Infima via the Completeness Axiom

The sup/inf duality proved here is not a one-off curiosity; it is the quiet engine behind large parts of analysis. The definitions of $\limsup$ and $\liminf$ of a sequence are literal infima and suprema of tail sets, and every theorem about $\limsup$ has a mirror-image theorem about $\liminf$ obtained by exactly this reflection trick. In measure theory, the outer measure of a set is defined as an infimum over covering sums, and existence of that infimum in $\mathbb{R}$ (or $[0,\infty]$) is what makes Lebesgue measure well-defined. In optimization and control theory, dual problems routinely convert a minimization (infimum) into a maximization (supremum) of a related quantity — the reflection $x \mapsto -x$ underlies "weak duality" arguments in linear and convex programming. In real analysis more broadly, the Bolzano–Weierstrass theorem, the extreme value theorem, and the construction of the Riemann integral (via infima of upper sums and suprema of lower sums) all lean on both halves of completeness being available without a second axiom. In computer science, algorithms for interval arithmetic and floating-point error bounds implicitly assume both bound types exist and interact via negation exactly as proved here.

Ask the tutor: Can you show me how the same reflection trick proves $\inf(A \cup B) = \min(\inf A, \inf B)$ from the analogous supremum fact?

Picture a nonempty set $S$ sitting on the real line, bounded below by some $m$. Its infimum, if it exists, is the point you'd get by "closing in" from the left as tightly as possible without passing any point of $S$ — the largest number that still stays $\le$ every element of $S$. Now imagine placing a mirror at $0$ and reflecting every point of $S$ to its negative. Each point $x \in S$ becomes $-x$, landing in the new set $-S$. A key structural fact of the order on $\mathbb{R}$ is that reflection reverses inequalities: $x \ge m$ for all $x\in S$ becomes $-x \le -m$ for all $-x \in -S$. So a lower bound $m$ of $S$ becomes an upper bound $-m$ of $-S$, and vice versa. Bounded-below sets become bounded-above sets under reflection. This means the completeness axiom — which only talks about suprema — can be applied to $-S$. It hands us $\sup(-S)$, the least upper bound of the mirrored set. Reflecting this single number back across the mirror should give exactly the greatest lower bound of the original $S$, because reflection is an order-reversing bijection and it turns "least upper bound" statements into "greatest lower bound" statements. That is the entire idea; the proof below simply verifies every inequality carefully.

Example 1 — computing an infimum by the reflection formula, asymmetric set. Let

So $S$ contains the isolated point $-3$, the increasing sequence $1, 1.5, 1.667,\dots \to 2$, and the decreasing sequence $8, 7.5, 7.333,\dots \to 7$. We compute $\inf S$ directly and cross-check with the reflection formula. Direct route. The element $-3 \in S$, and for every $n$, $2 - \tfrac1n > 1 \ge -3$ (in fact $2-\tfrac1n \ge 1 > -3$) and $7 + \tfrac1n > 7 > -3$. So $-3$ is a lower bound, and since $-3 \in S$ itself, no larger number can be a lower bound (a lower bound must be $\le$ every element, including $-3$ itself). Hence

attained as a minimum. Reflection route. Compute

The first family lies in $(-2,-1]$ approaching $-2$ from above (values $-1, -1.5, -1.667,\dots \to -2$), the third family lies below $-7$ approaching $-7$ from below, and $3 \in -S$. The largest element present is $3$ itself (since all $\tfrac1n - 2 \le -1 < 3$ and all $-7-\tfrac1n < -7 < 3$), so $3$ is an upper bound and, being attained, is the least one:

By the theorem, $\inf S = -\sup(-S) = -3$, matching the direct computation. Note also that $\inf S = -3$ is attained (it's a minimum), while $\sup S = 8$ is also attained; this example was chosen so that the bound is realized inside the set, to make direct verification airtight, while still having infinitely many "tail" elements to keep the set genuinely interesting.

Example 2 — an infimum that is not attained, requiring the full strength of completeness. Let

We claim $\inf T = 1$, and $1 \notin T$. Lower bound. For every $n \ge 1$, $\frac{n+1}{n} = 1 + \frac1n > 1$, so $1$ is a lower bound of $T$ (indeed a strict one, since equality never holds — if $1 + \frac1n = 1$ then $\frac1n = 0$, impossible for $n\in\mathbb Z^+$ since $n \cdot 0 = 0 \ne 1$). Greatest lower bound, via the theorem's method reflected back into a direct $\epsilon$-argument. Suppose $m$ is any lower bound of $T$ with $m > 1$. Then $m \le 1+\frac1n$ for every $n$ forces (taking the negation-free version) $\frac1n \ge m - 1 > 0$ for all $n\in\mathbb Z^+$, i.e. $n \le \frac{1}{m-1}$ for all $n \in \mathbb{Z}^+$ — but $\mathbb{Z}^+$ is unbounded above (this is exactly the content of the forthcoming Archimedean Property, which itself is proved using the least-upper-bound property), giving a contradiction. Hence no lower bound exceeds $1$, so $1$ is the greatest lower bound:

This example illustrates why the theorem needed a genuine existence proof rather than a guess: the infimum here is a limit point not belonging to the set, and pinning it down as "the" greatest lower bound uses the full deductive apparatus (here, ultimately, the Archimedean property, itself downstream of completeness) — not just inspection of the set's elements as in Example 1. We can also re-derive $\inf T = 1$ purely via the reflection formula, as a consistency check. We have $-T = \{-2, -\tfrac32, -\tfrac43, \dots\}$, an increasing sequence approaching $-1$ from below, with $-1$ never attained. By the completeness axiom applied to $-T$ (bounded above by, say, $0$), $\sup(-T)$ exists; the same $\epsilon$/Archimedean-style argument shows $\sup(-T) = -1$, and then $\inf T = -\sup(-T) = -(-1) = 1$, agreeing with the direct computation.