How is the relation “the smallest element is the same” reflexive?Need help counting equivalence classes.Finding the smallest relation that is reflexive, transitive, and symmetricSmallest relation for reflexive, symmetry and transitivityEquivalence relation example. How is this even reflexive?Is antisymmetric the same as reflexive?Finding the smallest equivalence relation containing a specific list of ordered pairsHow is this an equivalence relation?truefalse claims in relations and equivalence relationsWhat is the least and greatest element in symmetric but not reflexive relation over $1,2,3$?How is this case a reflexive relation?

Accidentally leaked the solution to an assignment, what to do now? (I'm the prof)

Example of a relative pronoun

Are tax years 2016 & 2017 back taxes deductible for tax year 2018?

Simulate Bitwise Cyclic Tag

Can a German sentence have two subjects?

XeLaTeX and pdfLaTeX ignore hyphenation

Can I make popcorn with any corn?

Could a US political party gain complete control over the government by removing checks & balances?

Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?

Why are 150k or 200k jobs considered good when there are 300k+ births a month?

DOS, create pipe for stdin/stdout of command.com(or 4dos.com) in C or Batch?

How do I create uniquely male characters?

cryptic clue: mammal sounds like relative consumer (8)

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

New order #4: World

Why Is Death Allowed In the Matrix?

How does one intimidate enemies without having the capacity for violence?

A Journey Through Space and Time

What is GPS' 19 year rollover and does it present a cybersecurity issue?

Copycat chess is back

Why is an old chain unsafe?

Can you lasso down a wizard who is using the Levitate spell?

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

How do you conduct xenoanthropology after first contact?



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


Need help counting equivalence classes.Finding the smallest relation that is reflexive, transitive, and symmetricSmallest relation for reflexive, symmetry and transitivityEquivalence relation example. How is this even reflexive?Is antisymmetric the same as reflexive?Finding the smallest equivalence relation containing a specific list of ordered pairsHow is this an equivalence relation?truefalse claims in relations and equivalence relationsWhat is the least and greatest element in symmetric but not reflexive relation over $1,2,3$?How is this case a reflexive relation?













8












$begingroup$


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



Prove that $mathcalR$ is an equivalence relation on $mathcalX$.



From my understanding, the definition of reflexive is:



$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



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.










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    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
    12 hours ago






  • 6




    $begingroup$
    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
    11 hours ago










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
    $endgroup$
    – qbuffer
    11 hours ago











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















8












$begingroup$


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



Prove that $mathcalR$ is an equivalence relation on $mathcalX$.



From my understanding, the definition of reflexive is:



$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



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.










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    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
    12 hours ago






  • 6




    $begingroup$
    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
    11 hours ago










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
    $endgroup$
    – qbuffer
    11 hours ago











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













8












8








8





$begingroup$


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



Prove that $mathcalR$ is an equivalence relation on $mathcalX$.



From my understanding, the definition of reflexive is:



$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



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.










share|cite|improve this question











$endgroup$




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



Prove that $mathcalR$ is an equivalence relation on $mathcalX$.



From my understanding, the definition of reflexive is:



$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 37 mins ago









Martin Sleziak

45k10122277




45k10122277










asked 12 hours ago









qbufferqbuffer

625




625







  • 4




    $begingroup$
    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
    12 hours ago






  • 6




    $begingroup$
    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
    11 hours ago










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
    $endgroup$
    – qbuffer
    11 hours ago











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












  • 4




    $begingroup$
    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
    12 hours ago






  • 6




    $begingroup$
    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
    11 hours ago










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
    $endgroup$
    – qbuffer
    11 hours ago











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







4




4




$begingroup$
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
12 hours ago




$begingroup$
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
12 hours ago




6




6




$begingroup$
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
11 hours ago




$begingroup$
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
11 hours ago












$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
11 hours ago





$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
11 hours ago













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




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










2 Answers
2






active

oldest

votes


















8












$begingroup$

Why are you testing reflexivity by looking at two different elements of $mathcalX$? 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, $x mathcalR x$ 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 $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.






share|cite|improve this answer











$endgroup$




















    4












    $begingroup$

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



    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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3178532%2fhow-is-the-relation-the-smallest-element-is-the-same-reflexive%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      8












      $begingroup$

      Why are you testing reflexivity by looking at two different elements of $mathcalX$? 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, $x mathcalR x$ 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 $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.






      share|cite|improve this answer











      $endgroup$

















        8












        $begingroup$

        Why are you testing reflexivity by looking at two different elements of $mathcalX$? 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, $x mathcalR x$ 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 $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.






        share|cite|improve this answer











        $endgroup$















          8












          8








          8





          $begingroup$

          Why are you testing reflexivity by looking at two different elements of $mathcalX$? 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, $x mathcalR x$ 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 $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.






          share|cite|improve this answer











          $endgroup$



          Why are you testing reflexivity by looking at two different elements of $mathcalX$? 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, $x mathcalR x$ 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 $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 10 hours ago

























          answered 12 hours ago









          Haris GusicHaris Gusic

          3,331525




          3,331525





















              4












              $begingroup$

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



              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









              $endgroup$

















                4












                $begingroup$

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



                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









                $endgroup$















                  4












                  4








                  4





                  $begingroup$

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



                  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 $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



                  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 11 hours ago









                  s0ulr3aper07s0ulr3aper07

                  658112




                  658112



























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3178532%2fhow-is-the-relation-the-smallest-element-is-the-same-reflexive%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

                      2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

                      Metrô de Los Teques Índice Linhas | Estações | Ver também | Referências Ligações externas | Menu de navegação«INSTITUCIÓN»«Mapa de rutas»originalMetrô de Los TequesC.A. Metro Los Teques |Alcaldía de Guaicaipuro – Sitio OficialGobernacion de Mirandaeeeeeee