Let $R$ be a binary relation on set X. Let us define the transitive closure of $R$, denoted by $S_R$ as follows:
$xS_Ry$, if $xRy$ or there exist $x_1,x_2,…, x_m\in X$ such that $xRx_1,x_1Rx_2, …, x_{m–1}Rx_m, x_mRy$.

Then how can we prove that

  1. $S_R$ is transitive
  2. $S_R$ is asymmetric if and only if it is irreflexive.


  • $\begingroup$ I changed the answer as well. $\endgroup$ Mar 26, 2017 at 9:11

1 Answer 1


If $xS_Ry$ and $yS_Rz$, so there are $\{x_1,..., x_m\}$ and $\{ y_1,...,y_n \}$ such that $xRx_1, x_1Rx_2,..., x_{m-1}Rx_m, x_mRy$ and $yRy_1, y_1Ry_2,..., y_{n-1}Ry_n, y_nRz$. Denote $x_{m+1}:= y$ and $x_{m+i+1}:=y_i$. Therefore, $xRx_1, x_1Rx_2,..., x_{m-1}Rx_m, x_mRx_{m+1}, ..., x_{m+n+1}Rz$. Hence, $xS_Rz$.


To solve 2, it's easy to see that asymmetric implies ireflexive. Because if you suppose that $S_R $ is not ireflexive, so there is $aS_Ra$. Therefore, choose $b=a $ and we have $aS_Rb $ and $bS_Ra $. So $S_R $ would not be asymmetric.

Now we are going to prove that if $S_R $ is not asymmetric implies $S_R $ is not ireflexive. If $S_R $ is not asymmetric, there are $a_0 $ and $b_0$ such that $a_0S_Rb_0$ and $b_0S_Ra_0$. Therefore, for transitiveness, $a_0 S_R a_0$. Hence $S_R$ is not ireflexive.

  • $\begingroup$ Thanks, but I don't understand the second proof. I.e. If $S_R$ is asymmetric, there are $a_0$ and $b_0$ such that $a_0S_Rb_0$ and $b_0S_Ra_0$? Here it seems symmetric. Do you mean if $S_R$ is $\textbf{not}$ $\textbf{asymmetric}$ instead? $\endgroup$
    – UserAb
    Mar 26, 2017 at 13:16
  • $\begingroup$ Sorry. I forgot the "not". I am going to correct that. $\endgroup$ Mar 26, 2017 at 14:36

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