Proof of: There exists a function which dominates all computable ones

Asked Aug 27 '21 at 22:19

Active Aug 28 '21 at 01:22

Viewed 106 times

Let $C \subset \mathbb{N} \times \mathbb{N} $ be the set of all (turing-)computable functions.

Then:

$$\exists f \in \mathbb{N}^\mathbb{N}: C \subset o(f)$$

I am looking for a proof and the name of a potential statement implying this statement. Personally, I feel this is very counter-intuitive...

This statement must be true. If you say that a problem $X \in P$, you usually say that $$ \exists M \in \mathcal P. \forall x \in X: \text{M accepts x with step counting function f} \in \mathcal{O(px^n)}$$

Now you can proof that such a problem is decidable by defining a Turing machines which halts after it has done at max $f(n)$ steps.

You can now show that this holds true for any function with a computable step counting function.(the complexity class might change, of course)

This is however does not hold true for an uncomputable step counting function $g$. You could try to contradict, saying that for g there is a bigger $f$, which is computable, such that $g \in \mathcal{O(f(n))}$. But this is known to be false.

edited Aug 27 '21 at 23:17

asked Aug 27 '21 at 22:19

TVSuchty

You have good reason to think it's true? – David C. Ullrich Aug 27 '21 at 22:33
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Wait. I bet it's true for any countable set $\mathcal C$ by a little diagonalization – David C. Ullrich Aug 27 '21 at 22:35
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$f(n)=n\max_{1\le j\le n}(c_j(n))$. – David C. Ullrich Aug 27 '21 at 22:36
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When you write $\mathbb{N}\times \mathbb{N}$, did you mean $\mathbb{N}^\mathbb{N}$? – Jason DeVito - on hiatus Aug 27 '21 at 22:39
It must be true, otherwise you could not show that a problem in NP by some definitions is necessarily decidable. (If you want I can add that in a more explained form)
@JasonDeVito Yes.
– TVSuchty Aug 27 '21 at 22:41
@DavidC.Ullrich This should work, I am 99% convinced. – TVSuchty Aug 27 '21 at 22:45
@DavidC.Ullrich Yes. Your argument is very smart. It holds true. thank you 100% convinced – TVSuchty Aug 27 '21 at 22:47
The argument is basically that for every computable function enumerated by n, the function will get dominated by f at n and higher – TVSuchty Aug 27 '21 at 22:50
Adding the bit about NP would be a very good idea, what they call "context" in these parts... Also you really should fix the $\Bbb N\times\Bbb N$ problem. – David C. Ullrich Aug 27 '21 at 23:02
Thank you. I tried improving the question accordingly. – TVSuchty Aug 27 '21 at 23:19

Proof of: There exists a function which dominates all computable ones

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