Computability 7: Exercises

 

　　Disjoint sets $A$, $B$ are said to be recursively inseparable if there is no recursive set $C$ such that $A\subseteq C$ and $B\subseteq C$.

　　Now we prove that $K_0=\{x\text{ | }\phi_x(x)=0\}$ and $K_1=\{x\text{ | }\phi_x(x)=1\}$ are recursively inseparable. Suppose there is a recursive set $C$ with characteristic function $\phi_e$ such that $K_0\subseteq C$ and $K_1\subseteq\overline{C}$. Then both

　　　　$e\in C\Leftrightarrow\phi_e(e)=1\Leftrightarrow e\in K_1\Rightarrow e\in\overline{C}$,　and　$e\in\overline{C}\Leftrightarrow\phi_e(e)=0\Leftrightarrow e\in K_0\Rightarrow e\in C$

are contradictory. Therefore, there is no such recursive set $C$.

 

　　Disjoint sets $A$, $B$ are said to be effectively recursively inseparable if there is a total computable function $f$ such that whenever $A\subseteq W_a$, $B\subseteq W_b$ and $W_a\cap W_b=\varnothing$, then $f(a,b)\notin W_a\cup W_b$.

　　For any $i\neq j$, we prove $K_i=\{x\text{ | }\phi_x(x)=i\}$ and $K_j=\{x\text{ | }\phi_x(x)=j\}$ are effectively recursively inseparable as follow. Consider any $a$ and $b$ such that $K_i\subseteq W_a$, $K_j\subseteq W_b$, $W_a\cap W_b=\varnothing$. The following function $g$ should be computable. $$g(x,y,z)\simeq\begin{cases}j & z\in W_a \\ i & z\in W_b \\ \uparrow & \text{otherwise } \end{cases}$$ According to the s-m-n theorem, there exists a total computable function $f$ such that $\phi_{f(x,y)}(z)\simeq g(x,y,z)$, and hence

　　　　$f(a,b)\in W_a\Rightarrow \phi_{f(a,b)}(f(a,b))=j\Leftrightarrow f(a,b)\in K_j\subseteq W_b\subseteq\overline{W_a}$,

　　　　$f(a,b)\in W_b\Rightarrow \phi_{f(a,b)}(f(a,b))=i\Leftrightarrow f(a,b)\in K_i\subseteq W_a\subseteq\overline{W_b}$,

both of which lead to a contradiction. Therefore, $f(a,b)\notin W_a\cup W_b$, and $A$, $B$ are effectively recursively inseparable.

 

　　If two sets are effectively recursively inseparable, then they must be recursively inseparable.

　　If two r.e. sets are effectively recursively inseparable, then both of them are proved to be creative.

 

　　Suppose $A$, $B$ are effectively recursively inseparable sets, and $A_1$, $B_1$ are disjoint sets. If there is a total computable function $f$ such that $x\in A\Leftrightarrow f(x)\in A$, $x\in B\Leftrightarrow f(x)\in B$, then $A_1$ and $B_1$ are effectively recursively inseparable.

　　Given $A$, $B$ are effectively recursively inseparable, we denote total computable function $g'$ satisfying this restriction. Consider any $a$ and $b$ such that $A_1\subseteq W_a$, $B_1\subseteq W_b$, $W_a\cap W_b=\varnothing$. According to the s-m-n theorem, there is a total computable function $\varphi$ such that $\phi_{\varphi(e)}\simeq\psi_U(e,f(x))$. We denote $a'=\varphi(a)$, $b'=\varphi(b)$, and hence $x\in W_{a'}\Leftrightarrow f(x)\in W_a$, $x\in W_{b'}\Leftrightarrow f(x)\in W_b$. It follows that $A\subseteq W_{a'}$, $B\subseteq W_{b'}$, $W_{a'}\cap W_{b'}=\varnothing$, and hence $g'(a,b)\notin W_{a'}\cup W_{b'}$. Then there is a total computable function $g(x,y)\simeq g'(\varphi(x),\varphi(y))$ such that $g(a,b)\notin W_a\cup W_b$. Therefore, $A_1$, $B_1$ are effectively recursively inseparable.

　　This is an extension of the reduction theorem regarding to productive sets.

 

 

P.S.　　补充两个构造 productive function 的例子。

 

　　1. Given $B$ is recursive and $A\oplus B$ is creative, prove $A$ is creative.

　　(1) $A$ is r.e. since $x\in A\Leftrightarrow 2x\in A\oplus B$ is partially decidable.

　　(2) There is a total computable function $S$ such that:

　　　　　$\phi_{S(x)}(y)\simeq\begin{cases}1 & (y\text{ is even }\wedge y/2\in W_x)\vee(y\text{ is odd }\wedge (y-1)/2\notin B) \\ \uparrow & \text{ otherwise}\end{cases}$

　　and hence $W_{S(x)}=W_x\oplus\overline{B}=\overline{\overline{W}_x\oplus B}$.

　　　Suppose $g$ is the productive function of $\overline{A\oplus B}$, and we have

　　　　$W_x\subseteq\overline{A}\Rightarrow W_{S(x)}\subseteq\overline{A\oplus B}\Rightarrow g(S(x))\in\overline{A\oplus B}\cap\overline{W}_{S(x)}\Rightarrow g(S(x))/2\in\overline{A}\cap\overline{W}_x$

　　to prove $\overline{A}$ is productive.

 

　　2. Given $B$ is recursive and $A\otimes B$ is creative, prove $A$ is creative.

　　　Just by the same token. The only difference is that

　　　　　$\phi_{S(x)}(y)\simeq\begin{cases}1 &\pi_1(y)\in W_x\vee\pi_2(y)\notin B\\ \uparrow & \text{ otherwise}\end{cases}$

　　such that $W_{S(x)}=\overline{\overline{W}_x\otimes B}$, and finally $g(S(x))\in\overline{A\otimes B}\cap\overline{W}_{S(x)}\Rightarrow\pi_1(g(S(x)))\in\overline{A}\cap\overline{W}_x$.

 

References:

　　1. Cutland, Nigel. Computability: an introduction to recursive function theory[M]. Cambridge: Cambridge University Press, 1980