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By definition, $R$, a relation in a set X, is reflexive if and only if $\forall x\in X$, $x\,R\,x$, and $R$ is symmetric if and only if $x\,R\,y\implies y\,R\,x$.

I think $x\,R\,x$ can also be symmetric when I read the definition, but I also feel there's something wrong or missing in my understanding.

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

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    $\begingroup$ $\leq$, $\geq$, $\subseteq$, ... $\endgroup$ – Pål GD Dec 27 '15 at 20:03
  • $\begingroup$ @BrianO The strict variants are not reflexive. $\endgroup$ – Ari Brodsky Dec 27 '15 at 22:54
  • $\begingroup$ @Ari Totally spaced out - somehow I translated 'reflexive' to "transitive". I'll delete that! Thanks. $\endgroup$ – BrianO Dec 27 '15 at 23:43
  • $\begingroup$ Somewhere, there's a list that shows relations can be any combination of reflexive, symmetric and transitive (despite the famous false proof that symmetric + transitive -> reflexive). $\endgroup$ – barrycarter Dec 28 '15 at 5:16
  • $\begingroup$ Not taller than. $\endgroup$ – Yves Daoust Dec 28 '15 at 9:40
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It's important to remember quantifiers. $R$ is symmetric if and only if $x R y \Rightarrow yRx$ for all $x,y$. Certainly $x Rx \Rightarrow x R x$, but this does not mean $R$ is symmetric.

An example of a relation that is reflexive but not symmetric is $\leq$. For all $x$, $x \leq x$. However, $x \leq y $ does not imply $ y \leq x$ - for example, $1 \leq 2$, but it's not the case that $2 \leq 1$.

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"Knows The Name of" is reflexive but not symmetric.

Everyone knows their own name.

$a \space R \space a$ and $b \space R \space b$

Alan knows Bob's name:

$a \space R \space b$

Bob does not know Alan's name- he is forgetful.

$Not$ $b \space {R} \space a$

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A relation is symmetric if $xRy \implies yRx$ for all $x,y$.

You always know that $xRx \implies xRx$, because it is not possible that $xRx$ is true and $xRx$ is false at the same time. This is regardless of whether $R$ is reflexive or not.

There are plenty of examples:

  • All ordenings on sets with more than one element, in particular $\leq$ on $\mathbb N, \mathbb Z, \mathbb Q, \mathbb R$.
  • $x \mid y$, that is, $x$ divides $y$, on $\mathbb N, \mathbb Z$.
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Somewhere, there's a list that shows relations can be any combination of reflexive, symmetric and transitive (despite the famous false proof that symmetric + transitive -> reflexive). – barrycarter 3 hours ago

Well, I couldn't find one to link to in a few minutes, so let me provide one here.

On the three-element set $\{a, b, c\}$, the following relations are:

  • Transitive, symmetric: $R_0 = \emptyset$
  • Transitive, not symmetric: $R_1 = \{(a,b)\}$
  • Not transitive, not symmetric: $R_2 = \{(a,b), (b,c)\}$
  • Not transitive, symmetric: $R_3 = \{(a,b), (b,a), (b,c), (c,b)\}$

None of the relations above are reflexive, but they can all be turned into reflexive relations, without affecting their transitivity or symmetry, by adding $R^* = \{(a,a), (b,b), (c,c)\}$ to them.

(In particular, $R_1 \cup R^* = \{(a,a), (a,b), (b,b), (c,c)\}$ is a reflexive, non-symmetric relation on the set $\{a, b, c\}$. Of course, the restriction of this relation to the two-element subset $\{a, b\}$ yields an even simpler example.)

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Examples are not that compelling because the conditions are so easy to meet that the general case can be constructed directly. The ones based on $\geq$ or other (partial) orderings to create asymmetry are misleading because they are transitive, a strong extra condition that is not typical of reflexive asymmetric relations.

General construction: take any asymmetric relation and add all the $xRx$ relations needed to make it reflexive. Any reflexive asymmetric relation is of that form.

The picture is of any directed graph, that has a loop at every vertex. To see examples, draw any directed graph, and put a loop at every vertex.

"Has seen" or "has telephoned the house of" are everyday examples.

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