An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity.

Doesn't symmetry and transitivity implies reflexivity? Consider the following argument.

For any $a$ and $b$, $a R b$ implies $b R a$ by symmetry. Using transitivity, we have $a R a$.

Source: Exercise 8.46, P195 of Mathematical Proofs, 2nd (not 3rd) ed. by Chartrand et al

  • 3
    This is a good problem to pose in an introductory Discrete Math/Logic course. – Srivatsan Jul 31 '11 at 16:23
  • @LePressentiment Why would you add a random source to a three and half year old question?? This is a standard exercise that you can find in many many books. – mrf Feb 7 '14 at 7:10
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    How do you know that $aRb$? maybe there is no such $b$. – LeeNeverGup Feb 7 '14 at 7:20
up vote 62 down vote accepted

Actually, without the reflexivity condition, the empty relation would count as an equivalence relation, which is non-ideal.

Your argument used the hypothesis that for each $a$, there exists $b$ such that $aRb$ holds. If this is true, then symmetry and transitivity imply reflexivity, but this is not true in general.

  • Why is the following not a counter-example that even with seriality, and transitivity and symmetry, a relation can fail to be reflexive on some set: Consider $S = \{1, 4, 5, 6\}$ and the relation $R$ where $(x,y) \in R$ if $x + y > 3$. On $S$, this relation is transitive, symmetric, and seriality holds. So reflexivity should too. Yet 1 + 1 $\not > 3$. – Muno Apr 6 '16 at 2:50
  • @Muno $R(1, x) \land R(x, 1) \implies R(1, 1)$ does not hold for $x \ne 1$ – DanielV Jun 11 '16 at 5:25

No.

The missing condition is sometimes called 'seriality' -- for any x there must be a y such that x R y.

If you add seriality to the symmetry and transitivity you get a reflexive relation again.

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    I googled seriality, not much came up. You sure it's spelled right? – Chao Xu Jul 22 '10 at 10:40
  • Perhaps I made up the noun form. If you google 'serial binary relation' it will come up. Although if you google 'seriality of binary relations' it will come up there too. – bryn Jul 22 '10 at 10:42
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    @ChaoXu ProofWiki uses the name serial relation. – Martin Sleziak Jul 10 '12 at 12:37
  • @ChaoXu, $R$ can also be called "total" or "entire". Note that $R$ doesn't have to be an endorelation for totality to make sense. – goblin Apr 21 '16 at 12:55

You can get kind of rid of reflexivity. Assume the $R$ is a symmetric relation which satisfies the property of "Drittengleichheit": $xRz\land yRz\Rightarrow xRy$. In this case $R$ is a equivalence relation; you can easily deduce transitivity and reflexibility of $R$.

Do you notice the difference between "Drittengleichheit" and transitivity?

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    How does this rule out the empty relation? More generally, how does this rule out relations for which some elements are not related to anything at all? – Monstrous Moonshiner Dec 29 '15 at 22:53

To answer Muno's question, $R$ on $S$ is not transitive: $x=z=1$ and $y=3$ is a counterexample (any $y$ other than $1$ will do). $x+y=y+z>3$ but $x+z<3$.

silvascientist: $R$ satisfies Drittengleichheit if the implication is true; it is vacuously true if $xRz$ and $yRz$ is false. This does not mean that $xRy$ is true (similarly $xRy$ and $xRy$ false does not mean $xRx$ is true, simply that it vacuously satisfies the property) for the empty relation on a non-empty set. The same applies if $x$ isn't related to anything, so in neither case is $R$ reflexive.

Consider the set $S =\{a,b,c\}$, with $a, b$, and $c$ distinct, and the relation $$R = \{(a,b),(b,a),(a,a),(b,b)\} \subset S \times S$$

It's symmetric and transitive but it's not reflexive.

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