# Empty Set is Well-Ordered

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Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: Q: What does $S$ have to do with anything?A: You can't have a relation without a set for it to be a relation on. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Theorem

Let $S$ be a set.

Let $\RR \subseteq S \times S$ be a relation on $S$.

Let $\O$ denote the empty set.

Let $\RR_\O$ denote the restriction of $\RR$ to $\O$.

Then $\struct {\O, \RR_\O}$ is a well-ordered set.

## Proof 1

We have that $\O$ is well-ordered under $\RR$ if and only if every non-empty subset of $\O$ has a smallest element under $\RR$.

But $\O$ has no non-empty subset.

Hence this condition is satisfied vacuously.

The result follows.

$\blacksquare$

## Proof 2

Let $V$ be a basic universe.

By definition of basic universe, $\O$ is an element of $V$.

By the Axiom of Transitivity, $\O$ is a class.

The result follows from Empty Class is Well-Ordered.

$\blacksquare$