How to express the powers of 10 with a first order sentence?

Question

A subset $A \subseteq \mathbb N$ is called arithmetical if there exists a formula $\varphi$ in the language of first order arithmetic with exactly one free variable $x$ such that $$x\in A \iff \varphi(x) = \texttt{true}$$

Here we take first order arithmetic to mean the language of well-formed sentences using only finitely many of the symbols $\forall$, $\exists$, $($, $)$, $x$, $y$, $z$ and other letters for variables, and the logical $\texttt{and}$, $\texttt{or}$, $\texttt{not}$, and $\implies$ operators, as well as the natural arithmetic symbols $0$, $1$, $+$, $\cdot$, and $=$. Variables range over the Peano naturals $\mathbb N = \{0,1,2,\ldots\}$.

Examples of arithmetical sets include

• Any finite or cofinite set, eg the singleton $\{5\}$ is expressed by the formula $\varphi(x) = (x = 5)$, where "$5$" is an abbreviation for $1+1+1+1+1$.
• The range of any polynomial with positive integer coefficients, eg the squares are expressed by the formula $\varphi(x) = \exists y(x = y \cdot y)$
• the primes, expressed by $\varphi(x) = (x > 1) \texttt{ and } \forall y(y | x \implies y = 1 \texttt{ or } y = x)$ where "$x|y$" is an abbreviation for $\exists z(y=x\cdot z)$
• for any prime $p$, the set of all powers of $p$, expressed by $\varphi(x) = \forall y(y|x \implies y = 1 \texttt{ or } p | y)$

Question:

I am looking for a formula $\varphi$ that will witness the set of all powers of $10$, or any other composite integer.

It is easy to generalize the above example and capture the set $\{2^m 5^n : m,n \geq 0 \}$, but for the life of me I cannot get down to $\{2^n 5^n : n \geq 0\}$. I know one must exist because according to Wikipedia any computable set (a set for which there exists a finite computer program that will, on input $\texttt{n}$, output in finite time the $n$-th element of the set) is an arithmetical set, but I can't think of anything clever to characterize the powers of a composite integer.

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More properties of arithmetical sets: if $A$, $B\subset \mathbb N$ are arithmetical, then so are

• $A\cup B$,
• $A\cap B$,
• $B\setminus A$,
• $A + B$,
• and $A \cdot B$.

There are only countably many arithmetical sets, and there are therefore uncountably many non-arithmetical sets. For any non-arithmetical set $X \subseteq \mathbb N$, one may find monotone sequences $(A_n)_n$ and $(B_n)_n$ of arithmetical sets such that

$$\bigcup_n A_n = X = \bigcap_n B_n$$

For instance, if $X = \{x_0,x_1,x_2,\dots\}$ is an enumeration of $X$ and $Y = \{y_0,y_1,y_2,\ldots\}$ is an enumeration of $Y = X^c$ (because $X$ is not arithmetical then it is not finite or cofinite therefore both of these sets are infinite), then let $A_n = \{x_0,x_1,\ldots,x_n\}$ and let $B_n = \mathbb N \setminus\{y_0, y_1,\ldots, y_n\}$.

Answer 1

You can use Godel's $\beta$ function to simulate quantification over arbitrary finite number sequences. From there, it's not too difficult to prove the recursion theorem, albeit it must be a theorem schema since Peano Arithmetic cannot simulate quantification over arbitrary infinite sequences. The theorem schema ranges over all arithmetical formulae, where the formulae are defining the functions used/created in the recursion. As a joke, I once outlined this entire procedure in explicit detail on this question. The resulting formula is extremely complicated, as you'd expect.

To express powers of $10$, we can produce an arithmetical formula $\psi(x,y)$ which obeys $\psi(x,y) \iff y=10^x$, and then the formula $\exists x, \psi(x,y)$ expresses that $y$ is a power of ten. The formula $\psi$ can be constructed using the recursion theorem, using the base case $10^0=1$ and recursive relation $10^{n+1}=10^n\cdot 10$. Reusing my work from that other question I mentioned, we let $a=1$ be our initial value, and let $\phi(x,y)\iff 10\cdot x=y$ define the recursively applied function, then $\psi(x,y)$ is definable as the grotesque monstrosity below.

$$\begin{align} \exists S_1,\exists S_2, &(\exists k_1, y+k_1+1=xS_2+S_2+1) \\ \land &\exists q_1, q_1\cdot(xS_2+S_2+1)+y = S_1 \\ \land &\exists k_2, 1+k_2+1=S_2+1 \\ \land &\exists q_2, q_2\cdot(S_2+1)+1=S_1 \\ \land &\forall n, (\exists k_0, n+k_0+1=x) \implies \big[\exists r_3, \exists r_4, (10\cdot r_3=r_4) \\ &\land \exists k_3, r_3+k_3+1=nS_2+S_2+1 \\ &\land \exists q_3, q_3\cdot(nS_2+S_2+1)+r_3 = S_1 \\ &\land \exists k_4, r_4+k_4+1=(n+1)S_2+S_2+1 \\ &\land \exists q_4, q_4\cdot((n+1)S_2+S_2+1)+r_4=S_1\big] \end{align}$$

Admittedly, this can be simplified somewhat, moreso if you simply use definitorial expansion. I intentionally eliminated any instances of "$<$" and bound quantifiers, since those are technically not primitive concepts in the language of arithmetic.

Honestly, the fact that this is even possible is basically just a mathematical coincidence. Integer arithmetic just so happens to be expressive enough to interpret finite sequences, at which point it's trivial to implement recursion. Recursion is basically just a subdued form of self-reference, and that kind of self-reference is where all of $\sf{PA}$'s power actually comes from.