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Symmetric property of an equivalence relation states that if $a$~$b$ then $b$~$a$; Transitive property states that if $a$~$b$ and $b$~$c$ then $a$~$c$. What is wrong with the following proof that symmetric and transitive property imply reflexive property ? Let $a$~$b$; then $b$~$a$, whence, by property of transitivity
(using $a = c$), $a$~$a$.(,where '~' denotes a binary relation ).

Is there any alternative of reflexive property which will insure that properties of symmetricity and transitivity imply property of reflexivity ?

Arthur
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    It depends on the existence of a $b$ such that $a\sim b$. – ancient mathematician Sep 29 '22 at 06:35
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    Another perspective: take any equivalence relation $\sim$ on any set $X$. Then, for any nonempty set $Y$ that's disjoint from $X$, form a relation on $X\cup Y$ by just using $\sim$ without modification—in particular, nothing in $Y$ is equivalent to anything. You can check that this new relation is still symmetric and transitive but no longer reflexive: the elements of $X$ are all related to themselves, but the elements of $Y$ are not. – Greg Martin Sep 29 '22 at 06:56

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