upper bound $M$ of $-S$ would make $-M$ a lower bound of $S$), so $\sup(-S) = +\infty$ and $-\sup(-S) = -\infty = \inf S$.

Tags: StandardSup & inf

Used in Exercise 5.7, Exercise 11.8

Exercise 5.5

Prove $\inf S \le \sup S$ for every nonempty subset $S$ of $\mathbb{R}$. Compare Exercise 4.6(a).

Hint

Pick $s \in S$. In every case $\inf S \le s \le \sup S$ in the extended order.

Solution

Pick $s \in S$. Since $\inf S$ is a lower bound of $S$ and $\sup S$ is an upper bound (this is also true when the value is $\mp\infty$), $\inf S \le s \le \sup S$ in the extended order, so $\inf S \le \sup S$.

Tags: Warm-upSup & inf

Used in Exercise 5.6

Exercise 5.6

Let $S$ and $T$ be nonempty subsets of $\mathbb{R}$ with $S \subseteq T$. Prove $\inf T \le \inf S \le \sup S \le \sup T$. Compare Exercise 4.7(a).

Hint

If $\sup T = +\infty$ there is nothing to prove on the right. Otherwise $T$, and hence $S$, is bounded above, and Exercise 4.7(a) applies. Treat the left side the same way.

Solution

Every upper bound of $T$ (in $\mathbb{R}$ or $+\infty$) is an upper bound of $S \subseteq T$, so $\sup S \le \sup T$: if $\sup T = +\infty$ this is trivial, and otherwise $T$ and hence $S$ is bounded above, and $\sup T$ is an upper bound of $S$, so $\sup S \le \sup T$ (Exercise 4.7(a)). In the same way every lower bound of $T$ is a lower bound of $S$, so $\inf T \le \inf S$. Exercise 5.5 gives the middle inequality $\inf S \le \sup S$.

Tags: StandardSup & inf

Exercise 5.7

Finish Example 1 by verifying the equality involving infimums.

Hint

Apply the supremum result to $-A$ and $-B$, using $-(A + B) = (-A) + (-B)$ and Exercise 5.4.

Solution

Let $A, B$ be nonempty subsets of $\mathbb{R}$. A nonempty set never has infimum $+\infty$, so $\inf A + \inf B$ is defined (it is $-\infty$ if either term is). Since $-(A+B) = (-A) + (-B)$, Exercise 5.4 and the supremum result of this example give $\inf(A+B) = -\sup\bigl((-A) + (-B)\bigr) = -\bigl[\sup(-A) + \sup(-B)\bigr] = -\sup(-A) - \sup(-B) = \inf A + \inf B,$ where the last two steps use $-(x + y) = (-x) + (-y)$ in the extended reals (neither term is $-\infty$) and Exercise 5.4 again.

Tags: StandardSup & inf