Let A and B be two subsets of ℕ. We say that A is reducible to B and write A ≤B if there exists

Let A and B be two subsets of ℕ. We say that A is reducible...

Let $A$ and $B$ be two subsets of $\mathbb{N}$. We say that $A$ is reducible to $B$ and write $A \leq B$ if there exists a (total) recursive function $f$ such that $x \in A \quad$ if and only if $f(x) \in B$. (a) Show that the relation $\leq$ is reflexive and transitive. (b) Assume that $A$ is reducible to $B$. Show that if $B$ is recursively enumerable, then $A$ is recursively enumerable; show also that if $B$ is recursive, then so is $A$. Set $$ \begin{aligned} X &=\left\{x: \phi^{1}(x, x) \text { is defined }\right\} \\ Y &=\left\{\alpha_{2}(x, y): \phi^{1}(x, y) \text { is defined }\right\} \end{aligned} $$ (c) Show that a set $A \subseteq \mathbb{N}$ is recursively enumerable if and only if $A \leq Y$. (d) Let $A$ and $B$ be two subsets of $\mathbb{N}$. Let $$ C=\{2 n: n \in A\} \cup\{2 n+1: n \in B\} . $$ Show that $A$ and $B$ are reducible to $C$ and that if $D$ is a subset of $\mathbb{N}$ such that $A$ and $B$ are both reducible to $D$, then $C$ is reducible to $D$. (e) We will say that $A$ is self-dual if $A \leq \mathbb{N}-A$. Show that for every $B \subseteq \mathbb{N}$, there exists a $C \subseteq \mathbb{N}$ that is self-dual and is such that $B \leq C$. (f) Let $\mathcal{T}$ be a set of partial recursive functions of one variable that is not empty and is not equal to the set of all partial recursive functions of one variable. Set $$ A=\left\{x: \phi_{x}^{1} \in \mathcal{T}\right\} . $$ (i) Show that if the partial.function whose domain is empty belongs to $\mathcal{T}$, then $X \leq \mathbb{N}-A$. (ii) Show that, in the opposite case, $X \leq A$. (iii) Show that $A$ is not self-dual. (g) Show that $Y \leq X$.

Let $A$ and $B$ be two subsets of $\mathbb{N}$. We say that $A$ is reducible to $B$ and write $A \leq B$ if there exists a (total) recursive function $f$ such that
$x \in A \quad$ if and only if $f(x) \in B$.
(a) Show that the relation $\leq$ is reflexive and transitive.
(b) Assume that $A$ is reducible to $B$. Show that if $B$ is recursively enumerable, then $A$ is recursively enumerable; show also that if $B$ is recursive, then so is $A$.
Set
$$
\begin{aligned}
X &=\left\{x: \phi^{1}(x, x) \text { is defined }\right\} \\
Y &=\left\{\alpha_{2}(x, y): \phi^{1}(x, y) \text { is defined }\right\}
\end{aligned}
$$
(c) Show that a set $A \subseteq \mathbb{N}$ is recursively enumerable if and only if $A \leq Y$.
(d) Let $A$ and $B$ be two subsets of $\mathbb{N}$. Let
$$
C=\{2 n: n \in A\} \cup\{2 n+1: n \in B\} .
$$
Show that $A$ and $B$ are reducible to $C$ and that if $D$ is a subset of $\mathbb{N}$ such that $A$ and $B$ are both reducible to $D$, then $C$ is reducible to $D$.
(e) We will say that $A$ is self-dual if $A \leq \mathbb{N}-A$. Show that for every $B \subseteq \mathbb{N}$, there exists a $C \subseteq \mathbb{N}$ that is self-dual and is such that $B \leq C$.
(f) Let $\mathcal{T}$ be a set of partial recursive functions of one variable that is not empty and is not equal to the set of all partial recursive functions of one variable. Set
$$
A=\left\{x: \phi_{x}^{1} \in \mathcal{T}\right\} .
$$
(i) Show that if the partial.function whose domain is empty belongs to $\mathcal{T}$, then $X \leq \mathbb{N}-A$.
(ii) Show that, in the opposite case, $X \leq A$.
(iii) Show that $A$ is not self-dual.
(g) Show that $Y \leq X$.

Key Concepts

Many-One Reducibility

This concept refers to a notion of reducibility between subsets of the natural numbers, where one set A is said to be reducible to another set B if there exists a total recursive function f such that membership in A can be decided by membership in B via f. It is a central idea in computability theory used to compare the computational complexity or decidability of sets.

Total Recursive Functions

Total recursive functions are functions from the natural numbers to the natural numbers that are defined for every input and are computable by some algorithm. They serve as the 'effective transformations' required in many-one reducibility and are foundational in defining reductions, recursive (decidable) sets, and other computability concepts.

Recursively Enumerable Sets

Recursively enumerable (r.e.) sets are those for which there exists a Turing machine that enumerates their elements, possibly without halting for non-members. They are crucial in computability theory as many problems, though not decidable, can be recognized by some algorithm; properties like reducibility often preserve the recursively enumerable aspect.

Recursive (Decidable) Sets

A recursive set is one whose membership can be decided by an algorithm in a finite amount of time. These sets are a strict subset of recursively enumerable sets and are central in understanding the boundaries of algorithmic decidability in computability theory, particularly in contexts where reductions preserve decidability.

Self-Duality in Reducibility

Self-duality refers to the property of a set being reducible to its complement. This concept is important in the study of the structure of degrees of unsolvability and helps in understanding how a set’s complexity relates to that of its complement, offering insights into symmetry properties within the hierarchy of undecidable or non-recursive sets.

Partial Recursive Functions

Partial recursive functions are functions that are computed by an algorithm that may not provide an output for every input, meaning they can be undefined for some arguments. They extend the notion of total recursive functions and are key in the study of computability, particularly when considering domains of function definitions and constructing sets based on these domains.

Transcript

00:01 We are given a subset of the set of order pairs of integers and a recursive definition.

00:09 S is the subset of the order pairs of integers defined by 0 is an s, and then if order pair ab is an s, an a plus 1, b plus 1 is an s, a plus 1, b plus 1 is an s, and a plus 2, b plus 1 is also an s.

00:30 In part a, we're asked to list the elements of s produced by the first four applications, of the recursive definition.

00:43 So to do this, for our first application, we're applying the steps on 0 -0.

00:55 So we get that the recursion formulas tell us that 0 -1 -1 and 2 -1 are all in s.

01:26 For the second application, we're going to apply to each of these three.

01:35 So we have applying to 0 .1, we get some repeats here.

01:43 We'll get 1 -1 again, or i guess i should say 0 -2, 1 -2, and 2 -2, r &s.

02:02 We also have that if we apply to 1 -1, we get 1 -2, which we already have, 2 -2, which we already have an s, and 3 -2.

02:21 This is new.

02:29 And finally we'll apply to 2 -1.

02:43 So applying to 2 -1, we get 2 -2, which is already in s, 3 -2, which is already in s, and 4 -2, which is a new element in s.

02:55 So we get all these new elements from the second iteration.

03:00 And in the third application we're now applying the recursive steps on 0 -2 -12, 2 -22, 3 -2, and 4 -2.

03:48 So applying to 0 -2, we get 1 -2, or sorry, 0 -3.

03:54 This is new.

03:56 We also have 1 -3, which is also new, and we have 2 -3, which is new, which is new, applying to 1 -2, we get 0, or 1 -3, which we already have, 2 -3, which we already have, and 3 -3, which is new.

04:32 Applying to 2 -2, we get 2 -2, we already have, 3 -3, which we already have, and 4 -3, which we already have, and 4 -3, which we already have, and 4 -3, which we already have, and 5 -3, which is new.

04:57 And applying to 4 -2 we get 4 -3 which we already have 5 -3 which we already have and then 6 -3 which is new.

05:12 And now for the fourth and final application we're going to apply to 0 -3 -13 -23 -23 -23 -4 -3 -3 and 6 -3.

06:01 So applying to 0 -3 we get 0 -4 which is new.

06:08 Which is new and two four which is new applying to one three you get one four which is old two four which is also old and three four which is new applying to two three we get two four which we already have three four which we already have and then four three which is new i'm sorry there should be four 4, which is new.

07:02 Applying to 3 -3, we get 3 -4, which is old, 4 -4 which is old, and 5 -4, which is new.

07:17 Applying to 4 -3, we get 4 -4 which is old, and 6 -4, which is old, and 6 -4, which is new.

07:29 Applying to 5 -3, we get 5 -4 which is old, 6 -4 which is old, and 7 -4, which is new...

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