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Say we have a set $X$ and an equivalence relation $C$ on $X$. Why do we need reflexivity?

Let $x,y \in X$ with $xCy $. By symmetry we obtain $yCx$.

Applying now transitivity, we have $xCx$.

So, we have reflexivity from symmetry and transitivity.

mrf
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ILoveMath
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    Does such a $y$ exist for all $x$? – user2520938 Nov 20 '14 at 07:35
  • @user2520938: Does this mean that reflexivity only deals with equivalence classes containing a single element? Without reflexivity those would not be equivalence classes at all ... – String Nov 20 '14 at 08:57

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You first need to assume there exists, $xCy$. Indeed it's not necessarily true. Think about an extreme cases that $X\neq \emptyset$ and define a relation $C$ on $X$ such that $C$ contains nothing, that is, no element of $X$ is in the relation, then clearly the relation is symmetry and transitive, but not reflexive.

John
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  • This is not true! There is a logical flaw in it. Any relation on the empty set is both symmetric, transitive AND reflexive: It DOES hold that for all $x\in\emptyset$ we have $xCx$. Just check all the possible $x$'s ... ;) – String Nov 20 '14 at 09:00
  • @String I think I specified that $X\neq \emptyset$. – John Nov 20 '14 at 11:05
  • Sorry, I misread that. I read $X=\emptyset$. The point seems to be, that if any element of $X$ is isolated (i.e. not equivalent to any other), then reflexivity is needed in order to make all elements of $X$ belong to an equivalence class. – String Nov 20 '14 at 11:08
  • @String Yes, you are right. – John Nov 20 '14 at 11:09