Zheng and Weihrauch (<http://www-sst.informatik.tu-cottbus.de/~wwwti/zheng/publications/1999/mfcs99.pdf>) define a real number $x$ to be $\Sigma\_n$ if and only if there is a computable function $f:\mathbb{N}^n\rightarrow\mathbb{Q}$ such that $x=\sup\_{i\_1}\inf\_{i\_2}...f(i\_1,...,i\_n)$. Another possible definition is that $x$ is $\Sigma\_n$ if and only if its Dedekind cut $\{q\in\mathbb{Q}|q<x\}$ is defined by a $\Sigma\_n$ formula (using an encoding of rational numbers with natural numbers).
Zheng and Weihrauch's definition is at least as strong as mine because $q<\sup\_{i\_1}\inf\_{i\_2}...f(i\_1,...,i\_n)\iff\exists i\_1 \forall i\_2 ... q<f(i\_1,...,i\_n)$. Are the two definitions equivalent, or is there a real number that is $\Sigma\_n$ by my definition but not by Zheng and Weihrauch's for some $n$?



Are these two definitions of arithmetical hierarchy of real numbers equivalent?
They are the same as the following induction proof shows.


*Base case:* For $\Sigma\_1$, if $x = \sup\_i f(i)$ for $f$ computable, then $q < x$ is equivalent to the $\Sigma\_1$ statement $\exists i [q<f(i)]$. Conversely, if $\{q\mid q < x\}$ is computably enumerable, then $x = \sup\_i f(i)$ where $f(i)$ is a computable enumeration of all $q < x$.


For $\Pi\_1$, if $x = \inf\_i f(i)$ for $f$ computable, then $q \leq x$ is equivalent to $\forall i [q \leq f(i)]$. Conversely, if $\{q\mid q \leq x\}$ is co-c.e., then $x = \inf\_i f(i)$ where $f(i)$ is a computable enumeration of ${q\mid q > x}$ (which is c.e., since it is the complement of $\{q\mid q \leq x\}$).


*Induction step:* For $\Sigma\_{n+1}$, assume $x = \sup\_i f(i)$ where $f(i)$ is $\Pi\_n[i]$. By the induction hypothesis, $\{q\mid q\leq f(i)\}$ is $\Pi^n[i]$. Then $q \leq x$ is equivalent to some $\Sigma\_{n+1}$ formula expressing $\exists i\ [q \leq f(i)]$.


Conversely, if $C = \{q\mid q < x\}$ is $\Sigma\_{n+1}$, then $q < x$ is equivalent to $\exists i\ \phi(i,q)$ where $\phi(i,q)$ is $\Pi\_n$. Since $C$ is downward closed, we may assume $\phi(i,q)$ is downward closed in $q$. (Replace $\phi(i,q)$ with $q\leq q\_{i\_0} \land \phi(i\_1,q\_{1\_0})$ where $i$ encodes the pair $(i\_0,i\_1)$ and $q\_i$ is the $i$th rational.) Now, let 
$$f(i) := \sup\{q\mid \phi(i,q)\}.$$
Then $x = \sup\_i f(i)$, and by the induction hypothesis, $f(i)$ is $\Pi\_1[i]$.


The $\Pi\_{n+1}$ case is similar. QED


