Every relation on a set has reflexive closure.












0












$begingroup$


Proof Attempt: Suppose $R$ is a relation on set $A$; $R=R_1 times R_2 subset A times A$. We assert $S = R cup i_{R_1 times R_1}$ to be the closure of $R$ such that $i_{R_1 times R_1}={(x,x) vert x in R_1}$. We want to show $S$ is reflexive. Let $(r,r')in S$. Then either $(r,r')in R_1 times R_2$ or $(r,r')in i_{R_1 times R_1}$. If $(r,r')in R_1times R_1$, then $rin R_1$ so $(r,r)in i_{R_1times R_1}$. If $(r,r')in i_{R_1times R_1}$, then $r=r'$. In any case, $(r,r)in S$ so $S$ is reflexive. Moreover, to show $S$ is the smallest reflexive relation that contains $R$, suppose $T=T_1times T_2 supseteq R$ and $T$ is reflexive. Then $(r,r')in R Rightarrow (r,r')in T_1times T_2 Rightarrow rin T_1 land r'in T_2$. Since $T$ is reflexive, then $(r,r)in T_1times T_2$ which means $rin T_2$ and that implies $T_1 subseteq T_2$. Furthermore, we have $R_1subseteq T_1$ so $R_1 subseteq T_2$. Thus, $i_{R_1times R_1}subseteq T_1times T_2$. Therefore, $S=R_1times R_2 cup i_{R_1 times R_2}subseteq T$. Clearly, $Rsubseteq S$ so S is the reflexive closure of $R$.



My concern: Is my argument correct? I was reading Velleman's text and he instead had $S=R cup i_A$, which is what I don't understand why.






proof-verification elementary-set-theory relations







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$endgroup$

















    0












    $begingroup$


    Proof Attempt: Suppose $R$ is a relation on set $A$; $R=R_1 times R_2 subset A times A$. We assert $S = R cup i_{R_1 times R_1}$ to be the closure of $R$ such that $i_{R_1 times R_1}={(x,x) vert x in R_1}$. We want to show $S$ is reflexive. Let $(r,r')in S$. Then either $(r,r')in R_1 times R_2$ or $(r,r')in i_{R_1 times R_1}$. If $(r,r')in R_1times R_1$, then $rin R_1$ so $(r,r)in i_{R_1times R_1}$. If $(r,r')in i_{R_1times R_1}$, then $r=r'$. In any case, $(r,r)in S$ so $S$ is reflexive. Moreover, to show $S$ is the smallest reflexive relation that contains $R$, suppose $T=T_1times T_2 supseteq R$ and $T$ is reflexive. Then $(r,r')in R Rightarrow (r,r')in T_1times T_2 Rightarrow rin T_1 land r'in T_2$. Since $T$ is reflexive, then $(r,r)in T_1times T_2$ which means $rin T_2$ and that implies $T_1 subseteq T_2$. Furthermore, we have $R_1subseteq T_1$ so $R_1 subseteq T_2$. Thus, $i_{R_1times R_1}subseteq T_1times T_2$. Therefore, $S=R_1times R_2 cup i_{R_1 times R_2}subseteq T$. Clearly, $Rsubseteq S$ so S is the reflexive closure of $R$.



    My concern: Is my argument correct? I was reading Velleman's text and he instead had $S=R cup i_A$, which is what I don't understand why.






    proof-verification elementary-set-theory relations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Proof Attempt: Suppose $R$ is a relation on set $A$; $R=R_1 times R_2 subset A times A$. We assert $S = R cup i_{R_1 times R_1}$ to be the closure of $R$ such that $i_{R_1 times R_1}={(x,x) vert x in R_1}$. We want to show $S$ is reflexive. Let $(r,r')in S$. Then either $(r,r')in R_1 times R_2$ or $(r,r')in i_{R_1 times R_1}$. If $(r,r')in R_1times R_1$, then $rin R_1$ so $(r,r)in i_{R_1times R_1}$. If $(r,r')in i_{R_1times R_1}$, then $r=r'$. In any case, $(r,r)in S$ so $S$ is reflexive. Moreover, to show $S$ is the smallest reflexive relation that contains $R$, suppose $T=T_1times T_2 supseteq R$ and $T$ is reflexive. Then $(r,r')in R Rightarrow (r,r')in T_1times T_2 Rightarrow rin T_1 land r'in T_2$. Since $T$ is reflexive, then $(r,r)in T_1times T_2$ which means $rin T_2$ and that implies $T_1 subseteq T_2$. Furthermore, we have $R_1subseteq T_1$ so $R_1 subseteq T_2$. Thus, $i_{R_1times R_1}subseteq T_1times T_2$. Therefore, $S=R_1times R_2 cup i_{R_1 times R_2}subseteq T$. Clearly, $Rsubseteq S$ so S is the reflexive closure of $R$.



      My concern: Is my argument correct? I was reading Velleman's text and he instead had $S=R cup i_A$, which is what I don't understand why.






      proof-verification elementary-set-theory relations







      share|cite|improve this question









      $endgroup$




      Proof Attempt: Suppose $R$ is a relation on set $A$; $R=R_1 times R_2 subset A times A$. We assert $S = R cup i_{R_1 times R_1}$ to be the closure of $R$ such that $i_{R_1 times R_1}={(x,x) vert x in R_1}$. We want to show $S$ is reflexive. Let $(r,r')in S$. Then either $(r,r')in R_1 times R_2$ or $(r,r')in i_{R_1 times R_1}$. If $(r,r')in R_1times R_1$, then $rin R_1$ so $(r,r)in i_{R_1times R_1}$. If $(r,r')in i_{R_1times R_1}$, then $r=r'$. In any case, $(r,r)in S$ so $S$ is reflexive. Moreover, to show $S$ is the smallest reflexive relation that contains $R$, suppose $T=T_1times T_2 supseteq R$ and $T$ is reflexive. Then $(r,r')in R Rightarrow (r,r')in T_1times T_2 Rightarrow rin T_1 land r'in T_2$. Since $T$ is reflexive, then $(r,r)in T_1times T_2$ which means $rin T_2$ and that implies $T_1 subseteq T_2$. Furthermore, we have $R_1subseteq T_1$ so $R_1 subseteq T_2$. Thus, $i_{R_1times R_1}subseteq T_1times T_2$. Therefore, $S=R_1times R_2 cup i_{R_1 times R_2}subseteq T$. Clearly, $Rsubseteq S$ so S is the reflexive closure of $R$.



      My concern: Is my argument correct? I was reading Velleman's text and he instead had $S=R cup i_A$, which is what I don't understand why.






      proof-verification elementary-set-theory relations




      proof-verification elementary-set-theory relations






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 25 at 2:53









      TheLast CipherTheLast Cipher

      708715




      708715






















          1 Answer
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          1












          $begingroup$

          There are a few problems here.



          First and foremost you seem to using a rather non-standard meaning of the word "reflexive". (Which is a polite way of saying I think you're misunderstanding what it means).



          The usual definition is




          A relation $R$ on $A$ is called reflexive if: For every $ain A$ it holds that $(a,a)in R$.




          Your proof looks like you think it is something like




          ... if: For every $(a,b)in R$ it holds that $(a,a)in R$.




          which is wrong. Do you see the difference? This misunderstanding dooms your attempt from the outset.



          But there's more: You're assuming that $R$ is an arbitrary relation and then immediately write $R=R_1times R_2$. Later you're writing $T=T_1times T_2$ for an arbitrary $T$. But you can't assume that a random relation can be written as a cartesian product!



          As a simple example, $R={(1,2),(3,4)}$ cannot be written as $R_1times R_2$ for any $R_1$ and $R_2$. Namely, $(1,2)in R_1times R_2$ can only be true $1in R_1$. And $(3,4)in R_1times R_2$ can only be true if $4in R_2$. But then certainly $(1,4)in R_1times R_2$, but $(1,4)$ is not in my $R$, so this $R$ cannot be $R_1times R_2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That definition is completely different! Thank you for the corrections!
            $endgroup$
            – TheLast Cipher
            Jan 25 at 4:07











          Your Answer





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          active

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          1












          $begingroup$

          There are a few problems here.



          First and foremost you seem to using a rather non-standard meaning of the word "reflexive". (Which is a polite way of saying I think you're misunderstanding what it means).



          The usual definition is




          A relation $R$ on $A$ is called reflexive if: For every $ain A$ it holds that $(a,a)in R$.




          Your proof looks like you think it is something like




          ... if: For every $(a,b)in R$ it holds that $(a,a)in R$.




          which is wrong. Do you see the difference? This misunderstanding dooms your attempt from the outset.



          But there's more: You're assuming that $R$ is an arbitrary relation and then immediately write $R=R_1times R_2$. Later you're writing $T=T_1times T_2$ for an arbitrary $T$. But you can't assume that a random relation can be written as a cartesian product!



          As a simple example, $R={(1,2),(3,4)}$ cannot be written as $R_1times R_2$ for any $R_1$ and $R_2$. Namely, $(1,2)in R_1times R_2$ can only be true $1in R_1$. And $(3,4)in R_1times R_2$ can only be true if $4in R_2$. But then certainly $(1,4)in R_1times R_2$, but $(1,4)$ is not in my $R$, so this $R$ cannot be $R_1times R_2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That definition is completely different! Thank you for the corrections!
            $endgroup$
            – TheLast Cipher
            Jan 25 at 4:07
















          1












          $begingroup$

          There are a few problems here.



          First and foremost you seem to using a rather non-standard meaning of the word "reflexive". (Which is a polite way of saying I think you're misunderstanding what it means).



          The usual definition is




          A relation $R$ on $A$ is called reflexive if: For every $ain A$ it holds that $(a,a)in R$.




          Your proof looks like you think it is something like




          ... if: For every $(a,b)in R$ it holds that $(a,a)in R$.




          which is wrong. Do you see the difference? This misunderstanding dooms your attempt from the outset.



          But there's more: You're assuming that $R$ is an arbitrary relation and then immediately write $R=R_1times R_2$. Later you're writing $T=T_1times T_2$ for an arbitrary $T$. But you can't assume that a random relation can be written as a cartesian product!



          As a simple example, $R={(1,2),(3,4)}$ cannot be written as $R_1times R_2$ for any $R_1$ and $R_2$. Namely, $(1,2)in R_1times R_2$ can only be true $1in R_1$. And $(3,4)in R_1times R_2$ can only be true if $4in R_2$. But then certainly $(1,4)in R_1times R_2$, but $(1,4)$ is not in my $R$, so this $R$ cannot be $R_1times R_2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That definition is completely different! Thank you for the corrections!
            $endgroup$
            – TheLast Cipher
            Jan 25 at 4:07














          1












          1








          1





          $begingroup$

          There are a few problems here.



          First and foremost you seem to using a rather non-standard meaning of the word "reflexive". (Which is a polite way of saying I think you're misunderstanding what it means).



          The usual definition is




          A relation $R$ on $A$ is called reflexive if: For every $ain A$ it holds that $(a,a)in R$.




          Your proof looks like you think it is something like




          ... if: For every $(a,b)in R$ it holds that $(a,a)in R$.




          which is wrong. Do you see the difference? This misunderstanding dooms your attempt from the outset.



          But there's more: You're assuming that $R$ is an arbitrary relation and then immediately write $R=R_1times R_2$. Later you're writing $T=T_1times T_2$ for an arbitrary $T$. But you can't assume that a random relation can be written as a cartesian product!



          As a simple example, $R={(1,2),(3,4)}$ cannot be written as $R_1times R_2$ for any $R_1$ and $R_2$. Namely, $(1,2)in R_1times R_2$ can only be true $1in R_1$. And $(3,4)in R_1times R_2$ can only be true if $4in R_2$. But then certainly $(1,4)in R_1times R_2$, but $(1,4)$ is not in my $R$, so this $R$ cannot be $R_1times R_2$.






          share|cite|improve this answer









          $endgroup$



          There are a few problems here.



          First and foremost you seem to using a rather non-standard meaning of the word "reflexive". (Which is a polite way of saying I think you're misunderstanding what it means).



          The usual definition is




          A relation $R$ on $A$ is called reflexive if: For every $ain A$ it holds that $(a,a)in R$.




          Your proof looks like you think it is something like




          ... if: For every $(a,b)in R$ it holds that $(a,a)in R$.




          which is wrong. Do you see the difference? This misunderstanding dooms your attempt from the outset.



          But there's more: You're assuming that $R$ is an arbitrary relation and then immediately write $R=R_1times R_2$. Later you're writing $T=T_1times T_2$ for an arbitrary $T$. But you can't assume that a random relation can be written as a cartesian product!



          As a simple example, $R={(1,2),(3,4)}$ cannot be written as $R_1times R_2$ for any $R_1$ and $R_2$. Namely, $(1,2)in R_1times R_2$ can only be true $1in R_1$. And $(3,4)in R_1times R_2$ can only be true if $4in R_2$. But then certainly $(1,4)in R_1times R_2$, but $(1,4)$ is not in my $R$, so this $R$ cannot be $R_1times R_2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 3:49









          Henning MakholmHenning Makholm

          241k17308549




          241k17308549












          • $begingroup$
            That definition is completely different! Thank you for the corrections!
            $endgroup$
            – TheLast Cipher
            Jan 25 at 4:07


















          • $begingroup$
            That definition is completely different! Thank you for the corrections!
            $endgroup$
            – TheLast Cipher
            Jan 25 at 4:07
















          $begingroup$
          That definition is completely different! Thank you for the corrections!
          $endgroup$
          – TheLast Cipher
          Jan 25 at 4:07




          $begingroup$
          That definition is completely different! Thank you for the corrections!
          $endgroup$
          – TheLast Cipher
          Jan 25 at 4:07


















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