Is there a partial recursive function mapping $e$ to the least element of $W_e$?












1












$begingroup$


Is there a partial recursive function $f$ that maps $e$ to the least element of $W_e$ if $W_eneq varnothing$?



Can we apply the $mu$-operator (minimization) to a partially decidable predicate to get a partial recursive function (as in defining $f(x)=mu z(zin W_e)$ if $W_eneqvarnothing$ and undefined otherwise, since $zin W_e$ is partially decidable)?



My guess was that there is no such $f$, but I wasn't sure how to show that.
Thanks in adavnce.






logic computability







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$endgroup$

















    1












    $begingroup$


    Is there a partial recursive function $f$ that maps $e$ to the least element of $W_e$ if $W_eneq varnothing$?



    Can we apply the $mu$-operator (minimization) to a partially decidable predicate to get a partial recursive function (as in defining $f(x)=mu z(zin W_e)$ if $W_eneqvarnothing$ and undefined otherwise, since $zin W_e$ is partially decidable)?



    My guess was that there is no such $f$, but I wasn't sure how to show that.
    Thanks in adavnce.






    logic computability







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Is there a partial recursive function $f$ that maps $e$ to the least element of $W_e$ if $W_eneq varnothing$?



      Can we apply the $mu$-operator (minimization) to a partially decidable predicate to get a partial recursive function (as in defining $f(x)=mu z(zin W_e)$ if $W_eneqvarnothing$ and undefined otherwise, since $zin W_e$ is partially decidable)?



      My guess was that there is no such $f$, but I wasn't sure how to show that.
      Thanks in adavnce.






      logic computability







      share|cite|improve this question









      $endgroup$




      Is there a partial recursive function $f$ that maps $e$ to the least element of $W_e$ if $W_eneq varnothing$?



      Can we apply the $mu$-operator (minimization) to a partially decidable predicate to get a partial recursive function (as in defining $f(x)=mu z(zin W_e)$ if $W_eneqvarnothing$ and undefined otherwise, since $zin W_e$ is partially decidable)?



      My guess was that there is no such $f$, but I wasn't sure how to show that.
      Thanks in adavnce.






      logic computability




      logic computability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 8 '18 at 20:26









      user500144user500144

      645




      645






















          1 Answer
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          active

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          1












          $begingroup$

          Your intuition is right: there is no such function.



          We can see this in a couple different ways:




          • Contradiction: Suppose there were such an $f$. Then - given an arbitrary (infinite) c.e. set $A$ - we could compute the sequence $$min(A), min(Asetminus{min(A)}), min(Asetminus{min(A), min(Asetminus{min(A)})}), ...$$ by iteratively applying $f$. But this precisely enumerates $A$ in order, and so makes $A$ computable. Since there are non-computable sets, this can't happen.



          • Direct diagonalization: Given a computable function $f$, we can find an $e$ such that $(i)$ $2$ enters $W_e$ at stage $0$, $(ii)$ $1$ enters $W_e$ at stage $n+1$ iff $f(e)$ halts and outputs "$2$" at stage exactly $n$, and $(iii)$ no other numbers enter $W_e$ under any other circumstances. Then we get $min(W_e)=2$ iff $f(e)not=2$, so $f$ doesn't do what we want it to.




            • This fits into the "game picture" of c.e. sets: I pick a computable $f$, and you start building my $W_e$. You put $2$ into it, which forces me to eventually decide what $f(e)$ is; as soon as I do, if I say "$2$" then you put $1$ into my set, and if I say anything other than "$2$" you do nothing. Either way, you've defeated my purported $f$.




          I haven't provided full details of either case, but the idea should be clear. Filling in the details (of the first one, especially) is a good exercise.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I haven't seen much of the "game" arguments, so could you explain what you mean by we "build $W_e$"? Isn't $W_e$ already determined for each $e$? Also, what do you mean by we can "find" $e$ such that 2 enters $W_e$ at stage 0... and so on?
            $endgroup$
            – user500144
            Dec 8 '18 at 21:02










          • $begingroup$
            @user500144 The two key ingredients here are Church's thesis and, more subtly, the recursion theorem. It's the latter which lets us seemingly "define $e$ while knowing already $e$."
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:05










          • $begingroup$
            Specifically, the argument has two pieces. Suppose $f$ has the property proposed. We begin with a Church's thesis argument: there is a total computable $h$ such that, for each $e$, we have $W_{h(e)}={1,2}$ if $f(e)downarrow=2$ and $W_{h(e)}={2}$ otherwise.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:16










          • $begingroup$
            (cont'd) The interesting step is applying the Recursion Theorem. Since $h$ is total computable, it has a fixed point - some $c$ such that $W_c=W_{h(c)}$. But then we have $W_{c}={1,2}$ if $f(c)downarrow=2$ and $W_c={2}$ otherwise - contradicting the assumption on $f$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:17










          • $begingroup$
            (contd) The recursion theorem aside (it is genuinely a technical result), all of this stuff becomes more intuitive once you think of c.e. sets as sets you can "enumerate effectively." E.g. intuitively we can effectively enumerate the theorems of a given (computably axiomatizable) theory: just search through all possible proofs and put sentences into the set we're building when we see them proved. This corresponds to the fact that the set of (Godel numbers of) theorems of a (computably axiomatizable) theory is c.e. So it's a useful intuitive device: if you can build it, it's $W_e$ for some $e$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:19











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          1 Answer
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          1












          $begingroup$

          Your intuition is right: there is no such function.



          We can see this in a couple different ways:




          • Contradiction: Suppose there were such an $f$. Then - given an arbitrary (infinite) c.e. set $A$ - we could compute the sequence $$min(A), min(Asetminus{min(A)}), min(Asetminus{min(A), min(Asetminus{min(A)})}), ...$$ by iteratively applying $f$. But this precisely enumerates $A$ in order, and so makes $A$ computable. Since there are non-computable sets, this can't happen.



          • Direct diagonalization: Given a computable function $f$, we can find an $e$ such that $(i)$ $2$ enters $W_e$ at stage $0$, $(ii)$ $1$ enters $W_e$ at stage $n+1$ iff $f(e)$ halts and outputs "$2$" at stage exactly $n$, and $(iii)$ no other numbers enter $W_e$ under any other circumstances. Then we get $min(W_e)=2$ iff $f(e)not=2$, so $f$ doesn't do what we want it to.




            • This fits into the "game picture" of c.e. sets: I pick a computable $f$, and you start building my $W_e$. You put $2$ into it, which forces me to eventually decide what $f(e)$ is; as soon as I do, if I say "$2$" then you put $1$ into my set, and if I say anything other than "$2$" you do nothing. Either way, you've defeated my purported $f$.




          I haven't provided full details of either case, but the idea should be clear. Filling in the details (of the first one, especially) is a good exercise.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I haven't seen much of the "game" arguments, so could you explain what you mean by we "build $W_e$"? Isn't $W_e$ already determined for each $e$? Also, what do you mean by we can "find" $e$ such that 2 enters $W_e$ at stage 0... and so on?
            $endgroup$
            – user500144
            Dec 8 '18 at 21:02










          • $begingroup$
            @user500144 The two key ingredients here are Church's thesis and, more subtly, the recursion theorem. It's the latter which lets us seemingly "define $e$ while knowing already $e$."
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:05










          • $begingroup$
            Specifically, the argument has two pieces. Suppose $f$ has the property proposed. We begin with a Church's thesis argument: there is a total computable $h$ such that, for each $e$, we have $W_{h(e)}={1,2}$ if $f(e)downarrow=2$ and $W_{h(e)}={2}$ otherwise.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:16










          • $begingroup$
            (cont'd) The interesting step is applying the Recursion Theorem. Since $h$ is total computable, it has a fixed point - some $c$ such that $W_c=W_{h(c)}$. But then we have $W_{c}={1,2}$ if $f(c)downarrow=2$ and $W_c={2}$ otherwise - contradicting the assumption on $f$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:17










          • $begingroup$
            (contd) The recursion theorem aside (it is genuinely a technical result), all of this stuff becomes more intuitive once you think of c.e. sets as sets you can "enumerate effectively." E.g. intuitively we can effectively enumerate the theorems of a given (computably axiomatizable) theory: just search through all possible proofs and put sentences into the set we're building when we see them proved. This corresponds to the fact that the set of (Godel numbers of) theorems of a (computably axiomatizable) theory is c.e. So it's a useful intuitive device: if you can build it, it's $W_e$ for some $e$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:19
















          1












          $begingroup$

          Your intuition is right: there is no such function.



          We can see this in a couple different ways:




          • Contradiction: Suppose there were such an $f$. Then - given an arbitrary (infinite) c.e. set $A$ - we could compute the sequence $$min(A), min(Asetminus{min(A)}), min(Asetminus{min(A), min(Asetminus{min(A)})}), ...$$ by iteratively applying $f$. But this precisely enumerates $A$ in order, and so makes $A$ computable. Since there are non-computable sets, this can't happen.



          • Direct diagonalization: Given a computable function $f$, we can find an $e$ such that $(i)$ $2$ enters $W_e$ at stage $0$, $(ii)$ $1$ enters $W_e$ at stage $n+1$ iff $f(e)$ halts and outputs "$2$" at stage exactly $n$, and $(iii)$ no other numbers enter $W_e$ under any other circumstances. Then we get $min(W_e)=2$ iff $f(e)not=2$, so $f$ doesn't do what we want it to.




            • This fits into the "game picture" of c.e. sets: I pick a computable $f$, and you start building my $W_e$. You put $2$ into it, which forces me to eventually decide what $f(e)$ is; as soon as I do, if I say "$2$" then you put $1$ into my set, and if I say anything other than "$2$" you do nothing. Either way, you've defeated my purported $f$.




          I haven't provided full details of either case, but the idea should be clear. Filling in the details (of the first one, especially) is a good exercise.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I haven't seen much of the "game" arguments, so could you explain what you mean by we "build $W_e$"? Isn't $W_e$ already determined for each $e$? Also, what do you mean by we can "find" $e$ such that 2 enters $W_e$ at stage 0... and so on?
            $endgroup$
            – user500144
            Dec 8 '18 at 21:02










          • $begingroup$
            @user500144 The two key ingredients here are Church's thesis and, more subtly, the recursion theorem. It's the latter which lets us seemingly "define $e$ while knowing already $e$."
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:05










          • $begingroup$
            Specifically, the argument has two pieces. Suppose $f$ has the property proposed. We begin with a Church's thesis argument: there is a total computable $h$ such that, for each $e$, we have $W_{h(e)}={1,2}$ if $f(e)downarrow=2$ and $W_{h(e)}={2}$ otherwise.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:16










          • $begingroup$
            (cont'd) The interesting step is applying the Recursion Theorem. Since $h$ is total computable, it has a fixed point - some $c$ such that $W_c=W_{h(c)}$. But then we have $W_{c}={1,2}$ if $f(c)downarrow=2$ and $W_c={2}$ otherwise - contradicting the assumption on $f$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:17










          • $begingroup$
            (contd) The recursion theorem aside (it is genuinely a technical result), all of this stuff becomes more intuitive once you think of c.e. sets as sets you can "enumerate effectively." E.g. intuitively we can effectively enumerate the theorems of a given (computably axiomatizable) theory: just search through all possible proofs and put sentences into the set we're building when we see them proved. This corresponds to the fact that the set of (Godel numbers of) theorems of a (computably axiomatizable) theory is c.e. So it's a useful intuitive device: if you can build it, it's $W_e$ for some $e$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:19














          1












          1








          1





          $begingroup$

          Your intuition is right: there is no such function.



          We can see this in a couple different ways:




          • Contradiction: Suppose there were such an $f$. Then - given an arbitrary (infinite) c.e. set $A$ - we could compute the sequence $$min(A), min(Asetminus{min(A)}), min(Asetminus{min(A), min(Asetminus{min(A)})}), ...$$ by iteratively applying $f$. But this precisely enumerates $A$ in order, and so makes $A$ computable. Since there are non-computable sets, this can't happen.



          • Direct diagonalization: Given a computable function $f$, we can find an $e$ such that $(i)$ $2$ enters $W_e$ at stage $0$, $(ii)$ $1$ enters $W_e$ at stage $n+1$ iff $f(e)$ halts and outputs "$2$" at stage exactly $n$, and $(iii)$ no other numbers enter $W_e$ under any other circumstances. Then we get $min(W_e)=2$ iff $f(e)not=2$, so $f$ doesn't do what we want it to.




            • This fits into the "game picture" of c.e. sets: I pick a computable $f$, and you start building my $W_e$. You put $2$ into it, which forces me to eventually decide what $f(e)$ is; as soon as I do, if I say "$2$" then you put $1$ into my set, and if I say anything other than "$2$" you do nothing. Either way, you've defeated my purported $f$.




          I haven't provided full details of either case, but the idea should be clear. Filling in the details (of the first one, especially) is a good exercise.






          share|cite|improve this answer









          $endgroup$



          Your intuition is right: there is no such function.



          We can see this in a couple different ways:




          • Contradiction: Suppose there were such an $f$. Then - given an arbitrary (infinite) c.e. set $A$ - we could compute the sequence $$min(A), min(Asetminus{min(A)}), min(Asetminus{min(A), min(Asetminus{min(A)})}), ...$$ by iteratively applying $f$. But this precisely enumerates $A$ in order, and so makes $A$ computable. Since there are non-computable sets, this can't happen.



          • Direct diagonalization: Given a computable function $f$, we can find an $e$ such that $(i)$ $2$ enters $W_e$ at stage $0$, $(ii)$ $1$ enters $W_e$ at stage $n+1$ iff $f(e)$ halts and outputs "$2$" at stage exactly $n$, and $(iii)$ no other numbers enter $W_e$ under any other circumstances. Then we get $min(W_e)=2$ iff $f(e)not=2$, so $f$ doesn't do what we want it to.




            • This fits into the "game picture" of c.e. sets: I pick a computable $f$, and you start building my $W_e$. You put $2$ into it, which forces me to eventually decide what $f(e)$ is; as soon as I do, if I say "$2$" then you put $1$ into my set, and if I say anything other than "$2$" you do nothing. Either way, you've defeated my purported $f$.




          I haven't provided full details of either case, but the idea should be clear. Filling in the details (of the first one, especially) is a good exercise.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 20:36









          Noah SchweberNoah Schweber

          127k10151290




          127k10151290












          • $begingroup$
            I haven't seen much of the "game" arguments, so could you explain what you mean by we "build $W_e$"? Isn't $W_e$ already determined for each $e$? Also, what do you mean by we can "find" $e$ such that 2 enters $W_e$ at stage 0... and so on?
            $endgroup$
            – user500144
            Dec 8 '18 at 21:02










          • $begingroup$
            @user500144 The two key ingredients here are Church's thesis and, more subtly, the recursion theorem. It's the latter which lets us seemingly "define $e$ while knowing already $e$."
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:05










          • $begingroup$
            Specifically, the argument has two pieces. Suppose $f$ has the property proposed. We begin with a Church's thesis argument: there is a total computable $h$ such that, for each $e$, we have $W_{h(e)}={1,2}$ if $f(e)downarrow=2$ and $W_{h(e)}={2}$ otherwise.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:16










          • $begingroup$
            (cont'd) The interesting step is applying the Recursion Theorem. Since $h$ is total computable, it has a fixed point - some $c$ such that $W_c=W_{h(c)}$. But then we have $W_{c}={1,2}$ if $f(c)downarrow=2$ and $W_c={2}$ otherwise - contradicting the assumption on $f$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:17










          • $begingroup$
            (contd) The recursion theorem aside (it is genuinely a technical result), all of this stuff becomes more intuitive once you think of c.e. sets as sets you can "enumerate effectively." E.g. intuitively we can effectively enumerate the theorems of a given (computably axiomatizable) theory: just search through all possible proofs and put sentences into the set we're building when we see them proved. This corresponds to the fact that the set of (Godel numbers of) theorems of a (computably axiomatizable) theory is c.e. So it's a useful intuitive device: if you can build it, it's $W_e$ for some $e$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:19


















          • $begingroup$
            I haven't seen much of the "game" arguments, so could you explain what you mean by we "build $W_e$"? Isn't $W_e$ already determined for each $e$? Also, what do you mean by we can "find" $e$ such that 2 enters $W_e$ at stage 0... and so on?
            $endgroup$
            – user500144
            Dec 8 '18 at 21:02










          • $begingroup$
            @user500144 The two key ingredients here are Church's thesis and, more subtly, the recursion theorem. It's the latter which lets us seemingly "define $e$ while knowing already $e$."
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:05










          • $begingroup$
            Specifically, the argument has two pieces. Suppose $f$ has the property proposed. We begin with a Church's thesis argument: there is a total computable $h$ such that, for each $e$, we have $W_{h(e)}={1,2}$ if $f(e)downarrow=2$ and $W_{h(e)}={2}$ otherwise.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:16










          • $begingroup$
            (cont'd) The interesting step is applying the Recursion Theorem. Since $h$ is total computable, it has a fixed point - some $c$ such that $W_c=W_{h(c)}$. But then we have $W_{c}={1,2}$ if $f(c)downarrow=2$ and $W_c={2}$ otherwise - contradicting the assumption on $f$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:17










          • $begingroup$
            (contd) The recursion theorem aside (it is genuinely a technical result), all of this stuff becomes more intuitive once you think of c.e. sets as sets you can "enumerate effectively." E.g. intuitively we can effectively enumerate the theorems of a given (computably axiomatizable) theory: just search through all possible proofs and put sentences into the set we're building when we see them proved. This corresponds to the fact that the set of (Godel numbers of) theorems of a (computably axiomatizable) theory is c.e. So it's a useful intuitive device: if you can build it, it's $W_e$ for some $e$.
            $endgroup$
            – Noah Schweber
            Dec 8 '18 at 21:19
















          $begingroup$
          I haven't seen much of the "game" arguments, so could you explain what you mean by we "build $W_e$"? Isn't $W_e$ already determined for each $e$? Also, what do you mean by we can "find" $e$ such that 2 enters $W_e$ at stage 0... and so on?
          $endgroup$
          – user500144
          Dec 8 '18 at 21:02




          $begingroup$
          I haven't seen much of the "game" arguments, so could you explain what you mean by we "build $W_e$"? Isn't $W_e$ already determined for each $e$? Also, what do you mean by we can "find" $e$ such that 2 enters $W_e$ at stage 0... and so on?
          $endgroup$
          – user500144
          Dec 8 '18 at 21:02












          $begingroup$
          @user500144 The two key ingredients here are Church's thesis and, more subtly, the recursion theorem. It's the latter which lets us seemingly "define $e$ while knowing already $e$."
          $endgroup$
          – Noah Schweber
          Dec 8 '18 at 21:05




          $begingroup$
          @user500144 The two key ingredients here are Church's thesis and, more subtly, the recursion theorem. It's the latter which lets us seemingly "define $e$ while knowing already $e$."
          $endgroup$
          – Noah Schweber
          Dec 8 '18 at 21:05












          $begingroup$
          Specifically, the argument has two pieces. Suppose $f$ has the property proposed. We begin with a Church's thesis argument: there is a total computable $h$ such that, for each $e$, we have $W_{h(e)}={1,2}$ if $f(e)downarrow=2$ and $W_{h(e)}={2}$ otherwise.
          $endgroup$
          – Noah Schweber
          Dec 8 '18 at 21:16




          $begingroup$
          Specifically, the argument has two pieces. Suppose $f$ has the property proposed. We begin with a Church's thesis argument: there is a total computable $h$ such that, for each $e$, we have $W_{h(e)}={1,2}$ if $f(e)downarrow=2$ and $W_{h(e)}={2}$ otherwise.
          $endgroup$
          – Noah Schweber
          Dec 8 '18 at 21:16












          $begingroup$
          (cont'd) The interesting step is applying the Recursion Theorem. Since $h$ is total computable, it has a fixed point - some $c$ such that $W_c=W_{h(c)}$. But then we have $W_{c}={1,2}$ if $f(c)downarrow=2$ and $W_c={2}$ otherwise - contradicting the assumption on $f$.
          $endgroup$
          – Noah Schweber
          Dec 8 '18 at 21:17




          $begingroup$
          (cont'd) The interesting step is applying the Recursion Theorem. Since $h$ is total computable, it has a fixed point - some $c$ such that $W_c=W_{h(c)}$. But then we have $W_{c}={1,2}$ if $f(c)downarrow=2$ and $W_c={2}$ otherwise - contradicting the assumption on $f$.
          $endgroup$
          – Noah Schweber
          Dec 8 '18 at 21:17












          $begingroup$
          (contd) The recursion theorem aside (it is genuinely a technical result), all of this stuff becomes more intuitive once you think of c.e. sets as sets you can "enumerate effectively." E.g. intuitively we can effectively enumerate the theorems of a given (computably axiomatizable) theory: just search through all possible proofs and put sentences into the set we're building when we see them proved. This corresponds to the fact that the set of (Godel numbers of) theorems of a (computably axiomatizable) theory is c.e. So it's a useful intuitive device: if you can build it, it's $W_e$ for some $e$.
          $endgroup$
          – Noah Schweber
          Dec 8 '18 at 21:19




          $begingroup$
          (contd) The recursion theorem aside (it is genuinely a technical result), all of this stuff becomes more intuitive once you think of c.e. sets as sets you can "enumerate effectively." E.g. intuitively we can effectively enumerate the theorems of a given (computably axiomatizable) theory: just search through all possible proofs and put sentences into the set we're building when we see them proved. This corresponds to the fact that the set of (Godel numbers of) theorems of a (computably axiomatizable) theory is c.e. So it's a useful intuitive device: if you can build it, it's $W_e$ for some $e$.
          $endgroup$
          – Noah Schweber
          Dec 8 '18 at 21:19


















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