Index Sets are Cylinders

116 Views Asked by At

A cylinder is a set of the form $C\times \mathbb{N}:=\{\langle c,n \rangle \ | \ c\in C\}$. I want to prove that

Every index set $C$ is computably isomorphic to a cylinder.

My attemp: Since $C\leq_1 C\times\mathbb{N}$ (1-1 reducible) by Myhill isomorphism theorem it is suffice to show that $C\times\mathbb{N}\leq_1 C$. Indeed I need to use the index set definition but here is where I'm stuck. I'd appreciate any help/hint.

1

There are 1 best solutions below

0
On BEST ANSWER

You need some way to "expand" $C$ so that you can turn a map to $C$ into an injective map to $C$. A good starting point is the following:

Let $[e]$ be the smallest index set containing $e$ (= the set of all indices $c$ such that $W_c=W_e$). There is a computable injection $i:\mathbb{N}\rightarrow [e]$.

This is basically just the padding lemma: we can build an infinite computable sequence $e_0<e_1<e_2<...$ of elements of $[e]$, basically by adding "junk code" to the Turing machine corresponding to $e$ itself. Now since $[e]$ is not computable, of course the set $\{e_i: i\in\mathbb{N}\}$ will be a proper subset of $[e]$, but that's fine.

To build a $1$-reduction of $C\times\mathbb{N}$ to $C$, we'll want to perform a similar "redundancy" construction. Basically, send $(a,b)$ to the $b$th "silly modification" of $a$. Of course we need to define this in such a way that "silly modifications don't overlap," and the usual statement of the padding lemma doesn't do this for us; instead, you will probably need to look at the proof of the padding lemma to prove this stronger version.