Equivalence relation generated by a relation: concrete characterization












0












$begingroup$


Let $R$ be a relation on a set $X$. The equivalence relation $sim_R$ generated by $R$ is defined as $bigcap A$ for $A$ being the set of all equivalence relations containing $R$ ($A$ is not empty as $Xtimes X in A$).



Different sources claim that it's possible to give a concrete definition of two elements $x$ and $y$ being in relation $sim_R$ to each other:




For any $x,y in X, x sim_R y$ if and only if there is a $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that




  • $x_1 = x$,


  • $x_n = y$,


  • for any $1 leq k < n$ we have either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$.





The "only if" part of the proof is carefully explained in this answer. However, I can't understand why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.






elementary-set-theory equivalence-relations







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $R$ be a relation on a set $X$. The equivalence relation $sim_R$ generated by $R$ is defined as $bigcap A$ for $A$ being the set of all equivalence relations containing $R$ ($A$ is not empty as $Xtimes X in A$).



    Different sources claim that it's possible to give a concrete definition of two elements $x$ and $y$ being in relation $sim_R$ to each other:




    For any $x,y in X, x sim_R y$ if and only if there is a $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that




    • $x_1 = x$,


    • $x_n = y$,


    • for any $1 leq k < n$ we have either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$.





    The "only if" part of the proof is carefully explained in this answer. However, I can't understand why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.






    elementary-set-theory equivalence-relations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $R$ be a relation on a set $X$. The equivalence relation $sim_R$ generated by $R$ is defined as $bigcap A$ for $A$ being the set of all equivalence relations containing $R$ ($A$ is not empty as $Xtimes X in A$).



      Different sources claim that it's possible to give a concrete definition of two elements $x$ and $y$ being in relation $sim_R$ to each other:




      For any $x,y in X, x sim_R y$ if and only if there is a $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that




      • $x_1 = x$,


      • $x_n = y$,


      • for any $1 leq k < n$ we have either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$.





      The "only if" part of the proof is carefully explained in this answer. However, I can't understand why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.






      elementary-set-theory equivalence-relations







      share|cite|improve this question









      $endgroup$




      Let $R$ be a relation on a set $X$. The equivalence relation $sim_R$ generated by $R$ is defined as $bigcap A$ for $A$ being the set of all equivalence relations containing $R$ ($A$ is not empty as $Xtimes X in A$).



      Different sources claim that it's possible to give a concrete definition of two elements $x$ and $y$ being in relation $sim_R$ to each other:




      For any $x,y in X, x sim_R y$ if and only if there is a $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that




      • $x_1 = x$,


      • $x_n = y$,


      • for any $1 leq k < n$ we have either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$.





      The "only if" part of the proof is carefully explained in this answer. However, I can't understand why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.






      elementary-set-theory equivalence-relations




      elementary-set-theory equivalence-relations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 27 '18 at 15:22









      Jxt921Jxt921

      976618




      976618






















          1 Answer
          1






          active

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          2





          +150







          $begingroup$


          Why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.




          This is easy. It suffices to show that $(x,y) in A$ for any all equivalence relation $A$ containing $R$. We have $(x_k,x_{k+1}) in A$ for any $1 leq k < n$, so $(x,y)=(x_1,x_n)in A$ by transitivity of $A$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But the problem is that it's not $(x_k,x_{k+1})$ for any $1 leq k < n$ and it's not $(x_{k+1},x_k)$ for any $1 leq k < n$, but it's $(x_k,x_{k+1})$ or $(x_{k+1},x_k)$ for any $1 leq k < n$.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 8:46






          • 1




            $begingroup$
            This is not a problem, because $A$ is symmetric. That is, if $(x_{k+1},x_k)in R$ then $(x_{k+1},x_k)in A$ and $(x_{k},x_{k+1})in A$.
            $endgroup$
            – Alex Ravsky
            Nov 30 '18 at 14:47












          • $begingroup$
            You are right, I'm sorry.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 17:53











          Your Answer





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          1 Answer
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          active

          oldest

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          active

          oldest

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          active

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          2





          +150







          $begingroup$


          Why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.




          This is easy. It suffices to show that $(x,y) in A$ for any all equivalence relation $A$ containing $R$. We have $(x_k,x_{k+1}) in A$ for any $1 leq k < n$, so $(x,y)=(x_1,x_n)in A$ by transitivity of $A$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But the problem is that it's not $(x_k,x_{k+1})$ for any $1 leq k < n$ and it's not $(x_{k+1},x_k)$ for any $1 leq k < n$, but it's $(x_k,x_{k+1})$ or $(x_{k+1},x_k)$ for any $1 leq k < n$.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 8:46






          • 1




            $begingroup$
            This is not a problem, because $A$ is symmetric. That is, if $(x_{k+1},x_k)in R$ then $(x_{k+1},x_k)in A$ and $(x_{k},x_{k+1})in A$.
            $endgroup$
            – Alex Ravsky
            Nov 30 '18 at 14:47












          • $begingroup$
            You are right, I'm sorry.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 17:53
















          2





          +150







          $begingroup$


          Why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.




          This is easy. It suffices to show that $(x,y) in A$ for any all equivalence relation $A$ containing $R$. We have $(x_k,x_{k+1}) in A$ for any $1 leq k < n$, so $(x,y)=(x_1,x_n)in A$ by transitivity of $A$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But the problem is that it's not $(x_k,x_{k+1})$ for any $1 leq k < n$ and it's not $(x_{k+1},x_k)$ for any $1 leq k < n$, but it's $(x_k,x_{k+1})$ or $(x_{k+1},x_k)$ for any $1 leq k < n$.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 8:46






          • 1




            $begingroup$
            This is not a problem, because $A$ is symmetric. That is, if $(x_{k+1},x_k)in R$ then $(x_{k+1},x_k)in A$ and $(x_{k},x_{k+1})in A$.
            $endgroup$
            – Alex Ravsky
            Nov 30 '18 at 14:47












          • $begingroup$
            You are right, I'm sorry.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 17:53














          2





          +150







          2





          +150



          2




          +150



          $begingroup$


          Why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.




          This is easy. It suffices to show that $(x,y) in A$ for any all equivalence relation $A$ containing $R$. We have $(x_k,x_{k+1}) in A$ for any $1 leq k < n$, so $(x,y)=(x_1,x_n)in A$ by transitivity of $A$.






          share|cite|improve this answer









          $endgroup$




          Why having $n in mathbb{N}_{>0}$ and $x_1,...,x_n in X$ so that $x_1 = x, x_n = y$ and for any $1 leq k < n$ having either $(x_k,x_{k+1}) in R$ or $(x_{k+1},x_k) in R$ implies that $(x,y) in {sim_R}$.




          This is easy. It suffices to show that $(x,y) in A$ for any all equivalence relation $A$ containing $R$. We have $(x_k,x_{k+1}) in A$ for any $1 leq k < n$, so $(x,y)=(x_1,x_n)in A$ by transitivity of $A$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 7:21









          Alex RavskyAlex Ravsky

          40.4k32282




          40.4k32282












          • $begingroup$
            But the problem is that it's not $(x_k,x_{k+1})$ for any $1 leq k < n$ and it's not $(x_{k+1},x_k)$ for any $1 leq k < n$, but it's $(x_k,x_{k+1})$ or $(x_{k+1},x_k)$ for any $1 leq k < n$.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 8:46






          • 1




            $begingroup$
            This is not a problem, because $A$ is symmetric. That is, if $(x_{k+1},x_k)in R$ then $(x_{k+1},x_k)in A$ and $(x_{k},x_{k+1})in A$.
            $endgroup$
            – Alex Ravsky
            Nov 30 '18 at 14:47












          • $begingroup$
            You are right, I'm sorry.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 17:53


















          • $begingroup$
            But the problem is that it's not $(x_k,x_{k+1})$ for any $1 leq k < n$ and it's not $(x_{k+1},x_k)$ for any $1 leq k < n$, but it's $(x_k,x_{k+1})$ or $(x_{k+1},x_k)$ for any $1 leq k < n$.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 8:46






          • 1




            $begingroup$
            This is not a problem, because $A$ is symmetric. That is, if $(x_{k+1},x_k)in R$ then $(x_{k+1},x_k)in A$ and $(x_{k},x_{k+1})in A$.
            $endgroup$
            – Alex Ravsky
            Nov 30 '18 at 14:47












          • $begingroup$
            You are right, I'm sorry.
            $endgroup$
            – Jxt921
            Nov 30 '18 at 17:53
















          $begingroup$
          But the problem is that it's not $(x_k,x_{k+1})$ for any $1 leq k < n$ and it's not $(x_{k+1},x_k)$ for any $1 leq k < n$, but it's $(x_k,x_{k+1})$ or $(x_{k+1},x_k)$ for any $1 leq k < n$.
          $endgroup$
          – Jxt921
          Nov 30 '18 at 8:46




          $begingroup$
          But the problem is that it's not $(x_k,x_{k+1})$ for any $1 leq k < n$ and it's not $(x_{k+1},x_k)$ for any $1 leq k < n$, but it's $(x_k,x_{k+1})$ or $(x_{k+1},x_k)$ for any $1 leq k < n$.
          $endgroup$
          – Jxt921
          Nov 30 '18 at 8:46




          1




          1




          $begingroup$
          This is not a problem, because $A$ is symmetric. That is, if $(x_{k+1},x_k)in R$ then $(x_{k+1},x_k)in A$ and $(x_{k},x_{k+1})in A$.
          $endgroup$
          – Alex Ravsky
          Nov 30 '18 at 14:47






          $begingroup$
          This is not a problem, because $A$ is symmetric. That is, if $(x_{k+1},x_k)in R$ then $(x_{k+1},x_k)in A$ and $(x_{k},x_{k+1})in A$.
          $endgroup$
          – Alex Ravsky
          Nov 30 '18 at 14:47














          $begingroup$
          You are right, I'm sorry.
          $endgroup$
          – Jxt921
          Nov 30 '18 at 17:53




          $begingroup$
          You are right, I'm sorry.
          $endgroup$
          – Jxt921
          Nov 30 '18 at 17:53


















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