Redundancy of reflexive property

Question

Why isn't the reflexive property redundant we defining an equivalence class? If $x \sim y$, then $y \sim x$ by the symmetric property. Using the transitive property we can deduce that $x \sim x$.

What if there is no $y$ with $x \sim y$? – Jun 01 '16 at 16:19 — , Jun 01 '16 at 16:19

score 3 · Answer 1 · answered Jun 01 '16 at 16:21

A counterexample: let $R$ be the relation on the set $X=\{ a, b , c\}$ defined by $$R= \{ (a,a), (b,b) , (a,b), (b,a)\}$$ This is not an equivalence relation because it is not reflexive ($(c,c) \notin R$!). By the way, $R$ is clearly symmetric and transitive.

Here what fails is the existence of some element $x \in X$ satisfying $(c,x) \in R$. Such an element would allow you to use the argument $cRx, xRc \Rightarrow cRc$; but there is none, so that we can't say anything.

score 1 · Answer 2 · answered Jun 01 '16 at 18:31

1

As long as every element is related to at least one element a symmetric, transitive relation is reflexive.

answered Jun 01 '16 at 18:31

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Redundancy of reflexive property

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