# Supremum is Unique

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## Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.

Let $T$ be a non-empty subset of $S$.

Then $T$ has at most one supremum in $S$.

## Proof

Let $c$ and $c'$ both be suprema of $T$ in $S$.

From the definition of supremum, $c$ and $c'$ are upper bounds of $T$ in $S$.

By that definition:

- $c$ is an upper bound of $T$ in $S$ and $c'$ is a supremum of $T$ in $S$ implies that $c' \preceq c$
- $c'$ is an upper bound of $T$ in $S$ and $c$ is a supremum of $T$ in $S$ implies that $c \preceq c'$.

So:

- $c' \preceq c \land c \preceq c'$

and thus by the antisymmetry of the ordering $\preceq$:

- $c = c'$

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 14$: Theorem $14.3$ - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $\S 1.1$: Real Numbers