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

Question

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.

Answer 1

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.

Answer 2

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 $X$.

Answer 3

$x$ 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.

Answer 4

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 $(2,2)\notin R$.

Answer 5

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).