# Example of a relation that is symmetric and transitive, but not reflexive

Can you give an example of a relation that is symmetric and transitive, but not reflexive?

By definition,

1. $$R$$, a relation in a set $$X$$, is reflexive if and only if $$\forall x\in X$$, $$x\,R\,x$$.

2. $$R$$ is symmetric if and only if $$\forall x, y\in X$$, $$x\,R\,y\implies y\,R\,x$$.

3. $$R$$ is transitive if and only if $$\forall x, y, z\in X$$, $$x\,R\,y\land y\,R\,z\implies x\,R\,z$$.

I can give a relation $$\leqslant$$, in a set of real numbers, as an example of reflexive and transitive, but not symmetric. But I can't think of a relation that is symmetric and transitive, but not reflexive.

• – Martin R Dec 29 '15 at 13:21
• What about the empty relation? – Improve Dec 29 '15 at 13:22
• Also related: math.stackexchange.com/questions/1482542/… – Aloizio Macedo Dec 29 '15 at 17:58
• If x R y then y R x (sym) so x R x (transitive). So to be symmetric and transitive but not reflexive no elements can be related at all. That's be the empty relationship. – fleablood Dec 30 '15 at 0:37
• @fleablood One element related to none, suffices, if R restricted to the rest is symmetric and transitive. – Did Dec 16 '16 at 7:59

Take $X=\{0,1,2\}$ and let the relation be $\{(0,0),(1,1),(0,1),(1,0)\}$

This is not reflexive because $(2,2)$ isn't in the relation.

Addendum: More generally, if we regard the relation $R$ as a subset of $X\times X$, then $R$ can't be reflexive if the projections $\pi_1(R)$ and $\pi_2(R)$ onto the two factors of $X\times X$ aren't both equal to $X$.

$x$ works at the same place as $y$ (defined on the set of all people).

Certainly it is symmetric and transitive, but it is not reflexive, as a guy without a job isn't related to himself.

• Doesn't reflexive relation "x works at the same place as x" hold true if x works? – buzzee Dec 29 '15 at 15:01
• @buzzee yes, and that is exactly the point! For the relation to be reflexive, it must be true that $x$ works the same place as $x$ for every person $x$. But it isn't true for every $x$. As you remark, it is only true, if $x$ works. – Mankind Dec 29 '15 at 15:43
• Then if we say every x works at the same place as x, it's true. – buzzee Dec 29 '15 at 16:32
• @buzzee if every person $x$ works, then my relation would be reflexive, yes. But it isn't, the reason being that there's lots of people, who don't work. But anyway, I like Mauro ALLEGRANZA's example even better: $x$ is a brother of $y$ is not reflexive. – Mankind Dec 29 '15 at 16:38
• (his answer seems to have disappeared now though) – Mankind Dec 29 '15 at 17:21

Take a set where no element is in relation with the other ones.

P.S. If $$xRy$$, then $$yRx$$ by symmetry, hence $$xRx$$ by transitivity. In particular, reflexivity holds in all points in relation with something other.

Or, for real numbers $x$ and $y$, let $xRy$ iff the product $xy$ is strictly positive, in other words for all $x,y\in\mathbb{R}$: $$xRy\quad\Longleftrightarrow\quad xy>0$$ This is not reflexive because the number zero does not produce a strictly positive product with itself.

If, in the spirit of BrianO's answer, we take away the set of points failing to relate to anything, we are left with $\mathbb{R}\setminus\{ 0 \}$ on which set this $R$ is an equivalence relation, the equivalence classes being the set of the two signs, positive and negative. An equivalent definition of the same $R$ is: $$xRy\quad\Longleftrightarrow\quad \frac{x}{|x|}=\frac{y}{|y|}$$ which can also be used in $\mathbb{R}^n$ (or other normed vector spaces if you know what that is).

• +1  I was reading the other answers, and I was just going to suggest the relation: for all $x,y\in\mathbb{R}$, $x\,R\,y\quad\Longleftrightarrow\quad x \times y \ne 0$.  Your answer is better because is can be described more succinctly ($x$ and $y$ have the same sign), and it’s less trivial, in that it yields two equivalence classes when applied to $\mathbb{R}\setminus\{0\}$  (with mine, all of $\mathbb{R}\setminus\{0\}$ is equivalent). – Scott Dec 29 '15 at 21:07

If $R$ is a relation that is transitive and symmetric, then $R$ is reflexive on $dom(R) = \{a\mid (\exists b)\, aRb\}$: if $a\in dom(R)$, then there is $b$ such that $aRb$, thus $bRa$ by symmetry, so $aRa$ by transitivity.

Note that if $R$ is symmetric, then $dom(R) = range(R) = \{b\mid (\exists a)\, aRb\}$.

Hence, to get an example of a relation $R$ on a set $A$ that is transitive and symmetric but not reflexive (on $A$), there has to be some $a\in A$ which is not $R$-related to any $b\in A$. There are many examples of this:

• $A = \{0, 1\}$ and $R= \{(0,0)\}$,

not reflexive on $A$ because $(1,1)\notin R$,

• $A = \{0, 1, 2\}$ and $R= \{(0,0), (0,1), (1,0), (1,1)\}$,

not reflexive on $A$ because $(2,2)\notin R$.

I had the same question as you did but as soon as I read yours I noticed where the confusion came from: it lied in the if and only if part of the definition of symmetry.

The point is that for a relation $R$ to be reflexive $aRa$ has to hold for each and every element just like you have stated in the definition. But the definition of of symmetry reads: if and only if $aRb \implies bRa$ which means if no such relation exists, you can't use it. So if $\exists a \vert \forall b \in X, a\not R b$ (which means there is at least one element that is not in relation with any other) then you can't use the definition of symmetry to prove $aRa$.

This doesn't necessarily mean the relation is not symmetric, it only means you can't prove it using the given facts. If only we were given the fact that $R$ is a serial relation we would be able to prove it. So if you take @MPW's case and add $(2,2)$ to it you still wouldn't be able to prove reflexivity. That's because $2$ is not in relation with any other element.

This is called a “partial equivalence relation (PER)”. PERs can be used to simultaneously quotient a set and imbue the quotiented set with a notion of equivalence.

A genuinely useful example (copied straight from the linked page) is functions that respect equivalence relations of the domain and codomain. Suppose we have sets $X$ and $Y$, with (partial or otherwise) equivalence relations $\approx_X$ and $\approx_Y$, respectively. We define the partial equivalence relation $\approx_{X \to Y}$ by $$f \approx_{X \to Y} f' := \forall x,x' \in X. x \approx_X x' \implies f(x) \approx_Y f'(x')$$ Then, $\{~f~|~f \approx_{X \to Y} f~\}$ are exactly the functions that preserve equivalence. Furthermore, $\approx_{X \to Y}$ is an equivalence relation on that set, equating $f$ and $f'$ exactly when they map the same inputs (up to $\approx_X$) to the same outputs (up to $\approx_Y$). This is true even if equality of functions in the underlying set theory is more intensional – hence the use of this technique in type theory.