Properties of Suprema and Infima

Let $A, B \subseteq \mathbb{R}$ be non-empty with $A \subseteq B$ and $B$ bounded above. Then $A$ is also bounded above and $$\sup A \leq \sup B.$$ Dually: if $B$ is bounded below, then $\inf B \leq \inf A$.

Proof sketch. Every upper bound of $B$ is also an upper bound of $A$ (since $A \subseteq B$), so the least upper bound of $A$ cannot exceed that of $B$. $\square$

For any non-empty $E \subseteq \mathbb{R}$ bounded below: $$\inf E = -\sup(-E), \qquad \text{where } {-E} = \{-x : x \in E\}.$$ Equivalently, $\sup E = -\inf(-E)$ whenever $E$ is bounded above.

Proof sketch. $b$ is a lower bound of $E$ iff $-b$ is an upper bound of $-E$; the greatest lower bound of $E$ corresponds to the negation of the least upper bound of $-E$. $\square$

Let $A, B \subseteq \mathbb{R}$ be non-empty and bounded above. Then $A \cup B$ is bounded above and $$\sup(A \cup B) = \max(\sup A,\, \sup B).$$ Dually: $\inf(A \cup B) = \min(\inf A, \inf B)$ when $A, B$ are bounded below.

Proof sketch. Let $M = \max(\sup A, \sup B)$. Every element of $A \cup B$ belongs to $A$ or $B$, so is $\leq \sup A \leq M$ or $\leq \sup B \leq M$; thus $M$ is an upper bound. For any $\varepsilon > 0$ there exists $a \in A$ with $a > \sup A - \varepsilon$, so $a \in A \cup B$ and $a > M - \varepsilon$ (if $\sup A = M$); the case $\sup B = M$ is symmetric. Hence $M = \sup(A \cup B)$. $\square$

Let $A, B \subseteq \mathbb{R}$ be non-empty and bounded above. Define $A + B = \{a + b : a \in A,\, b \in B\}$. Then $$\sup(A + B) = \sup A + \sup B.$$ Dually: $\inf(A + B) = \inf A + \inf B$ when both are bounded below.

Proof sketch. For all $a \in A$ and $b \in B$: $a + b \leq \sup A + \sup B$, so $\sup A + \sup B$ is an upper bound. For any $\varepsilon > 0$, choose $a \in A$ with $a > \sup A - \varepsilon/2$ and $b \in B$ with $b > \sup B - \varepsilon/2$; then $a + b > \sup A + \sup B - \varepsilon$. $\square$

Let $E \subseteq \mathbb{R}$ be non-empty and bounded above, $c \in \mathbb{R}$. Define $cE = \{cx : x \in E\}$. Then: $$\sup(cE) = \begin{cases} c \cdot \sup E & \text{if } c > 0, \\ c \cdot \inf E & \text{if } c < 0, \\ 0 & \text{if } c = 0. \end{cases}$$

Key insight. When $c < 0$, multiplication reverses the order, so every upper bound becomes a lower bound and vice versa — which is why $\sup(cE) = c \cdot \inf E$ (not $c \cdot \sup E$).

Let $E \subseteq \mathbb{R}$ be non-empty and bounded above. Then for every $\varepsilon > 0$ there exists $x_0 \in E$ such that $$\sup E - \varepsilon < x_0 \leq \sup E.$$ In particular, if $\sup E \notin E$, then $\sup E - \varepsilon < x_0 < \sup E$.

This is exactly the ε-characterisation of supremum rephrased as an existence statement — it is the workhorse of nearly every convergence and continuity proof in analysis.

If $A_1 \subseteq A_2 \subseteq A_3 \subseteq \cdots$ are non-empty sets bounded above, then $$\sup\!\left(\bigcup_{n=1}^{\infty} A_n\right) = \lim_{n \to \infty} \sup A_n = \sup_{n \geq 1} \sup A_n.$$

Proof sketch. By monotonicity (Property 1), $\sup A_n$ is non-decreasing. The union $\bigcup A_n$ contains every $A_n$, so its supremum is at least $\sup A_n$ for all $n$; and every element of $\bigcup A_n$ lies in some $A_n$ and so is bounded by $\sup A_n \leq \sup_n \sup A_n$. $\square$

The crisis of the 1860s. Before Weierstrass, analysts relied on geometric intuition when asserting that a continuous function on a closed interval attains its maximum. Weierstrass was the first to replace this intuition with a purely arithmetic proof — using exactly Properties 1 and 6 above to show that the supremum of $\{f(x) : x \in [a,b]\}$ is actually attained. This 1861 breakthrough launched what historians call the Arithmetisation of Analysis.

The sup-of-union formula. Property 3 seems obvious but its careful proof requires the ε-characterisation. Cauchy had implicitly used it without proof; Weierstrass's student Heine first stated it explicitly in 1872 in the context of uniform continuity.