Computability of function of two arguments.












1












$begingroup$


I understand that this question shouldn't be hard but still.




Let $U: mathbb N times mathbb N rightarrow mathbb N$ be a function such that for any fixed $c$ function $D_c(y) = U(c, y)$ is computable. Does it imply that $U$ is computable?




I believe that answer is no, but I can't find an example to demonstrate it.



Thanks!






computability







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$endgroup$

















    1












    $begingroup$


    I understand that this question shouldn't be hard but still.




    Let $U: mathbb N times mathbb N rightarrow mathbb N$ be a function such that for any fixed $c$ function $D_c(y) = U(c, y)$ is computable. Does it imply that $U$ is computable?




    I believe that answer is no, but I can't find an example to demonstrate it.



    Thanks!






    computability







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I understand that this question shouldn't be hard but still.




      Let $U: mathbb N times mathbb N rightarrow mathbb N$ be a function such that for any fixed $c$ function $D_c(y) = U(c, y)$ is computable. Does it imply that $U$ is computable?




      I believe that answer is no, but I can't find an example to demonstrate it.



      Thanks!






      computability







      share|cite|improve this question









      $endgroup$




      I understand that this question shouldn't be hard but still.




      Let $U: mathbb N times mathbb N rightarrow mathbb N$ be a function such that for any fixed $c$ function $D_c(y) = U(c, y)$ is computable. Does it imply that $U$ is computable?




      I believe that answer is no, but I can't find an example to demonstrate it.



      Thanks!






      computability




      computability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 22:50









      Gleb ChiliGleb Chili

      43728




      43728






















          1 Answer
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          $begingroup$

          No. Let $C$ be the set of computable functions from $Bbb{N}$. Then $C$ is (countably) infinite, and there are uncountably many functions $D : Bbb{N} to C$. Each such function $D$ corresponds to a unique function $U : Bbb{N} times Bbb{N} to Bbb{N}$ such that $U(c, y) = D_c(y)$ for all $c$ and $y$ with $D_c$ computable. But there are only countably many computable functions from $Bbb{N}times Bbb{N}$ to $Bbb{N}$, so uncountably many of those functions $U$ are not computable.



          For a specific counter-example: let $H$ be the set of codes of partial recursive functions (or Turing machines) $x$, such that ${x}(x)$ is defined (or terminates). Define
          $$
          U(c, y) = left{
          begin{array}{cl}
          0 &mbox{if $c in H$} \
          1 & mbox{if $c notin H$}
          end{array}right.
          $$



          Then $D_c(y) = U(c, y)$ is a computable function of $y$ for every $c$, but $U$ is not computable because $H$ is not recursive.






          share|cite|improve this answer











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            $begingroup$

            No. Let $C$ be the set of computable functions from $Bbb{N}$. Then $C$ is (countably) infinite, and there are uncountably many functions $D : Bbb{N} to C$. Each such function $D$ corresponds to a unique function $U : Bbb{N} times Bbb{N} to Bbb{N}$ such that $U(c, y) = D_c(y)$ for all $c$ and $y$ with $D_c$ computable. But there are only countably many computable functions from $Bbb{N}times Bbb{N}$ to $Bbb{N}$, so uncountably many of those functions $U$ are not computable.



            For a specific counter-example: let $H$ be the set of codes of partial recursive functions (or Turing machines) $x$, such that ${x}(x)$ is defined (or terminates). Define
            $$
            U(c, y) = left{
            begin{array}{cl}
            0 &mbox{if $c in H$} \
            1 & mbox{if $c notin H$}
            end{array}right.
            $$



            Then $D_c(y) = U(c, y)$ is a computable function of $y$ for every $c$, but $U$ is not computable because $H$ is not recursive.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              No. Let $C$ be the set of computable functions from $Bbb{N}$. Then $C$ is (countably) infinite, and there are uncountably many functions $D : Bbb{N} to C$. Each such function $D$ corresponds to a unique function $U : Bbb{N} times Bbb{N} to Bbb{N}$ such that $U(c, y) = D_c(y)$ for all $c$ and $y$ with $D_c$ computable. But there are only countably many computable functions from $Bbb{N}times Bbb{N}$ to $Bbb{N}$, so uncountably many of those functions $U$ are not computable.



              For a specific counter-example: let $H$ be the set of codes of partial recursive functions (or Turing machines) $x$, such that ${x}(x)$ is defined (or terminates). Define
              $$
              U(c, y) = left{
              begin{array}{cl}
              0 &mbox{if $c in H$} \
              1 & mbox{if $c notin H$}
              end{array}right.
              $$



              Then $D_c(y) = U(c, y)$ is a computable function of $y$ for every $c$, but $U$ is not computable because $H$ is not recursive.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                No. Let $C$ be the set of computable functions from $Bbb{N}$. Then $C$ is (countably) infinite, and there are uncountably many functions $D : Bbb{N} to C$. Each such function $D$ corresponds to a unique function $U : Bbb{N} times Bbb{N} to Bbb{N}$ such that $U(c, y) = D_c(y)$ for all $c$ and $y$ with $D_c$ computable. But there are only countably many computable functions from $Bbb{N}times Bbb{N}$ to $Bbb{N}$, so uncountably many of those functions $U$ are not computable.



                For a specific counter-example: let $H$ be the set of codes of partial recursive functions (or Turing machines) $x$, such that ${x}(x)$ is defined (or terminates). Define
                $$
                U(c, y) = left{
                begin{array}{cl}
                0 &mbox{if $c in H$} \
                1 & mbox{if $c notin H$}
                end{array}right.
                $$



                Then $D_c(y) = U(c, y)$ is a computable function of $y$ for every $c$, but $U$ is not computable because $H$ is not recursive.






                share|cite|improve this answer











                $endgroup$



                No. Let $C$ be the set of computable functions from $Bbb{N}$. Then $C$ is (countably) infinite, and there are uncountably many functions $D : Bbb{N} to C$. Each such function $D$ corresponds to a unique function $U : Bbb{N} times Bbb{N} to Bbb{N}$ such that $U(c, y) = D_c(y)$ for all $c$ and $y$ with $D_c$ computable. But there are only countably many computable functions from $Bbb{N}times Bbb{N}$ to $Bbb{N}$, so uncountably many of those functions $U$ are not computable.



                For a specific counter-example: let $H$ be the set of codes of partial recursive functions (or Turing machines) $x$, such that ${x}(x)$ is defined (or terminates). Define
                $$
                U(c, y) = left{
                begin{array}{cl}
                0 &mbox{if $c in H$} \
                1 & mbox{if $c notin H$}
                end{array}right.
                $$



                Then $D_c(y) = U(c, y)$ is a computable function of $y$ for every $c$, but $U$ is not computable because $H$ is not recursive.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 13 '18 at 23:25

























                answered Dec 13 '18 at 23:18









                Rob ArthanRob Arthan

                29.1k42866




                29.1k42866






























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