Supremum & Infimum

You know that $\mathbb{R}$ is complete — every Cauchy sequence converges. But completeness has an equivalent face that is often more convenient in practice: every non-empty set of reals that is bounded above has a smallest upper bound. That bound is called the supremum, and its mirror image for lower bounds is the infimum. These two concepts underpin the entire structure of real analysis.

Bounds

Let $S \subseteq \mathbb{R}$ be a non-empty set.

A number $M \in \mathbb{R}$ is an upper bound of $S$ if

$$\forall s \in S,\quad s \leq M.$$

A number $m \in \mathbb{R}$ is a lower bound of $S$ if

$$\forall s \in S,\quad m \leq s.$$

If $S$ has at least one upper bound it is bounded above; if it has at least one lower bound it is bounded below. A set that is bounded above and below is simply called bounded.

Note that bounds are not unique. If $M$ is an upper bound of $S$, so is $M + 1$, $M + 100$, and any larger number. The interesting object is the tightest bound.

Supremum and infimum

The supremum of $S$, written $\sup S$, is the least upper bound of $S$: it is an upper bound of $S$ and is $\leq$ every other upper bound of $S$.

Unpacking the definition: $\sup S = \alpha$ means

1. $\forall s \in S,\; s \leq \alpha$ (upper bound), and
2. $\forall \varepsilon > 0,\; \exists\, s \in S \text{ such that } s > \alpha - \varepsilon$ (no smaller upper bound works).

Condition 2 is the characteristic property of the supremum: you cannot lower $\alpha$ by even $\varepsilon$ without losing the upper-bound property.

The infimum of $S$, written $\inf S$, is the greatest lower bound of $S$: it is a lower bound of $S$ and is $\geq$ every other lower bound. Dually, $\inf S = \beta$ means

1. $\forall s \in S,\; \beta \leq s$, and
2. $\forall \varepsilon > 0,\; \exists\, s \in S \text{ such that } s < \beta + \varepsilon$.

Both $\sup S$ and $\inf S$ are unique when they exist: if $\alpha$ and $\alpha'$ are both least upper bounds then $\alpha \leq \alpha'$ and $\alpha' \leq \alpha$, so $\alpha = \alpha'$.

The least upper bound property

The existence of $\sup S$ is not automatic — it depends on the number system. In $\mathbb{Q}$, the set $\{q \in \mathbb{Q} : q^2 < 2\}$ is bounded above (by $2$, for instance) yet has no supremum in $\mathbb{Q}$, because $\sqrt{2} \notin \mathbb{Q}$.

In $\mathbb{R}$, the situation is perfect:

Theorem (Least Upper Bound Property). Every non-empty subset of $\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$.

Proof sketch. Because $\mathbb{R}$ is complete, every Cauchy sequence of reals converges in $\mathbb{R}$. From completeness one can show: for a non-empty $S$ bounded above, consider the binary search sequence that tightens an interval $[a_n, b_n]$ where $a_n \in S$ (or near $S$) and $b_n$ is always an upper bound. The interval lengths shrink to zero, so the endpoints form Cauchy sequences converging to a common limit, which is the supremum. $\square$

By symmetry (negate all elements), every non-empty subset bounded below has an infimum in $\mathbb{R}$.

The least upper bound property is logically equivalent to the completeness of $\mathbb{R}$: either can be taken as the defining axiom and the other derived. Together they express the same fact — $\mathbb{R}$ has no holes.

Examples

Closed interval $[a, b]$

$$\sup [a, b] = b, \qquad \inf [a, b] = a.$$

Both bounds are attained: $b \in [a, b]$ and $a \in [a, b]$.

Open interval $(a, b)$

$$\sup (a, b) = b, \qquad \inf (a, b) = a.$$

Neither bound is attained: $b \notin (a, b)$ and $a \notin (a, b)$. The supremum exists in $\mathbb{R}$ even though no element of the set reaches it — the set approaches $b$ arbitrarily closely but never touches it.

Harmonic sequence

Let $S = \left\{\dfrac{1}{n} : n \in \mathbb{N},\; n \geq 1\right\} = \left\{1,\, \tfrac{1}{2},\, \tfrac{1}{3},\, \ldots\right\}$.

$$\sup S = 1 \in S, \qquad \inf S = 0 \notin S.$$

The infimum $0$ is not achieved: every element $1/n > 0$, but for any $\varepsilon > 0$ you can choose $n > 1/\varepsilon$ to get $1/n < \varepsilon$, confirming $0$ is the greatest lower bound.

Unbounded sets

The set $\mathbb{N}$ is bounded below ($\inf \mathbb{N} = 0$) but unbounded above: no finite $M$ is an upper bound, so $\sup \mathbb{N}$ does not exist in $\mathbb{R}$. It is conventional to write $\sup \mathbb{N} = +\infty$.

Sup and inf in the set vs outside it

A key insight from the examples: the supremum of $S$ may or may not belong to $S$.

• $\sup S \in S$ if and only if $S$ has a maximum (greatest element).
• $\inf S \in S$ if and only if $S$ has a minimum (least element).

Every set has at most one maximum and one minimum; if either exists it equals the corresponding sup or inf. But a set can have a supremum without having a maximum — as $(a, b)$ illustrates.

The Archimedean property

A classical consequence of the least upper bound property is:

Theorem (Archimedean property). For every $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ with $n > x$.

Proof. Suppose for contradiction that some $x \in \mathbb{R}$ is an upper bound for $\mathbb{N}$. Then $\mathbb{N}$ is a non-empty bounded-above subset of $\mathbb{R}$, so by the LUB property it has a supremum $\alpha = \sup \mathbb{N}$. Since $\alpha - 1$ is not an upper bound of $\mathbb{N}$, there exists $n \in \mathbb{N}$ with $n > \alpha - 1$, i.e.\ $n + 1 > \alpha$. But $n + 1 \in \mathbb{N}$, contradicting $\alpha$ being an upper bound. $\square$

An equivalent formulation: for any $\varepsilon > 0$, there exists $n \in \mathbb{N}$ with $1/n < \varepsilon$. This is used constantly in analysis to show that quantities can be made arbitrarily small.

Useful reformulations

The following are all equivalent to $\alpha = \sup S$, and each is useful in different proofs:

• $\alpha$ is an upper bound, and for every $\varepsilon > 0$ the interval $(\alpha - \varepsilon, \alpha]$ contains a point of $S$.
• $\alpha$ is an upper bound, and there is a sequence $(s_n)$ in $S$ with $s_n \to \alpha$.
• $\alpha = \min\{M \in \mathbb{R} : M \text{ is an upper bound of } S\}$.

The second reformulation — a sequence in $S$ converging to the supremum — is often the most useful in proofs, connecting the sup/inf language back to sequences and limits.

Summary

• An upper bound of $S$ satisfies $s \leq M$ for all $s \in S$; a lower bound satisfies $m \leq s$ for all $s \in S$.
• The supremum $\sup S$ is the least upper bound; the infimum $\inf S$ is the greatest lower bound. Both are unique when they exist.
• The characterisation: $\alpha = \sup S$ iff $\alpha$ is an upper bound and for every $\varepsilon > 0$ some $s \in S$ satisfies $s > \alpha - \varepsilon$.
• The least upper bound property: every non-empty $S \subseteq \mathbb{R}$ bounded above has $\sup S \in \mathbb{R}$. This is equivalent to the completeness of $\mathbb{R}$.
• The supremum may or may not lie in $S$: $\sup S \in S$ iff $S$ has a maximum.
• The Archimedean property follows: for any $x \in \mathbb{R}$ there exists $n \in \mathbb{N}$ with $n > x$, so $\mathbb{N}$ is unbounded above and $1/n$ can be made arbitrarily small.