---
tag: 04ETX30HT2
kind: statement
role: proposition
name: Uniqueness of the supremum
msc: [06A06]
uses: [0BTMQ04556]
sources:
  - url: https://en.wikipedia.org/wiki/Infimum_and_supremum
    locator: Existence and uniqueness
---

## Statement

Let $(P, \le)$ be a partially ordered set and let $S$ be a subset of $P$.
If $b$ and $b'$ are both [[suprema|0BTMQ04556]] of $S$, then $b = b'$.

## Remarks

Proof: $b$ is an upper bound and $b'$ is a least upper bound, thus $b' \le b$. In the same way $b \le b'$, and antisymmetry gives $b = b'$.
The same argument shows that a set has at most one infimum.

## Explanations

### L1 human

A set cannot have two different least upper bounds.
If it had two, each one would be at most the other one, so they would be equal.