Is there a proof of the proposition that $equiv$ is an equivalence relation?












0












$begingroup$


Would anyone happen to have a proof that the following use of $equiv$ is an equivalence relation:



Let $mathcal{F}(mathbb{N})$ be an ultrafilter on $mathbb{N}$, constructed by choosing subsets of $mathbb{N}$ with large cardinality. Suppose $lbrace s_nrbrace$ and $lbrace r_nrbrace$ are real-valued sequences. Now, let $A$ be the set of all $ninmathbb{N}$ such that $r_n=s_n$. Then, $lbrace s_nrbrace$ and $lbrace r_nrbrace$ satisfy $lbrace s_nrbraceequivlbrace r_nrbrace$ if and only if $Ainmathcal{F}(mathbb{N})$.



Thank you in advance.






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    0












    $begingroup$


    Would anyone happen to have a proof that the following use of $equiv$ is an equivalence relation:



    Let $mathcal{F}(mathbb{N})$ be an ultrafilter on $mathbb{N}$, constructed by choosing subsets of $mathbb{N}$ with large cardinality. Suppose $lbrace s_nrbrace$ and $lbrace r_nrbrace$ are real-valued sequences. Now, let $A$ be the set of all $ninmathbb{N}$ such that $r_n=s_n$. Then, $lbrace s_nrbrace$ and $lbrace r_nrbrace$ satisfy $lbrace s_nrbraceequivlbrace r_nrbrace$ if and only if $Ainmathcal{F}(mathbb{N})$.



    Thank you in advance.






    reference-request filters







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Would anyone happen to have a proof that the following use of $equiv$ is an equivalence relation:



      Let $mathcal{F}(mathbb{N})$ be an ultrafilter on $mathbb{N}$, constructed by choosing subsets of $mathbb{N}$ with large cardinality. Suppose $lbrace s_nrbrace$ and $lbrace r_nrbrace$ are real-valued sequences. Now, let $A$ be the set of all $ninmathbb{N}$ such that $r_n=s_n$. Then, $lbrace s_nrbrace$ and $lbrace r_nrbrace$ satisfy $lbrace s_nrbraceequivlbrace r_nrbrace$ if and only if $Ainmathcal{F}(mathbb{N})$.



      Thank you in advance.






      reference-request filters







      share|cite|improve this question









      $endgroup$




      Would anyone happen to have a proof that the following use of $equiv$ is an equivalence relation:



      Let $mathcal{F}(mathbb{N})$ be an ultrafilter on $mathbb{N}$, constructed by choosing subsets of $mathbb{N}$ with large cardinality. Suppose $lbrace s_nrbrace$ and $lbrace r_nrbrace$ are real-valued sequences. Now, let $A$ be the set of all $ninmathbb{N}$ such that $r_n=s_n$. Then, $lbrace s_nrbrace$ and $lbrace r_nrbrace$ satisfy $lbrace s_nrbraceequivlbrace r_nrbrace$ if and only if $Ainmathcal{F}(mathbb{N})$.



      Thank you in advance.






      reference-request filters




      reference-request filters






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 12 at 12:57









      joshuaheckroodtjoshuaheckroodt

      1,205622




      1,205622






















          1 Answer
          1






          active

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          1












          $begingroup$

          "constructed by chooising subsets of large cardinality" is somewhat inexact and misleading. You probably mean to say it is a so-called free ultrafilter, which is not of the form $mathcal{F}_n={Asubseteq mathbb{N}: n in A}$, for some $n in mathbb{N}$, which can be "constructed" (not really) by extending the filter of co-finite subsets (all subsets of $mathbb{N}$ with finite complement) to an ultrafilter (which uses a form of the axiom of choice, and is far from constructive).



          So let $mathcal{F}$ be non-empty a(n ultra)filter (free or not, it does not matter for the proof) and define $(a_n)_n equiv (b_n)_n$ iff ${n : a_n = b_n} in mathcal{F}$.



          This is easily an equivalence relation: $(a_n) equiv (a_n)_n$ follows from $mathbb{N} in mathcal{F}$ for all non-empty filters (from closedness under supersets). Symmetry is clear from symmetry of $=$ itself:



          $${n: a_n = b_n} = {n: b_n = a_n}$$



          And moreover if ${n: a_n = b_n } in mathcal{F}$ and ${n: b_n = c_n} in mathcal{F}$ then



          $${n: a_n = b_n } cap {n: b_n = c_n } subseteq {n: a_n = c_n }$$



          so again the latter set is in the filter too. This shows transitivity.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for this. I just want to make sure I understand the following correctly, since I'm still quite new to this theory: if $A$ is a cofinite subset of $mathbb{N}$, then $A$ is infinite, correct? Which means that whichever filter is on $A$ can be either free or principal?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:04






          • 1




            $begingroup$
            @joshuaheckroodt yes, cofinite sets are infinite but the reverse does not hold, e.g. the even numbers are infinite but not cofinite. An ultrafilter is principal (of the form $mathcal{F}_n$ iff it contains a finite set. So it is free iff it contains all cofinite subsets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:17










          • $begingroup$
            So a free ultrafilter only contains infinite sets as its elements?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:22






          • 1




            $begingroup$
            @joshuaheckroodt yes and in particular all cofinite sets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:22






          • 1




            $begingroup$
            Thank's so much for the clarification, and again for the proof that $equiv$ is an equivalence relation.
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:24











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          $begingroup$

          "constructed by chooising subsets of large cardinality" is somewhat inexact and misleading. You probably mean to say it is a so-called free ultrafilter, which is not of the form $mathcal{F}_n={Asubseteq mathbb{N}: n in A}$, for some $n in mathbb{N}$, which can be "constructed" (not really) by extending the filter of co-finite subsets (all subsets of $mathbb{N}$ with finite complement) to an ultrafilter (which uses a form of the axiom of choice, and is far from constructive).



          So let $mathcal{F}$ be non-empty a(n ultra)filter (free or not, it does not matter for the proof) and define $(a_n)_n equiv (b_n)_n$ iff ${n : a_n = b_n} in mathcal{F}$.



          This is easily an equivalence relation: $(a_n) equiv (a_n)_n$ follows from $mathbb{N} in mathcal{F}$ for all non-empty filters (from closedness under supersets). Symmetry is clear from symmetry of $=$ itself:



          $${n: a_n = b_n} = {n: b_n = a_n}$$



          And moreover if ${n: a_n = b_n } in mathcal{F}$ and ${n: b_n = c_n} in mathcal{F}$ then



          $${n: a_n = b_n } cap {n: b_n = c_n } subseteq {n: a_n = c_n }$$



          so again the latter set is in the filter too. This shows transitivity.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for this. I just want to make sure I understand the following correctly, since I'm still quite new to this theory: if $A$ is a cofinite subset of $mathbb{N}$, then $A$ is infinite, correct? Which means that whichever filter is on $A$ can be either free or principal?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:04






          • 1




            $begingroup$
            @joshuaheckroodt yes, cofinite sets are infinite but the reverse does not hold, e.g. the even numbers are infinite but not cofinite. An ultrafilter is principal (of the form $mathcal{F}_n$ iff it contains a finite set. So it is free iff it contains all cofinite subsets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:17










          • $begingroup$
            So a free ultrafilter only contains infinite sets as its elements?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:22






          • 1




            $begingroup$
            @joshuaheckroodt yes and in particular all cofinite sets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:22






          • 1




            $begingroup$
            Thank's so much for the clarification, and again for the proof that $equiv$ is an equivalence relation.
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:24
















          1












          $begingroup$

          "constructed by chooising subsets of large cardinality" is somewhat inexact and misleading. You probably mean to say it is a so-called free ultrafilter, which is not of the form $mathcal{F}_n={Asubseteq mathbb{N}: n in A}$, for some $n in mathbb{N}$, which can be "constructed" (not really) by extending the filter of co-finite subsets (all subsets of $mathbb{N}$ with finite complement) to an ultrafilter (which uses a form of the axiom of choice, and is far from constructive).



          So let $mathcal{F}$ be non-empty a(n ultra)filter (free or not, it does not matter for the proof) and define $(a_n)_n equiv (b_n)_n$ iff ${n : a_n = b_n} in mathcal{F}$.



          This is easily an equivalence relation: $(a_n) equiv (a_n)_n$ follows from $mathbb{N} in mathcal{F}$ for all non-empty filters (from closedness under supersets). Symmetry is clear from symmetry of $=$ itself:



          $${n: a_n = b_n} = {n: b_n = a_n}$$



          And moreover if ${n: a_n = b_n } in mathcal{F}$ and ${n: b_n = c_n} in mathcal{F}$ then



          $${n: a_n = b_n } cap {n: b_n = c_n } subseteq {n: a_n = c_n }$$



          so again the latter set is in the filter too. This shows transitivity.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for this. I just want to make sure I understand the following correctly, since I'm still quite new to this theory: if $A$ is a cofinite subset of $mathbb{N}$, then $A$ is infinite, correct? Which means that whichever filter is on $A$ can be either free or principal?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:04






          • 1




            $begingroup$
            @joshuaheckroodt yes, cofinite sets are infinite but the reverse does not hold, e.g. the even numbers are infinite but not cofinite. An ultrafilter is principal (of the form $mathcal{F}_n$ iff it contains a finite set. So it is free iff it contains all cofinite subsets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:17










          • $begingroup$
            So a free ultrafilter only contains infinite sets as its elements?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:22






          • 1




            $begingroup$
            @joshuaheckroodt yes and in particular all cofinite sets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:22






          • 1




            $begingroup$
            Thank's so much for the clarification, and again for the proof that $equiv$ is an equivalence relation.
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:24














          1












          1








          1





          $begingroup$

          "constructed by chooising subsets of large cardinality" is somewhat inexact and misleading. You probably mean to say it is a so-called free ultrafilter, which is not of the form $mathcal{F}_n={Asubseteq mathbb{N}: n in A}$, for some $n in mathbb{N}$, which can be "constructed" (not really) by extending the filter of co-finite subsets (all subsets of $mathbb{N}$ with finite complement) to an ultrafilter (which uses a form of the axiom of choice, and is far from constructive).



          So let $mathcal{F}$ be non-empty a(n ultra)filter (free or not, it does not matter for the proof) and define $(a_n)_n equiv (b_n)_n$ iff ${n : a_n = b_n} in mathcal{F}$.



          This is easily an equivalence relation: $(a_n) equiv (a_n)_n$ follows from $mathbb{N} in mathcal{F}$ for all non-empty filters (from closedness under supersets). Symmetry is clear from symmetry of $=$ itself:



          $${n: a_n = b_n} = {n: b_n = a_n}$$



          And moreover if ${n: a_n = b_n } in mathcal{F}$ and ${n: b_n = c_n} in mathcal{F}$ then



          $${n: a_n = b_n } cap {n: b_n = c_n } subseteq {n: a_n = c_n }$$



          so again the latter set is in the filter too. This shows transitivity.






          share|cite|improve this answer











          $endgroup$



          "constructed by chooising subsets of large cardinality" is somewhat inexact and misleading. You probably mean to say it is a so-called free ultrafilter, which is not of the form $mathcal{F}_n={Asubseteq mathbb{N}: n in A}$, for some $n in mathbb{N}$, which can be "constructed" (not really) by extending the filter of co-finite subsets (all subsets of $mathbb{N}$ with finite complement) to an ultrafilter (which uses a form of the axiom of choice, and is far from constructive).



          So let $mathcal{F}$ be non-empty a(n ultra)filter (free or not, it does not matter for the proof) and define $(a_n)_n equiv (b_n)_n$ iff ${n : a_n = b_n} in mathcal{F}$.



          This is easily an equivalence relation: $(a_n) equiv (a_n)_n$ follows from $mathbb{N} in mathcal{F}$ for all non-empty filters (from closedness under supersets). Symmetry is clear from symmetry of $=$ itself:



          $${n: a_n = b_n} = {n: b_n = a_n}$$



          And moreover if ${n: a_n = b_n } in mathcal{F}$ and ${n: b_n = c_n} in mathcal{F}$ then



          $${n: a_n = b_n } cap {n: b_n = c_n } subseteq {n: a_n = c_n }$$



          so again the latter set is in the filter too. This shows transitivity.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered Jan 13 at 17:33









          Henno BrandsmaHenno Brandsma

          107k347114




          107k347114












          • $begingroup$
            Thanks for this. I just want to make sure I understand the following correctly, since I'm still quite new to this theory: if $A$ is a cofinite subset of $mathbb{N}$, then $A$ is infinite, correct? Which means that whichever filter is on $A$ can be either free or principal?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:04






          • 1




            $begingroup$
            @joshuaheckroodt yes, cofinite sets are infinite but the reverse does not hold, e.g. the even numbers are infinite but not cofinite. An ultrafilter is principal (of the form $mathcal{F}_n$ iff it contains a finite set. So it is free iff it contains all cofinite subsets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:17










          • $begingroup$
            So a free ultrafilter only contains infinite sets as its elements?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:22






          • 1




            $begingroup$
            @joshuaheckroodt yes and in particular all cofinite sets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:22






          • 1




            $begingroup$
            Thank's so much for the clarification, and again for the proof that $equiv$ is an equivalence relation.
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:24


















          • $begingroup$
            Thanks for this. I just want to make sure I understand the following correctly, since I'm still quite new to this theory: if $A$ is a cofinite subset of $mathbb{N}$, then $A$ is infinite, correct? Which means that whichever filter is on $A$ can be either free or principal?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:04






          • 1




            $begingroup$
            @joshuaheckroodt yes, cofinite sets are infinite but the reverse does not hold, e.g. the even numbers are infinite but not cofinite. An ultrafilter is principal (of the form $mathcal{F}_n$ iff it contains a finite set. So it is free iff it contains all cofinite subsets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:17










          • $begingroup$
            So a free ultrafilter only contains infinite sets as its elements?
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:22






          • 1




            $begingroup$
            @joshuaheckroodt yes and in particular all cofinite sets.
            $endgroup$
            – Henno Brandsma
            Jan 13 at 18:22






          • 1




            $begingroup$
            Thank's so much for the clarification, and again for the proof that $equiv$ is an equivalence relation.
            $endgroup$
            – joshuaheckroodt
            Jan 13 at 18:24
















          $begingroup$
          Thanks for this. I just want to make sure I understand the following correctly, since I'm still quite new to this theory: if $A$ is a cofinite subset of $mathbb{N}$, then $A$ is infinite, correct? Which means that whichever filter is on $A$ can be either free or principal?
          $endgroup$
          – joshuaheckroodt
          Jan 13 at 18:04




          $begingroup$
          Thanks for this. I just want to make sure I understand the following correctly, since I'm still quite new to this theory: if $A$ is a cofinite subset of $mathbb{N}$, then $A$ is infinite, correct? Which means that whichever filter is on $A$ can be either free or principal?
          $endgroup$
          – joshuaheckroodt
          Jan 13 at 18:04




          1




          1




          $begingroup$
          @joshuaheckroodt yes, cofinite sets are infinite but the reverse does not hold, e.g. the even numbers are infinite but not cofinite. An ultrafilter is principal (of the form $mathcal{F}_n$ iff it contains a finite set. So it is free iff it contains all cofinite subsets.
          $endgroup$
          – Henno Brandsma
          Jan 13 at 18:17




          $begingroup$
          @joshuaheckroodt yes, cofinite sets are infinite but the reverse does not hold, e.g. the even numbers are infinite but not cofinite. An ultrafilter is principal (of the form $mathcal{F}_n$ iff it contains a finite set. So it is free iff it contains all cofinite subsets.
          $endgroup$
          – Henno Brandsma
          Jan 13 at 18:17












          $begingroup$
          So a free ultrafilter only contains infinite sets as its elements?
          $endgroup$
          – joshuaheckroodt
          Jan 13 at 18:22




          $begingroup$
          So a free ultrafilter only contains infinite sets as its elements?
          $endgroup$
          – joshuaheckroodt
          Jan 13 at 18:22




          1




          1




          $begingroup$
          @joshuaheckroodt yes and in particular all cofinite sets.
          $endgroup$
          – Henno Brandsma
          Jan 13 at 18:22




          $begingroup$
          @joshuaheckroodt yes and in particular all cofinite sets.
          $endgroup$
          – Henno Brandsma
          Jan 13 at 18:22




          1




          1




          $begingroup$
          Thank's so much for the clarification, and again for the proof that $equiv$ is an equivalence relation.
          $endgroup$
          – joshuaheckroodt
          Jan 13 at 18:24




          $begingroup$
          Thank's so much for the clarification, and again for the proof that $equiv$ is an equivalence relation.
          $endgroup$
          – joshuaheckroodt
          Jan 13 at 18:24


















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