I was leafing through the "Introduction to Classical Real Analysis" (Stromberg), and while reading the paragraph "Equivalence Relations" in the "Preliminaries" section, I saw:

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> One checks that any such $R$ [i.e. which is reflexive, symmetric and transitive] satisfies $R^{-1}=R=R \circ R$ and that, conversely, any relation $R$ satisfying these two equalities is an equivalence relation on its domain.
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I am not able to prove that $R^{-1}=R=R \circ R$ implies $R$ being reflexive, that is, why if $R^{-1}=R=R \circ R$ then $$\newcommand{\dom}{\operatorname{dom }} x \in \dom R \implies (x,x) \in R \;?$$

Here the domain of the binary relation $R$ over a set $X$ is defined as $$\dom R = \{x \in X : (x,y) \in R \textrm{ for some } y \in X\}.$$