---
tag: K5787SF394
kind: proof
name: Proof of the existence of the infimum by negation
msc: [26A03]
proves: 0EDH403M21
uses: [0AXWET80YW, 04KFTA34C5, 0BTMQ04556, 0B5H2V4JW8, 0QSPNASE50]
sources:
  - url: https://en.wikipedia.org/wiki/Infimum_and_supremum
    locator: Duality
---

## Proof

1. Let $S \subseteq \R$ be nonempty and [[bounded below|0B5H2V4JW8]]. Let $m$ be a lower bound of $S$.
2. In the [[ordered field|0QSPNASE50]] $\R$, $u \le v$ implies $-v \le -u$. Indeed, add $-u - v$ to both sides.
3. Let $-S = \set{-s : s \in S}$. It is nonempty, because $S$ is nonempty.
4. For each $s \in S$ we have $m \le s$, so $-s \le -m$ by step 2. Thus $-m$ is an upper bound of $-S$.
5. By the [[completeness axiom|0AXWET80YW]], $-S$ has a [[supremum|0BTMQ04556]] $b$. Let $a = -b$.
6. The number $a$ is a lower bound of $S$. For $s \in S$, we have $-s \in -S$, so $-s \le b$. By step 2, $a = -b \le s$.
7. Let $\ell$ be any lower bound of $S$. As in step 4, $-\ell$ is an upper bound of $-S$.
8. Since $b$ is the least upper bound, $b \le -\ell$. By step 2, $\ell \le -b = a$.
9. By steps 6 and 8, $a$ is a lower bound of $S$ that is larger than or equal to each lower bound. So $a$ is an [[infimum|04KFTA34C5]] of $S$.

## Idea

Negation reverses the order, so it turns lower bounds into upper bounds. The supremum of the negated set exists by completeness, and its negative is the infimum.