Let $x \sim y$ be defined as meaning that the ordered tuple $(x, y)$ is in some set $S$. If this is the case, an equivalence relation on the set $S$ is defined as a subset of $S \times S$ with the following properties:

  1. For all $x$ in $S$, $x \sim x$.
  2. If $x \sim y$, then $y \sim x$.
  3. If $x \sim y$ and $y \sim z$, then $x \sim z$.

I think this is an over-determined definition, because (1) is a consequence of (2) and (3), as long as it contains anything in addition to the empty set:

Assume that the equivalence relation contains at least point, $x \sim y$. By (2), it contains $y \sim x$. By (3), since $x \sim y$ and $y \sim x$, we also have $x \sim x$. Since we put no constraints on $x$ except that it be a part of the relation, this implies that for any point $x \sim y$, we have $x \sim x$. (We also have $y \sim y$, since $x \sim y$ implies $y \sim x$.) So why do we need to specify (1)?

  • 3
    $\begingroup$ What you have shown is that if $\sim$ satisfies (2) and (3), then for every $x$ such that $x\sim y$ holds for some $y$ we have $x\sim x$. But the statement (1)': "For all $x$ there is some $y$ such that $x\sim y$" is a nontrivial hypothesis on $\sim$. FWIW, on its own (1)' is implied by but not implied by (1), but under the assumptions (2) and (3) they are equivalent. $\endgroup$ – Noah Schweber Sep 27 '19 at 17:49
  • $\begingroup$ Any subset of the diagonal can be recognized as a symmetric and transitive relation. $\endgroup$ – drhab Sep 27 '19 at 18:35

Your error comes when you assume the relation is nonempty. The empty relation satisfies $(2)$ and $(3)$ vacuously, but not $(1)$. In fact the relation could be nonempty and still fail. Let the $S=\{1,2,3\}$ and the relation be $\{(1,1),(2,2)\}$. Again the relation satisfies $(2)$ and $(3)$, but not $(1)$

  • $\begingroup$ After reading your explanation, I see my error: by using x ~ y in the proof, I've only proven that reflexitivity occurs for elements of S that are also first elements in the relation, whereas I would need to prove that reflexitivity occurs for all elements of S. As for the empty set, that's more difficult for me to wrap my head around, since I'm not sure what the empty set looks like in the context of a set crossed with itself. Is it still "phi", or is it {phi, phi}? I'm almost positive it's the former, but it still boggles my mind. $\endgroup$ – JoshG Sep 27 '19 at 18:47
  • $\begingroup$ The empty set is empty. It contains none of the pairs of $S \times S$, so no element is related to any other. Symmetry is stated as $x \sim y \implies y \sim x$, so if the relation is empty, the antecedent is always false and the implication is always true. $\endgroup$ – Ross Millikan Sep 27 '19 at 18:49

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