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Why can't {(a,a)} be an equivalence relation for the set {a,b,c}?

{(a,a)} is reflexive, symmetric, and transitive.

user328442
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kgui
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2 Answers2

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It's not reflexive because it doesn't contain (b,b) and (c,c)

Alex Mathers
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It's not reflexive unless it contains $(x,x)$ for all $x \in\{a,b,c\}$. It doesn't contain either $(b,b)$ or $(c,c)$ so it's not reflexive.

Brian Moehring
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  • So it has to contain all elements from the original set but, it doesn't have to contain for e.g. (a,b), (b,a)? – kgui Nov 30 '19 at 03:06
  • @crt Right. An equivalence relation is a generalization of $=$. In that case, we may have $a=a, b=b, c=c$ but all other pairs are unequal. – Brian Moehring Nov 30 '19 at 03:10
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    Correct, in a sense. For a relation $R$ on a set $S$ to be reflexive, we do require that $(s,s) \in R$ for every single $s\in S$. However, symmetry doesn't have that same requisite: $R$ is symmetric when, if $(a,b) \in R$, then $(b,a) \in R$. This doesn't mean that every possible pair $(a,b)$ and its reversal $(b,a)$ have to be in there, only that if there is a pair $(a,b)$ in $R$, then so is its reversal $(b,a)$. – PrincessEev Nov 30 '19 at 03:11