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Based on the articles I've read, given any binary relation $R\subseteq X\times X$ we say:

$$R\text{ is reflexive }\iff\forall x\in X:xRx\iff (\forall a\in X\exists b\in X:aRb)\land (aRb\implies aRa\land bRb)$$

Why not define reflexive so that:

$$R\text{ is reflexive }\iff (aRb\implies aRa\land bRb)$$

The standard definition requires one first specify an external set $X$ but it seems the only useful property one is trying to capture is just that every element related to anything is related to itself.

nathan71
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  • Your definition of a reflexive relation is incorrect: $$R \text{ is reflexive }; \iff \forall a \in X (x R x)$$ meaning (x, x)\in r – amWhy Jun 08 '17 at 23:17
  • @amWhy I understand my definition is not equivalent to the standard definition, my question was why define it the first way if the second definition captures one of the same key properties. – nathan71 Jun 08 '17 at 23:23
  • But your first definition of a reflexive relation is wrong, that's what I'm saying, so anything based on that incorrect definition of reflexive relations is going to be wrong. $R = {(a, a), (b, b), (c, c)}$ is a reflexive relation on $X={a, b, c}$ (i.e.if $aRa. bRb. cRc$, then R is a reflexive.). a, nor b, nor c, needs to be related to any other element than itself. – amWhy Jun 08 '17 at 23:29
  • I did not specify $a=b=c$ though? – nathan71 Jun 08 '17 at 23:33
  • You have not defined the relation. What I wrote is an example of what a reflexive relation might look like. The smallest equivalence relation on a set of $n$ elements is when : $x_1Rx_1, x_2R x_2, \cdots, x_nRx_n$, i.e.$R={(x_1, x_1), (x_2, x_2), \cdots, (x_n, x_n)}$ – amWhy Jun 08 '17 at 23:44
  • A concept which is somewhat related to this question: https://en.wikipedia.org/wiki/Partial_equivalence_relation – Daniel Schepler Jun 08 '17 at 23:46
  • nathan71 Check out this link on equivalence relations. An equivalence relation on a set must satisfy reflexivity, symmetry, and transitivity. The reflexivity requirement means $\forall x \in X(xRx)$. Period. Reflexiivity requires nothing more or less than that every element in a set X is related to itself. – amWhy Jun 09 '17 at 00:01
  • You are proposing is a definition of what we might call subreflexive because it is reflexive on the subset of all elements that are related to something. This is a valid idea, but just don't call it reflexive. – Somos Jun 09 '17 at 01:43

3 Answers3

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Your definition would include the empty relation, and thus it would become an equivalence relation. The nice thing about equivalence relations is that every equivalence relation partitions a set (and conversely, every partition defines an equivalence relation). But the empty relation doesn't partition anything, so we would lose what is probably the most important result about equivalence relations.

Demophilus
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  • But... the empty relation is an equivalence relation on the empty set. –  Jun 15 '17 at 13:21
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Because often we care about the external set. E.g. a partial ordering of a set $X$ is a transitive, reflexive, and asymmetric ($x\le y, y\le x\implies x=y$) relation on $X$. If I only use your "local" version of reflexivity, then in no sense need this partially order $X$! Consider the empty relation (which is clearly transitive and asymmetric, and would be reflexive in your sense): does this "partially order" $X$ in any way?

The fact is that we rarely care about relations on their own, but rather together with the "external" set they're living on; so most of the time we don't want to throw that data away.

Noah Schweber
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It would be nice to capture the definition of reflexivity without referring to the base set. However, reflexivity is actually trying to capture something important about the base set:

A relation $R$ is reflexive $\Leftrightarrow$ for every element $x$ in the base set $X$, we have that $xRx$, or in other words $\langle x, x\rangle \in R$.

This definition is slightly different from "every element that's related to anything is also related to itself". Here's an example that shows why:

Let $X_{all}$ be the set of all positive integers, and let $X_{even}$ be the set of even numbers. Define a relation

$$\begin{eqnarray*} R_1 & \equiv \{\langle x, x\rangle \in X_{all} : x\text{ is even}\}\\ R_2 & \equiv \{\langle x, x\rangle \in X_{even} : x\text{ is even}\}\\ \end{eqnarray*}$$

Now, $R_2 \subseteq X_{even}\times X_{even}$ is reflexive because for every element $x$ in $X_{even}$, $\langle x, x\rangle \in R_2$.

However, $R_1 \subseteq X_{all} \times X_{all}$ is not reflexive: there are elements such as $3\in X_{all}$ which belong to the base set $X_{all}$ but which are not related to themselves.

$R_1$ is a relation where "every element that is related to something is also related to itself", but $R_1$ fails to be a reflexive relation because not every element in the base set is related to itself.

Oh, I think I see now. You're wondering why your alternative definition doesn't capture the idea of reflexivity better than the standard definition—that is, "the relation mentions a few elements that are paired together, and all of the ones mentioned must be related to each other". I think that's a nice minimal definition, actually, because it refers only to the relation $R$ as a set and not to the ambient set $X$ in which it's embedded (as you point out). The standard definition of reflexivity has nice properties, but it could be that relations with your suggested definition have nice properties, too.
user326210
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