Equivalence relations on natural numbers












4












$begingroup$


How many equivalence relations are there on $mathbb{N}$, where $mathbb{N}$ is the set of natural numbers? How can one compute it?






equivalence-relations set-partition







share|cite|improve this question











$endgroup$












  • $begingroup$
    Uncountably infinitely many, obviously. Is this the answer you are after?
    $endgroup$
    – Did
    Jan 6 '13 at 14:15










  • $begingroup$
    The answer is a particular case of my answer here.
    $endgroup$
    – Asaf Karagila♦
    Jan 6 '13 at 14:25
















4












$begingroup$


How many equivalence relations are there on $mathbb{N}$, where $mathbb{N}$ is the set of natural numbers? How can one compute it?






equivalence-relations set-partition







share|cite|improve this question











$endgroup$












  • $begingroup$
    Uncountably infinitely many, obviously. Is this the answer you are after?
    $endgroup$
    – Did
    Jan 6 '13 at 14:15










  • $begingroup$
    The answer is a particular case of my answer here.
    $endgroup$
    – Asaf Karagila♦
    Jan 6 '13 at 14:25














4












4








4





$begingroup$


How many equivalence relations are there on $mathbb{N}$, where $mathbb{N}$ is the set of natural numbers? How can one compute it?






equivalence-relations set-partition







share|cite|improve this question











$endgroup$




How many equivalence relations are there on $mathbb{N}$, where $mathbb{N}$ is the set of natural numbers? How can one compute it?






equivalence-relations set-partition




equivalence-relations set-partition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 26 '17 at 15:58









Peter Taylor

8,79312341




8,79312341










asked Jan 6 '13 at 14:04









BilboBilbo

270623




270623












  • $begingroup$
    Uncountably infinitely many, obviously. Is this the answer you are after?
    $endgroup$
    – Did
    Jan 6 '13 at 14:15










  • $begingroup$
    The answer is a particular case of my answer here.
    $endgroup$
    – Asaf Karagila♦
    Jan 6 '13 at 14:25


















  • $begingroup$
    Uncountably infinitely many, obviously. Is this the answer you are after?
    $endgroup$
    – Did
    Jan 6 '13 at 14:15










  • $begingroup$
    The answer is a particular case of my answer here.
    $endgroup$
    – Asaf Karagila♦
    Jan 6 '13 at 14:25
















$begingroup$
Uncountably infinitely many, obviously. Is this the answer you are after?
$endgroup$
– Did
Jan 6 '13 at 14:15




$begingroup$
Uncountably infinitely many, obviously. Is this the answer you are after?
$endgroup$
– Did
Jan 6 '13 at 14:15












$begingroup$
The answer is a particular case of my answer here.
$endgroup$
– Asaf Karagila♦
Jan 6 '13 at 14:25




$begingroup$
The answer is a particular case of my answer here.
$endgroup$
– Asaf Karagila♦
Jan 6 '13 at 14:25










2 Answers
2






active

oldest

votes


















2












$begingroup$

There are $2^{aleph_0}$ many. As mentioned above, equivalence relation corresponds to partition of $mathbb{N}$. For each $f in 2^{mathbb{N}}$, i.e. a function $f : mathbb{N} rightarrow {0,1}$. Define a partition, $P_f$, such that, $P_f$ has a subset of size $n$ if and only if $f(n) = 1$. Then if $f neq g$, then $P_f neq P_g$. Hence the number of equivalence relations on $mathbb{N}$ is at least as many as $2^{aleph_0}$. It can not be bigger because the power set $mathcal{P}(mathbb{N})$ has cardinality $2^{aleph_0}$.




Explicitly, you can have construct $P_f$ as follows: If $f(i) = 1$, then put into $P_f$, the subset $(sum_{k < i st f(k) = 1} k + 1, sum_{k < i st f(k) = 1} k + 1 + i)$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    An equivalence relation on any set partitions that set.



    So we can approach your question by asking, equivalently: How many ways are there to partition $mathbb{N}$?




    • A partition of $X$ with is a set P of nonempty subsets of X, such that every element of X is an element of a single element of P. Each element of P is a cell of the partition. Moreover, the elements of P are pairwise disjoint and their union is X.


    • $|mathbb{N}| = aleph_0$. There are uncountably many possible distinct partitions of $mathbb{N}: 2^{aleph_0}$ distinct partitions.
      (See this post for a nicely detailed proof of this - not just a sketch of a proof.)


    • Since there is a natural bijection from the set of all possible equivalence
      relations on a set $mathbb{N}$ and the set of all partitions of $mathbb{N}$, there are $2^{aleph_0}$ equivalence relations on $mathbb{N}$.





    • For counting the number of partitions on a finite sets, see Counting Partitions.


    • For an elaboration on the relationship between an equivalence relation on a set and the partition of the set induced by that relation, vice versa, see this previous post







    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%2f271540%2fequivalence-relations-on-natural-numbers%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









      2












      $begingroup$

      There are $2^{aleph_0}$ many. As mentioned above, equivalence relation corresponds to partition of $mathbb{N}$. For each $f in 2^{mathbb{N}}$, i.e. a function $f : mathbb{N} rightarrow {0,1}$. Define a partition, $P_f$, such that, $P_f$ has a subset of size $n$ if and only if $f(n) = 1$. Then if $f neq g$, then $P_f neq P_g$. Hence the number of equivalence relations on $mathbb{N}$ is at least as many as $2^{aleph_0}$. It can not be bigger because the power set $mathcal{P}(mathbb{N})$ has cardinality $2^{aleph_0}$.




      Explicitly, you can have construct $P_f$ as follows: If $f(i) = 1$, then put into $P_f$, the subset $(sum_{k < i st f(k) = 1} k + 1, sum_{k < i st f(k) = 1} k + 1 + i)$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        There are $2^{aleph_0}$ many. As mentioned above, equivalence relation corresponds to partition of $mathbb{N}$. For each $f in 2^{mathbb{N}}$, i.e. a function $f : mathbb{N} rightarrow {0,1}$. Define a partition, $P_f$, such that, $P_f$ has a subset of size $n$ if and only if $f(n) = 1$. Then if $f neq g$, then $P_f neq P_g$. Hence the number of equivalence relations on $mathbb{N}$ is at least as many as $2^{aleph_0}$. It can not be bigger because the power set $mathcal{P}(mathbb{N})$ has cardinality $2^{aleph_0}$.




        Explicitly, you can have construct $P_f$ as follows: If $f(i) = 1$, then put into $P_f$, the subset $(sum_{k < i st f(k) = 1} k + 1, sum_{k < i st f(k) = 1} k + 1 + i)$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          There are $2^{aleph_0}$ many. As mentioned above, equivalence relation corresponds to partition of $mathbb{N}$. For each $f in 2^{mathbb{N}}$, i.e. a function $f : mathbb{N} rightarrow {0,1}$. Define a partition, $P_f$, such that, $P_f$ has a subset of size $n$ if and only if $f(n) = 1$. Then if $f neq g$, then $P_f neq P_g$. Hence the number of equivalence relations on $mathbb{N}$ is at least as many as $2^{aleph_0}$. It can not be bigger because the power set $mathcal{P}(mathbb{N})$ has cardinality $2^{aleph_0}$.




          Explicitly, you can have construct $P_f$ as follows: If $f(i) = 1$, then put into $P_f$, the subset $(sum_{k < i st f(k) = 1} k + 1, sum_{k < i st f(k) = 1} k + 1 + i)$.






          share|cite|improve this answer











          $endgroup$



          There are $2^{aleph_0}$ many. As mentioned above, equivalence relation corresponds to partition of $mathbb{N}$. For each $f in 2^{mathbb{N}}$, i.e. a function $f : mathbb{N} rightarrow {0,1}$. Define a partition, $P_f$, such that, $P_f$ has a subset of size $n$ if and only if $f(n) = 1$. Then if $f neq g$, then $P_f neq P_g$. Hence the number of equivalence relations on $mathbb{N}$ is at least as many as $2^{aleph_0}$. It can not be bigger because the power set $mathcal{P}(mathbb{N})$ has cardinality $2^{aleph_0}$.




          Explicitly, you can have construct $P_f$ as follows: If $f(i) = 1$, then put into $P_f$, the subset $(sum_{k < i st f(k) = 1} k + 1, sum_{k < i st f(k) = 1} k + 1 + i)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 '13 at 14:35

























          answered Jan 6 '13 at 14:28









          WilliamWilliam

          17.1k22156




          17.1k22156























              2












              $begingroup$

              An equivalence relation on any set partitions that set.



              So we can approach your question by asking, equivalently: How many ways are there to partition $mathbb{N}$?




              • A partition of $X$ with is a set P of nonempty subsets of X, such that every element of X is an element of a single element of P. Each element of P is a cell of the partition. Moreover, the elements of P are pairwise disjoint and their union is X.


              • $|mathbb{N}| = aleph_0$. There are uncountably many possible distinct partitions of $mathbb{N}: 2^{aleph_0}$ distinct partitions.
                (See this post for a nicely detailed proof of this - not just a sketch of a proof.)


              • Since there is a natural bijection from the set of all possible equivalence
                relations on a set $mathbb{N}$ and the set of all partitions of $mathbb{N}$, there are $2^{aleph_0}$ equivalence relations on $mathbb{N}$.





              • For counting the number of partitions on a finite sets, see Counting Partitions.


              • For an elaboration on the relationship between an equivalence relation on a set and the partition of the set induced by that relation, vice versa, see this previous post







              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                An equivalence relation on any set partitions that set.



                So we can approach your question by asking, equivalently: How many ways are there to partition $mathbb{N}$?




                • A partition of $X$ with is a set P of nonempty subsets of X, such that every element of X is an element of a single element of P. Each element of P is a cell of the partition. Moreover, the elements of P are pairwise disjoint and their union is X.


                • $|mathbb{N}| = aleph_0$. There are uncountably many possible distinct partitions of $mathbb{N}: 2^{aleph_0}$ distinct partitions.
                  (See this post for a nicely detailed proof of this - not just a sketch of a proof.)


                • Since there is a natural bijection from the set of all possible equivalence
                  relations on a set $mathbb{N}$ and the set of all partitions of $mathbb{N}$, there are $2^{aleph_0}$ equivalence relations on $mathbb{N}$.





                • For counting the number of partitions on a finite sets, see Counting Partitions.


                • For an elaboration on the relationship between an equivalence relation on a set and the partition of the set induced by that relation, vice versa, see this previous post







                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  An equivalence relation on any set partitions that set.



                  So we can approach your question by asking, equivalently: How many ways are there to partition $mathbb{N}$?




                  • A partition of $X$ with is a set P of nonempty subsets of X, such that every element of X is an element of a single element of P. Each element of P is a cell of the partition. Moreover, the elements of P are pairwise disjoint and their union is X.


                  • $|mathbb{N}| = aleph_0$. There are uncountably many possible distinct partitions of $mathbb{N}: 2^{aleph_0}$ distinct partitions.
                    (See this post for a nicely detailed proof of this - not just a sketch of a proof.)


                  • Since there is a natural bijection from the set of all possible equivalence
                    relations on a set $mathbb{N}$ and the set of all partitions of $mathbb{N}$, there are $2^{aleph_0}$ equivalence relations on $mathbb{N}$.





                  • For counting the number of partitions on a finite sets, see Counting Partitions.


                  • For an elaboration on the relationship between an equivalence relation on a set and the partition of the set induced by that relation, vice versa, see this previous post







                  share|cite|improve this answer











                  $endgroup$



                  An equivalence relation on any set partitions that set.



                  So we can approach your question by asking, equivalently: How many ways are there to partition $mathbb{N}$?




                  • A partition of $X$ with is a set P of nonempty subsets of X, such that every element of X is an element of a single element of P. Each element of P is a cell of the partition. Moreover, the elements of P are pairwise disjoint and their union is X.


                  • $|mathbb{N}| = aleph_0$. There are uncountably many possible distinct partitions of $mathbb{N}: 2^{aleph_0}$ distinct partitions.
                    (See this post for a nicely detailed proof of this - not just a sketch of a proof.)


                  • Since there is a natural bijection from the set of all possible equivalence
                    relations on a set $mathbb{N}$ and the set of all partitions of $mathbb{N}$, there are $2^{aleph_0}$ equivalence relations on $mathbb{N}$.





                  • For counting the number of partitions on a finite sets, see Counting Partitions.


                  • For an elaboration on the relationship between an equivalence relation on a set and the partition of the set induced by that relation, vice versa, see this previous post








                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 13 '17 at 12:21









                  Community♦

                  1




                  1










                  answered Jan 6 '13 at 14:16









                  amWhyamWhy

                  192k28225439




                  192k28225439






























                      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%2f271540%2fequivalence-relations-on-natural-numbers%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

                      Le Mesnil-Réaume

                      Image
                      Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more

                      web3.py web3.isConnected() returns false always

                      Image
                      1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
                      Read more