-1

Consider the following set of arithmetic functions,

$a_0,a_1,a_2.....a_m(n)=a_0+a_1n+a_2n^2+...a_mn^m$

$a_0,a_1,a_2.....a_n$ can be any subset of naturals. So clearly, the number of these arithmetic functions is uncountable and can't be mapped to the naturals. What am I missing?

EDIT-$f(n)=\sum_{k=0}^{\infty} \frac{n^k}{a_kk!}$. What if we use this function? Now we can allow infinite subsets.

Mauro ALLEGRANZA
  • 94,169
Ryder Rude
  • 1,417
  • Your computer just did it to render that text to your screen. See, for instance, ASCII. – Eric Towers Apr 03 '20 at 07:22
  • "Now we can allow infinite subsets." no, that series need not converge. Annd in cases where it does converge it doesn't define an "arithmetic function"... – David C. Ullrich Apr 03 '20 at 11:47
  • @DavidC.Ullrich It always converges as it is upper bounded by $e^n$. And why is it not an arithmetic function? If we want the output to be an integer, we can apply Greatest Integer Function on the RHS. – Ryder Rude Apr 03 '20 at 14:25

2 Answers2

6

But the set of finite sequences of numbers $a_0, a_1, a_2, \ldots, a_n$ is not uncountable! So the set of polynomials of the form $f(n) = a_0+a_1n+a_2n^2+...a_mn^m$ isn't uncountable either.

Take the mapping that sends the finite sequence $a_0, a_1, a_2, \ldots, a_n$ to $\pi_0^{a_0}\pi_1^{a_1}\pi_2^{a_2}\cdot\cdot\cdot\pi_n^{a_n}$ where $\pi_n$ is the $n+1$th prime. That's an injective map into the naturals, establishing countability! (And of course this is an example of the sort of mapping we use in Gödel codings ...)

Peter Smith
  • 54,743
  • By that argument, we can map any element of $P(N)$ to $N$. Then why do we say that $P(N)$ is uncountable? – Ryder Rude Apr 03 '20 at 08:17
  • 1
    No we can't. The powerset of $\mathbb{N}$ contains infinite subsets of the naturals. You can't count all the subsets of $\mathbb{N}$, finite and infinite. But you can count (as we've shown) the finite subsets. And it is the finite ones that are relevant in your original question. – Peter Smith Apr 03 '20 at 08:20
  • 1
    @RyderRude No, because a general subset of $\mathbb{N}$ may contain infinitely many elements. For example, try it with the even numbers or the whole of $\mathbb{N}$. There is a big difference between the set of finite subsets and all subsets . – badjohn Apr 03 '20 at 08:21
  • $f(n)=\sum_{k=0}^{\infty} \frac{n^k}{a_kk!}$. What if we use this function? Now we can allow infinite subsets. – Ryder Rude Apr 03 '20 at 08:33
  • In proving the first incompleteness theorem we are only interested in the representability of recursive functions (not the expressibility of everything that is in a broad sense to do with natural numbers). So ask yourself: is that function recursive? If it is, it is representable. If it isn't it is irrelevant to the first theorem. – Peter Smith Apr 03 '20 at 08:37
  • @PeterSmith What does recursive function mean in this context? – Ryder Rude Apr 03 '20 at 10:24
  • 1
    If you need to ask that, you really need to do some homework first. Try e.g. the notes you can download from https://www.logicmatters.net/igt/godel-without-tears/ Or read a good book -- there's a reading guide in §5.2 of https://www.logicmatters.net/wp-content/uploads/2019/12/TeachYourselfLogic2020.pdf – Peter Smith Apr 03 '20 at 10:56
2

Addressing your new comment:

Yes, we can assign to each $A\subseteq\mathbb{N}$ a function $f_A:\mathbb{N}\rightarrow\mathbb{N}$ in an injective way - that is, such that $f_A=f_B$ iff $A=B$. (I'm ignoring the question of whether your specific construction works, that's beside the point.) More snappily, $\vert 2^\mathbb{N}\vert=\vert\mathbb{N}^\mathbb{N}\vert$.

However, this has nothing to do with Godel's incompleteness theorem! In the incompleteness theorem we're not saying that we have a way to represent every $g:\mathbb{N}\rightarrow\mathbb{N}$ by a natural number; all we're saying is that we have a way to represent every "simple" $g:\mathbb{N}\rightarrow\mathbb{N}$ by a natural number. And the set of "simple" functions is indeed countable, so there's no contradiction.

Specifically, "simple" for Godel means "computable" (or "recursive"). That said, it turns out that's overkill: for example, it's enough to just look at the primitive recursive functions (and indeed this is what Godel originally did).

Now a bit of an editorial comment:

Before tackling the incompleteness theorem you really need to be familiar with the basics of computable functions. Fully understanding the following facts is a good starting point:

  • The set of computable functions is countable.

  • Every computable function is definable in the structure $\mathfrak{N}=(\mathbb{N};+,\times)$, but not conversely.

    • The former is basically Kleene's $T$ predicate; for the latter, note that the set of $\mathfrak{N}$-definable functions is also countable so a cardinality argument won't help.

  • However, a broad class of functions are computable - for example, the characteristic function of the set of powers of $10$, or of the set of Godel numbers of well-formed formulas.

The proof of Godel's theorem then really kicks off properly with the result that all computable functions are representable in the theory in question (usually PA or similar).

Noah Schweber
  • 245,398
  • I think the proof of a statement involving computable functions might involve the use of uncomputable functions and their properties as the intermediate steps. We might be ruling out such proofs by not being able to encode non-computable functions. – Ryder Rude Apr 04 '20 at 02:32
  • @RyderRude I'm not really sure what you're getting at with that, but for Godel at least there's no need to talk about non-computable functions at all. It really is the case that everything is happening in the realm of computability, there. – Noah Schweber Apr 04 '20 at 03:01
  • For example, finding the real roots of a real cubic equation involves going through the complex numbers as the middle step. That's what I'm talking about. But if it's known that uncomputable functions have no use in the proofs of statements involving computable functions, then it's fine. – Ryder Rude Apr 04 '20 at 03:10
  • @RyderRude Non-computable functions sometimes are, and sometimes are not, relevant to given proofs about computable functions. But in the specific case of Godel's incompleteness theorem, they're not things we need to worry about. And the fact that merely numbering the computable functions is sufficient will become clear by reading the proof. – Noah Schweber Apr 04 '20 at 03:14