Why is the identity relation a subset of an equivalence relation?

[evinda]

Hi again! (Smile)

If $\rho$ is an equivalence relation, could you explain me why the following relations stand? (Thinking)

• $I_A \subset \rho$
• $\rho^{-1}=\rho$
• $\rho \circ \rho \subset \rho$

 

[Evgeny.Makarov]

These three conditions correspond to reflexivity, symmetry and transitivity, respectively. Again, I would recommend taking some concrete relations (ideally, those where these properties hold and where they don't) and determine whether the properties you listed hold. This will help build intuition.

For example, let $m\mathrel{\rho}n\iff m\le n$ on natural numbers. Then $\langle 3,5\rangle\in\rho$, but $\rho^{-1}\ni\langle 5,3\rangle\notin\rho$; therefore, $\rho^{-1}\not\subseteq\rho$. The same fact is expressed by saying that $3\le 5$, but $5\not\le 3$. Both these statements mean that $\rho$ is not symmetric. Note, by the way, that $\rho^{-1}\subseteq\rho$ is equivalent to $\rho^{-1}=\rho$ for any $\rho$ because $\rho^{-1}\subseteq\rho$ implies that $(\rho^{-1})^{-1}\subseteq\rho^{-1}$, i.e., $\rho\subseteq\rho^{-1}$.