# Infimum/Supremum

The **infimum** and **supremum** are concepts in mathematical analysis that generalize the notions of **minimum** and **maximum** of finite sets. They are extensively used in real analysis, including the axiomatic construction of the real numbers and the formal definition of the Riemann integral. The limits of the infimum and supremum of parts of sequences of real numbers are used in some convergence tests and, in particular, in computations of domains of convergence of power series.

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## Definition

The infimum and supremum can be defined in general contexts (e.g. partially ordered sets), but they are most commonly used in the context of subsets and functions of real numbers.

Let $S$ be a subset of the real numbers $\mathbb R$. Then $\text{inf } S$ is the

greatest lower boundof the elements of $S$, if it exists; that is, $\text{inf } S$ is the largest real number $t$ such that $t \le s$ for all $s \in S$. Similarly, $\text{sup } S$ is theleast upper boundof the elements of $S$, if it exists; that is, $\text{sup } S$ is the smallest real number $r$ such that $s \le r$ for all $s \in S$.

**Remarks:**

(1) If $S$ is finite, then $\text{inf } S$ and $\text{sup } S$ are just the minimum and maximum elements of $S,$ respectively.

(2) Note that $\text{inf } S$ and $\text{sup } S$ may not lie in $S$ in general. For instance, if $S = (0,1)$, then $\text{inf } S = 0$ and $\text{sup } S = 1.$

(3) The infimum and supremum are unique if they exist.

(4) If $S$ does not have a lower bound, it is reasonable to write $\text{inf } S = -\infty.$ If $S$ does not have an upper bound, it is reasonable to write $\text{sup } S = \infty.$

(5) The infimum and supremum are related via $\text{inf } S = -\text{sup } (-S),$ where $-S = \{ -s \colon s \in S \}.$ This is often convenient for proofs.

## Properties

The following property is a useful characterization of the infimum and supremum of a set of real numbers.

Let $S$ be a set of real numbers.

Suppose $x$ is a lower bound for $S.$ Then $x = \text{inf } S$ if and only if, for every $\epsilon > 0$, there is an $s \in S$ such that $s < x+\epsilon$.

Suppose $y$ is an upper bound for $S.$ Then $y = \text{sup } S$ if and only if, for every $\epsilon > 0$, there is an $s \in S$ such that $s > y-\epsilon$.

The proofs of the two statements are more or less identical $($and can be formally translated to each other by remark (5) above$).$ Here is the proof of the first statement. If $x = \text{inf } S,$ then $x+\epsilon$ cannot be a lower bound for $S$, so there must be an element of $S$ that is bigger than it. On the other hand, if $x$ is a lower bound that is not the infimum, then there is a larger lower bound $x'$ for $S$. Let $\epsilon = x'-x$; then there is no $s \in S$ such that $s< x+\epsilon = x'.\ _\square$

The concepts of infimum and supremum can be extended to functions on the real numbers:

If $f(x)$ is a real-valued function defined on a subset $D$ of the real numbers, define

$\begin{aligned} \text{inf}(f) &= \text{inf } \{f(x) \colon x \in D \} \\ \text{sup}(f) &= \text{sup } \{f(x) \colon x \in D \}. \end{aligned}$

Then $\text{inf}(f+g) \ge \text{inf}(f)+\text{inf}(g)$ and $\text{sup}(f+g) \le \text{sup}(f)+\text{sup}(g).$

(The proof is left as an exercise.)

## Completeness of the Real Numbers

Constructing the real numbers from scratch is a standard topic in introductory analysis. One way is to start with the integers, then create the rational numbers, and then pass to the real numbers by viewing them as limits of certain types of sequences of rational numbers. This process is called *completion,* as in "$\mathbb R$ is the completion of $\mathbb Q$."

The fundamental property that the real numbers satisfy is called *completeness.* There are several formulations of this property that are logically equivalent. One of them is the **least upper bound property**:

Every nonempty subset of the real numbers with an upper bound has a supremum.

If one uses the notation $\text{sup } S = \infty$ for sets with no upper bound as in remark (4) above, this can be restated "every nonempty subset of the real numbers has a supremum (which may be $\infty$)."

Note that this property is not true for the rational numbers: the set of all rational numbers less than $\sqrt{2}$ has an upper bound that is rational (e.g. 2), but there is no least rational upper bound $\big($there are rational numbers less than $\sqrt{2} + \epsilon$ for any $\epsilon > 0\big).$

The least upper bound property implies many of the basic facts about the real numbers that are used in analysis.

The

intermediate value theoremstates that if $f$ is a continuous function on $[a,b]$ and $y$ is any number between $f(a)$ and $f(b),$ then there is some $c \in [a,b]$ such that $f(c) = y.$To see that this theorem follows from the least upper bound property, suppose without loss of generality that $f(a) \le f(b),$ and consider $S = \{ x \in [a,b] \colon f(z) \le y \text{ for all } z \in [a,x] \}.$ Then $S$ is nonempty since $a \in S,$ and it has an upper bound $($namely $b),$ so there is a least upper bound. Call that least upper bound $c.$

Suppose $f(c) > y.$ Then let $\epsilon = f(c) - y > 0.$ By continuity, there is a $\delta >0$ such that $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon.$ But $|f(x)-f(c)|<\epsilon$ implies $f(x) > y$ for all $x$ in that range, so no $x$'s in that range lie in $S.$ So $c-\delta$ is an upper bound for $S$ as well, which is a contradiction of the "leastness" of $c.$

Suppose $f(c) < y$. Then let $\epsilon = y-f(c) > 0.$ By continuity, there is a $\delta>0$ such that $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon.$ But $|f(x)-f(c)|<\epsilon$ implies $f(x) < y$ for all $x$ in that range, so every $x$ in that range lies in $S.$ So, for instance, $x = c+\frac{\delta}2$ is in $S$, which is a contradiction since $c$ is an upper bound.

The conclusion is that $f(c)=y,$ as desired.

## Lim inf and Lim sup

Let $x_n$ be a sequence of real numbers. For any $k,$ let $S_k = \{ x_n \colon n \ge k \}.$ Then

$\begin{aligned} \liminf_{n\to\infty} x_n &= \lim_{k\to\infty} (\text{inf } S_k) \\\\ \limsup_{n\to\infty} x_n &= \lim_{k\to\infty} (\text{sup } S_k). \end{aligned}$

Let $x_n$ be the sequence $\frac12, -\frac13, \frac14, -\frac15, \ldots$. Then $\text{inf } S_k = -\frac1{2\ell+1},$ where $\ell = \left\lceil \frac k2 \right\rceil,$ and $\text{sup } S_k = \frac1{2m},$ where $m = \left\lfloor \frac{k+1}2 \right\rfloor.$ The limit of both of these expressions as $k\to\infty$ is $0,$ which is also the limit of the sequence. So

$\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n = \lim_{n\to\infty} x_n = 0.$

Note that $\text{inf } x_n = -\frac13$ and $\text{sup } x_n = \frac12.$

Let $x_n$ be the sequence

$2,\ -2^{1/2},\ 2^{1/3},\ -2^{1/4},\ 2^{1/5},\ -2^{1/6},\ \ldots.$

What are the values of

$\inf x_n, \ \ \liminf_{n\to\infty} x_n, \ \ \lim_{n\to\infty} x_n, \ \ \limsup_{n\to\infty} x_n, \ \ \sup x_n \, ?$

**Notation:** In the choices, $\text{DNE}$ means "does not exist."

**Properties:**

(1) Unlike the limit of the sequence, the $\liminf$ and $\limsup$ always exist, if we allow $-\infty$ and $+\infty$ as possible values. This is because the sequence $t_k = \text{inf } S_k$ is a non-decreasing sequence $($similarly the $\text{sup}$ sequence is non-increasing$),$ so its limit either exists or equals $\pm \infty$.

(2) The limit $\lim\limits_{n\to\infty} x_n$ exists if and only if $\liminf\limits_{n\to\infty} x_n = \limsup\limits_{n\to\infty} x_n;$ if the limit exists, all three values are equal.

(3) If $\limsup\limits_{n\to\infty} x_n \ne \infty,$ then it is the smallest real number $s$ such that, for any $\epsilon > 0,$ only finitely many elements of the sequence are $> s+\epsilon.$ Note that it may not be the case that only finitely many elements of the sequence are $> s.$ $\big($For instance, $\limsup\limits_{n\to\infty} \frac1{n} = 0.\big)$ So "every number larger than the $\limsup$ is an eventual upper bound." Similarly, every number smaller than the $\liminf$ is an eventual lower bound.

Let $p_n$ be the $n^\text{th}$ prime number. Then

$\limsup\limits_{n\to\infty} (p_{n+1}-p_n) = \infty.$

Proof:$k! + 2, k!+3, \ldots, k!+k$ are all composite, for any $k \ge 2.$ So for any $k,$ there are infinitely many values of $n$ such that $p_{n+1}-p_n \ge k-1$. $($Take $p_n$ to be the largest prime less than $r!+2$, where $r\ge k$; then the next prime is at least $r-1$ integers away.$)$ So $k-1$ cannot be an eventual upper bound, so it is not larger than the $\limsup.$ Since this is true for all $k$, the result follows.On the other hand,

$\liminf\limits_{n\to\infty} (p_{n+1}-p_n)$

is still unknown. The twin primes conjecture is equivalent to the statement that it equals $2,$ but currently all that is known is that it is at most $246.$ (Until 2013, it was not even known that it was finite!)

## Application to Power Series

Let $\sum\limits_{n=0}^\infty a_nz^n$ be a power series with complex coefficients. Let

$R = \frac1{\limsup_{n\to\infty} |a_n|^{1/n}}.$

$($If the denominator is $0,$ let $R=\infty.$ If the denominator is $\infty,$ let $R=0.)$ Then the series converges if $|z|<R$ and diverges if $|z|>R$. $($If $R=\infty,$ the series always converges.$)$

In this context, $R$ is called the "radius of convergence" for the series.

The series

$z-z^2+z^4-z^8+z^{16}-\cdots$

has radius of convergence $1,$ because $\limsup\limits_{n\to\infty} |a_n|^{1/n}$ equals

$\limsup_{n\to\infty} (0,1,1,0,1,0,0,0,1,0,0,0,0,0,0,0,1,\ldots) = 1.$