Smallest Element is Unique

Theorem

Let $\struct {S, \preceq}$ be an ordered set.

If $S$ has a smallest element, then it can have only one.

That is, if $a$ and $b$ are both smallest elements of $S$, then $a = b$.

Let $V$ be a basic universe.

Let $\RR \subseteq V \times V$ be an ordering.

Let $A$ be a subclass of the field of $\RR$.

Suppose $A$ has a smallest element $s$ with respect to $\RR$.

Then $s$ is unique.

Let $a$ and $b$ both be smallest elements of $S$.

Then by definition:

$\forall y \in S: a \preceq y$

$\forall y \in S: b \preceq y$

Thus it follows that:

$a \preceq b$

$b \preceq a$

But as $\preceq$ is an ordering, it is antisymmetric.

Hence by definition of antisymmetric, $a = b$.

$\blacksquare$

1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 14$: Order
1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 2.3$
1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $5$
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): null element
2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations
2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.11$
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): null element