How is the relation “the smallest element is the same” reflexive?












9












$begingroup$


Let $mathcal{X}$ be the set of all nonempty subsets of the set ${1,2,3,...,10}$. Define the relation $mathcal{R}$ on $mathcal{X}$ by: $forall A, B in mathcal{X}, A mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, ${1,2,3} mathcal{R} {1,3,5,8}$ because the smallest element of ${1,2,3}$ is $1$ which is also the smallest element of ${1,3,5,8}$.



Prove that $mathcal{R}$ is an equivalence relation on $mathcal{X}$.



From my understanding, the definition of reflexive is:



$$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



However, for this problem, you can have the relation with these two sets:



${1}$ and ${1,2}$



Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?



I'm having trouble seeing how this is reflexive. Getting confused by the definition here.






discrete-mathematics elementary-set-theory relations equivalence-relations







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that ${ 1 } mathcal R { 1,2 }$ but we have also ${ 1 } mathcal R { 1 }$ and ${ 1,2 } mathcal R { 1,2 }$
    $endgroup$
    – Mauro ALLEGRANZA
    yesterday






  • 7




    $begingroup$
    Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
    $endgroup$
    – Arturo Magidin
    yesterday










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
    $endgroup$
    – qbuffer
    yesterday












  • $begingroup$
    @qbuffer Have a look at the updated version of my answer.
    $endgroup$
    – Haris Gusic
    yesterday
















9












$begingroup$


Let $mathcal{X}$ be the set of all nonempty subsets of the set ${1,2,3,...,10}$. Define the relation $mathcal{R}$ on $mathcal{X}$ by: $forall A, B in mathcal{X}, A mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, ${1,2,3} mathcal{R} {1,3,5,8}$ because the smallest element of ${1,2,3}$ is $1$ which is also the smallest element of ${1,3,5,8}$.



Prove that $mathcal{R}$ is an equivalence relation on $mathcal{X}$.



From my understanding, the definition of reflexive is:



$$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



However, for this problem, you can have the relation with these two sets:



${1}$ and ${1,2}$



Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?



I'm having trouble seeing how this is reflexive. Getting confused by the definition here.






discrete-mathematics elementary-set-theory relations equivalence-relations







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that ${ 1 } mathcal R { 1,2 }$ but we have also ${ 1 } mathcal R { 1 }$ and ${ 1,2 } mathcal R { 1,2 }$
    $endgroup$
    – Mauro ALLEGRANZA
    yesterday






  • 7




    $begingroup$
    Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
    $endgroup$
    – Arturo Magidin
    yesterday










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
    $endgroup$
    – qbuffer
    yesterday












  • $begingroup$
    @qbuffer Have a look at the updated version of my answer.
    $endgroup$
    – Haris Gusic
    yesterday














9












9








9





$begingroup$


Let $mathcal{X}$ be the set of all nonempty subsets of the set ${1,2,3,...,10}$. Define the relation $mathcal{R}$ on $mathcal{X}$ by: $forall A, B in mathcal{X}, A mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, ${1,2,3} mathcal{R} {1,3,5,8}$ because the smallest element of ${1,2,3}$ is $1$ which is also the smallest element of ${1,3,5,8}$.



Prove that $mathcal{R}$ is an equivalence relation on $mathcal{X}$.



From my understanding, the definition of reflexive is:



$$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



However, for this problem, you can have the relation with these two sets:



${1}$ and ${1,2}$



Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?



I'm having trouble seeing how this is reflexive. Getting confused by the definition here.






discrete-mathematics elementary-set-theory relations equivalence-relations







share|cite|improve this question











$endgroup$




Let $mathcal{X}$ be the set of all nonempty subsets of the set ${1,2,3,...,10}$. Define the relation $mathcal{R}$ on $mathcal{X}$ by: $forall A, B in mathcal{X}, A mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, ${1,2,3} mathcal{R} {1,3,5,8}$ because the smallest element of ${1,2,3}$ is $1$ which is also the smallest element of ${1,3,5,8}$.



Prove that $mathcal{R}$ is an equivalence relation on $mathcal{X}$.



From my understanding, the definition of reflexive is:



$$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



However, for this problem, you can have the relation with these two sets:



${1}$ and ${1,2}$



Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?



I'm having trouble seeing how this is reflexive. Getting confused by the definition here.






discrete-mathematics elementary-set-theory relations equivalence-relations




discrete-mathematics elementary-set-theory relations equivalence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 21 hours ago









Martin Sleziak

45k10122277




45k10122277










asked yesterday









qbufferqbuffer

675




675








  • 5




    $begingroup$
    Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that ${ 1 } mathcal R { 1,2 }$ but we have also ${ 1 } mathcal R { 1 }$ and ${ 1,2 } mathcal R { 1,2 }$
    $endgroup$
    – Mauro ALLEGRANZA
    yesterday






  • 7




    $begingroup$
    Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
    $endgroup$
    – Arturo Magidin
    yesterday










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
    $endgroup$
    – qbuffer
    yesterday












  • $begingroup$
    @qbuffer Have a look at the updated version of my answer.
    $endgroup$
    – Haris Gusic
    yesterday














  • 5




    $begingroup$
    Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that ${ 1 } mathcal R { 1,2 }$ but we have also ${ 1 } mathcal R { 1 }$ and ${ 1,2 } mathcal R { 1,2 }$
    $endgroup$
    – Mauro ALLEGRANZA
    yesterday






  • 7




    $begingroup$
    Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
    $endgroup$
    – Arturo Magidin
    yesterday










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
    $endgroup$
    – qbuffer
    yesterday












  • $begingroup$
    @qbuffer Have a look at the updated version of my answer.
    $endgroup$
    – Haris Gusic
    yesterday








5




5




$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that ${ 1 } mathcal R { 1,2 }$ but we have also ${ 1 } mathcal R { 1 }$ and ${ 1,2 } mathcal R { 1,2 }$
$endgroup$
– Mauro ALLEGRANZA
yesterday




$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that ${ 1 } mathcal R { 1,2 }$ but we have also ${ 1 } mathcal R { 1 }$ and ${ 1,2 } mathcal R { 1,2 }$
$endgroup$
– Mauro ALLEGRANZA
yesterday




7




7




$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
yesterday




$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
yesterday












$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
$endgroup$
– qbuffer
yesterday






$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
$endgroup$
– qbuffer
yesterday














$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
yesterday




$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
yesterday










2 Answers
2






active

oldest

votes


















10












$begingroup$

Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcal{R} x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.






share|cite|improve this answer











$endgroup$





















    5












    $begingroup$

    A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



    Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

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      active

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      10












      $begingroup$

      Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



      To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcal{R} x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



      You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.






      share|cite|improve this answer











      $endgroup$


















        10












        $begingroup$

        Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



        To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcal{R} x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



        You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.






        share|cite|improve this answer











        $endgroup$
















          10












          10








          10





          $begingroup$

          Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



          To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcal{R} x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



          You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.






          share|cite|improve this answer











          $endgroup$



          Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



          To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcal{R} x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



          You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          Haris GusicHaris Gusic

          3,431627




          3,431627























              5












              $begingroup$

              A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



              Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



                Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



                  Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.






                  share|cite|improve this answer









                  $endgroup$



                  A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$



                  Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  s0ulr3aper07s0ulr3aper07

                  673112




                  673112






























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