Let $R$ be a relation on a set $A$. Define $T(R)=R\cup R^{-1} \cup \{(x,x)\mid x \in A \}$. Show that $T(R)$ is reflexive and symmetric.

Question

Let $R$ be a relation on a set $A$. Define $T(R)=R\cup R^{-1} \cup \{(x,x)\mid x \in A \}$. Show that $T(R)$ is reflexive and symmetric.

Let $\triangle = \{(x,x) \mid x\in A\}$

I only know that $R \cup \triangle$ is the reflexive closure and $R \cup R^{-1}$ is the symmetric closure.

Any idea?

Answer 1

You almost have it! Let me properly write the idea:

• $T(R)$ is reflexive since if $x \in A$, then $(x,x) \in \Delta$ and $\Delta \subseteq T(R)$ by definition of $T(R)$.
• $T(R)$ is symmetric since if $(x,y) \in T(R)$, then $(x,y) \in R$ or $(x,y) \in R^{-1}$ or $(x,y) \in \Delta$. In the case that $(x,y) \in R$ we have $(y,x) \in R^{-1} \subseteq T(R)$; in the case that $(x,y) \in R^{-1}$ we have $(x,y) \in R \subseteq T(R)$; and in the case that $(x,y) \in \Delta$ we have $(y,x) = (x,y) \in T(R)$. Thus, $(x,y) \in T(R)$ always implies that $(y,x) \in T(R)$.

Answer 2

Let $x \in A$. Then, $(x,x) \in \triangle \subseteq T(R)$. Hence, $T(R)$ is reflexive.

Let $x,y \in A$ and let $(x,y) \in T(R)$. Then, $(x,y) \in R$ or $(x,y) \in R^{-1}$ or $(x,y) \in \triangle$.

If $(x,y) \in R$, then by definition, we have $(y,x) \in R^{-1} \subseteq T(R)$. Hence, $T(R)$ is symmetric.

If $(x,y) \in R^{-1}$, then by definition, we have $(y,x) \in R \subseteq T(R)$. Hence, $T(R)$ is symmetric.

If $(x,y) \in \triangle$, then we must have $y = x$. Thus, $(x,y)=(y,x)$. Hence, $T(R)$ is symmetric.

That's it, $T(R)$ is reflexive and symmetric as desired.