Equivalence relations on natural numbers












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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







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  • $begingroup$
    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











    $endgroup$













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      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




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                  answered Jan 6 '13 at 14:16









                  amWhyamWhy

                  192k28225439




                  192k28225439






























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