What is the difference between Turing-reducibility and m-reducibility?












1












$begingroup$


The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.



We shall write $A ≤_{m} B$



Let be $A, B subset mathbb{N}$



is it true that $A leq_{m} B implies Aleq_{T}B$?



Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.



If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?






logic computability







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$endgroup$

















    1












    $begingroup$


    The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
    total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.



    We shall write $A ≤_{m} B$



    Let be $A, B subset mathbb{N}$



    is it true that $A leq_{m} B implies Aleq_{T}B$?



    Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.



    If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?






    logic computability







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
      total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.



      We shall write $A ≤_{m} B$



      Let be $A, B subset mathbb{N}$



      is it true that $A leq_{m} B implies Aleq_{T}B$?



      Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.



      If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?






      logic computability







      share|cite|improve this question











      $endgroup$




      The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
      total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.



      We shall write $A ≤_{m} B$



      Let be $A, B subset mathbb{N}$



      is it true that $A leq_{m} B implies Aleq_{T}B$?



      Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.



      If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?






      logic computability




      logic computability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 22:18







      Nicolás Castro

















      asked Dec 10 '18 at 22:11









      Nicolás CastroNicolás Castro

      83




      83






















          1 Answer
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          1












          $begingroup$

          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)




          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23













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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)




          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23


















          1












          $begingroup$

          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)




          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23
















          1












          1








          1





          $begingroup$

          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)




          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).






          share|cite|improve this answer









          $endgroup$



          It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.



          An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.



          This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)




          Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 22:18









          Noah SchweberNoah Schweber

          124k10150287




          124k10150287












          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23




















          • $begingroup$
            Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
            $endgroup$
            – Nicolás Castro
            Dec 10 '18 at 22:34












          • $begingroup$
            @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
            $endgroup$
            – Noah Schweber
            Dec 10 '18 at 23:23


















          $begingroup$
          Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
          $endgroup$
          – Nicolás Castro
          Dec 10 '18 at 22:34






          $begingroup$
          Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
          $endgroup$
          – Nicolás Castro
          Dec 10 '18 at 22:34














          $begingroup$
          @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
          $endgroup$
          – Noah Schweber
          Dec 10 '18 at 23:23






          $begingroup$
          @NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
          $endgroup$
          – Noah Schweber
          Dec 10 '18 at 23:23




















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