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[This is a self-answered question. Please read both the question and the answer before thinking about closing as a duplicate.]

I've seen the statement below, which says that the reflexivity assumption in the definition of an equivalence relation is unnecessary:

Proposition: Let $S$ be a set, and $R$ a relation on $S$ that is both symmetric and transitive. Then $R$ is reflexive.

The "proof" goes: Take an arbitrary $x \in S$, and take $y \in S$ such that $xRy$. Then by symmetry $yRx$. And transitivity says that $xRy$ and $yRx$ imply that $xRx$. Since $x$ was arbitrary, $R$ is reflexive.

What is wrong with this proof? And how do I find a counter-example?

JonathanZ
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  • Note: The two existing question that are the best duplicate candidates that I found are this and this. I thought the characterization in the answer was nice enough that posting a new Q+A was worth it. (And I didn't find it in either of them, although I could have missed it). I guess I'll find out if others feel that way too. – JonathanZ May 25 '23 at 02:40
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    The answer in your first link explains the issue clearly and succinctly. I don’t see anything new in your answer except the $xy>0$ example. Is there something else in your answer you think is new? – Michael Weiss May 25 '23 at 02:53
  • @MichaelWeiss - I think the characterization of all such counter examples as an equivalence unioned with an empty relation is not in any of those answers. I didn't know how, or where, best to add this, so i wrote this up. I obviously think it's pretty neat, but it may just be the pleasure of discovery. – JonathanZ May 25 '23 at 03:11
  • I felt this was obvious from the second paragraph of that answer, which states exactly when and why we can fail to have $xRx$ with $R$ transitive and symmetric. Just my opinion. I’m not downvoting or voting to close. – Michael Weiss May 25 '23 at 03:58

1 Answers1

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The error in the proof lies in the statement "... and take $y \in S$ such that $xRy$", as it assumes there is such a $y$, which we have no guarantee of.

As the rest of the proof is okay, this gives us guidance as to how to build a counterexample. There must be an $x$ that is not $R$-related to any other element.

Once you realize this, it's easy to build counterexamples, but the one that I think has the most simple, elegant statement (say for $\mathbb Z$ or $\mathbb R$) is "$xRy \iff x\cdot y \gt 0$." It's probably easier to see that this is a counterexample if you think of the condition "$x\cdot y\gt 0$" as "$x$ and $y$ have the same sign, but $0$ doesn't even have a sign".

A nice thing to notice is that if you remove $0$ from the set, you do have an equivalence relation. An even nicer thing to observe is that this is the only way such counterexamples arise. To wit,

Proposition: Let $S$ be a set, and $R$ a relation on $S$ that is both symmetric and transitive. Then we can write $S$ as a disjoint union, $ S = S_0 \cup S_1$, so that

  • $R$ restricted to $S_0$ is the empty relation, and

  • $R$ restricted to $S_1$ is an equivalence.

(By "$R$ restricted to $S_0$" I mean $R \cap (S_0\times S_0)$, when we consider the relation $R$ as a subset of $S \times S$).

Proof: Define $S_0$ as $\{x \in S \mid \text{there doesn't exist }y \text{ such that } x R y\}$. As $R$ is symmetric, $S_0$ is also equal to $\{x \in S \mid \text{there doesn't exist }y \text{ such that } y R x\}$, and thus $R$ is the empty relation on $S_0$.

Setting $S_1 = S \setminus S_0$, we can see that for elements in $S_1$ the invalid "proof" presented in the question actually goes through, as elements in $S_1$ are $R$-related to at least one element. Q.E.D.

So we can generate all the counterexamples we want (and all the counterexamples possible) by taking an equivalence relation on some set, and then adding a bunch of new elements to the set and saying they are not related to any other elements, or even themselves.

JonathanZ
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