Well-Ordering Minimal Elements are Unique

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Theorem

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

Then every non-empty subset of $S$ has a unique minimal element.


Proof

The proof consists of a uniqueness and an existence part.

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


Existence

As $S$ is well-ordered by $\preceq$, we have by definition existence of a minimal elements $s$ of $S'$ (with respect to $\preceq$).


Uniqueness

Let $s, s'$ be minimal elements of $S'$.

By Well-Ordering is Total Ordering, we have at least one of $s \preceq s'$, $s'\preceq s$.

Since both $s$ and $s'$ are minimal elements, we conclude that necessarily $s = s'$.

$\blacksquare$


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