MHB Equivalence Relations: Explaining $I_A$, $\rho^{-1}$ and $\rho \circ \rho$

AI Thread Summary
In the discussion on equivalence relations, the properties of reflexivity, symmetry, and transitivity are examined through the relations $I_A \subset \rho$, $\rho^{-1} = \rho$, and $\rho \circ \rho \subset \rho$. It is emphasized that using concrete examples can clarify whether these properties hold. For instance, the relation defined by $m \mathrel{\rho} n \iff m \le n$ on natural numbers demonstrates that $\rho^{-1} \not\subseteq \rho$, indicating a lack of symmetry. The discussion also notes that $\rho^{-1} = \rho$ is equivalent to $\rho^{-1} \subseteq \rho$. Understanding these properties is crucial for grasping the nature of equivalence relations.
evinda
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Hi again! (Smile)

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

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