Why isn't reflexivity redundant in the definition of equivalence relation?

Question

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

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 07 '14 at 07:10
I have created a canonical Q&A, which is meant to address this question, and others like it. — Xander Henderson, Aug 26 '20 at 22:16
Does this answer your question? Examples and Counterexamples of Relations which Satisfy Certain Properties — , Aug 28 '20 at 14:21
Also problem 12 of section 1.1 of I.N. Herstein's 'Topics in Algebra'. — Saeed, Sep 30 '22 at 04:00

score 99 · Accepted Answer · answered Jul 22 '10 at 04:28

99

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.

answered Jul 22 '10 at 04:28

Akhil Mathew

32,250

2

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 06 '16 at 02:50
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@Muno $R(1, x) \land R(x, 1) \implies R(1, 1)$ does not hold for $x \ne 1$ – DanielV Jun 11 '16 at 05:25
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@Muno I guess your relation is not transitive? – High GPA Dec 03 '20 at 20:41

bryn · Answer 2 · 2024-12-28T05:19:01.013

40

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.

edited Dec 28 '24 at 05:19

answered Jul 22 '10 at 04:36

bryn

10,045

1

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 GONE Apr 21 '16 at 12:55

Steven Alexis Gregory · Answer 3 · 2021-03-04T12:24:19.297

19

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

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

Added 3/4/2021

The argument $aRb \implies bRa \implies aRa$ is compelling except for one thing. What if there is no $b$ such that $aRb$?

edited Mar 04 '21 at 12:24

answered Nov 05 '17 at 23:55

Steven Alexis Gregory

27,619

Shouldn't that say "What if there is no b such that aRb?" say "What if there is no a such that aRb?" Note: the 'a'. – nickalh Aug 13 '22 at 07:55
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No, $a$ is given and you want to prove $aRa$. He argued $aRb \implies bRa \implies aRa$. I asked what if there is no $b$ such that $aRb$? – Steven Alexis Gregory Aug 17 '22 at 15:31

Zarathustra · Answer 4 · 2016-04-21T12:50:27.833

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.

score 2 · Answer 5 · answered Dec 29 '15 at 22:45

2

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?

answered Dec 29 '15 at 22:45

Michael Hoppe

18,614

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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? – silvascientist Dec 29 '15 at 22:53

score 1 · Answer 6 · answered Sep 02 '18 at 14:01

Reflexivity always requires for the binary relation to include every element of the set, which it may not even if transitivity and symmetry are both valid for the relation. However, it may be redundant if every element of the set is said to have a relative or the relation is redefined to be only on the subset of current set leaving out the excluded members for whome the relation doesn't map to anywhere. Although sometimes the definition of the relation itself might restrict the reflexivity, for example: the pair (a,b) where 'a' is b's brother.

Why isn't reflexivity redundant in the definition of equivalence relation?

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