Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
How do you know that $aRb$? maybe there is no such $b$. – LeeNeverGup Feb 7 '14 at 7:20
up vote 43 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.

share|cite|improve this answer
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 at 2:50


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.

share|cite|improve this answer
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
@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 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?

share|cite|improve this answer
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? – silvascientist 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.