The bounded variable hierarchy collapses to a fixed finite level, because of the existence of definable pairing functions.

In more detail, I will use the Cantor pairing function
$$\def\p#1{\langle#1\rangle}\def\N{\mathbb N}\let\eq\leftrightarrow\p{x,y}=\frac{(x+y)(x+y+1)}2+x,$$
which is a bijection $\N^2\to\N$. Let $l(z)$, $r(z)$ denote the corresponding projections, so that
$$l(\p{x,y})=x,\qquad r(\p{x,y})=y,\qquad\p{l(z),r(z)}=z.$$
Since $\p{x,y}$ is a polynomial with rational coefficients, $\p{x,y}=z$ is equivalent to a quantifier-free (hence $3$-variable) formula, namely
$$\p{x,y}=z\eq z+z=(x+y)\cdot(x+y+1)+x,$$
and consequently the functions $l$ and $r$ are definable by $\Delta_0$ formulas using $3$ distinct variables:
$$l(z)=x\eq\exists y\le z\,\p{x,y}=z,$$
and similarly for $r$.

**Proposition:** Every $\Delta_0$ formula in one free variable is equivalent to a $\Delta_0$ formula using only $3$ distinct variables.

**Proof:**

It will be convenient to preprocess the formulas first. Let us call a bounded quantifier $\exists y\le t(\vec x)\,\theta(\vec x,y)$ *safe* if
$$\bigl(\exists y\le t(\vec x)\,\theta(\vec x,y)\bigr)\eq\bigl(\exists y\,\theta(\vec x,y)\bigr),$$
in which case we can replace the bound $t(\vec x)$ with any larger bound without affecting the truth of the formula. Similarly for universal quantifiers. By replacing $\exists y\le t(\vec x)\,\theta$ with $\exists y\le t(\vec x)\,(y\le t(\vec x)\land\theta)$, and unwinding terms with the help of extra existential quantifiers, it is easy to see that any $\Delta_0$ formula is equivalent to a $\Delta_0$ formula $\theta(\vec x)$ such that

all quantifiers in $\theta$ are safe,

the index $i$ of any variable $x_i$ quantified in any subformula $\xi$ of $\theta$ is higher than the indices of all free variables of $\xi$, and

all atomic subformulas of $\theta$ (except quantifier bounds) are of the forms $x_i=x_j$, $x_i+x_j=x_k$, or $x_i\cdot x_j=x_k$.

Let us call such formulas *special*.

**Claim:** For any special formula $\theta(x_0,\dots,x_n)$, there is a formula $\xi(z)$ using only $3$ distinct variables such that
$$\xi(z)\eq\theta\bigl(r(l^n(z)),\dots,r(l(z)),r(z)\bigr).$$

Note that the Claim implies the Proposition: if $\xi(z)\eq\theta(r(z))$ can be written using $3$ variables, then so can
$$\theta(x)\eq\exists z\le x^2\,\bigl(z=\p{0,x}\land\xi(z)\bigr).$$

We will prove the Claim by induction on the complexity of $\theta$.

The induction steps for Boolean connectives are trivial. The step for bounded quantifiers is also easy: if $\theta(x_0,\dots,x_n)$ is $\exists x_{n+1}\le t(\vec x)\,\theta'(x_0,\dots,x_{n+1})$, where the quantifier is safe, and $\xi'$ is a $3$-variable formula equivalent to $\theta'(r(l^{n+1}(z)),\dots,r(l(z)),r(z))$, then
$$\begin{align*}
\theta(r(l^n(z)),\dots,r(z))&\eq\exists x_{n+1}\,\xi'(\p{z,x_{n+1}})\\
&\eq\exists w\,(l(w)=z\land\xi'(w))\\
&\eq\exists w\le s(z)\,(l(w)=z\land\xi'(w))
\end{align*}$$
for a suitable term $s(z)$.

It remains to prove the Claim for special atomic formulas. First, notice that since $l$ and $r$ are definable using $3$ variables, the same holds for any finite composition of these functions: for example,
$$x=l^{n+1}(z)\eq\exists y\le z\,(x=l(y)\land y=l^n(z)),$$
where we proceed to expand the inner formulas. This immediately takes care of atomic formulas $x_i=x_j$, as
$$r(l^i(z))=r(l^j(z))\eq\exists x\le z\,(x=r(l^i(z))\land x=r(l^j(z))).$$
For the formulas $x_i+x_j=x_k$ (and similarly $x_i\cdot x_j=x_k$), we can proceed as follows:
$$r(l^i(z))+r(l^j(z))=r(l^k(z))\eq\exists w\le t(z)\,(l(l(w))=r(l^i(z))\land r(l(w))=r(l^j(z))\land r(w)=r(l^k(z))\land\exists x,z\le w\,(w=\p{\p{x,z},x+z})).$$
(Note that we recycled the $z$ variable.) Here, $l(l(w))=r(l^i(z))$ can be written as $\exists x\le w\,(x=l(l(w))\land x=r(l^i(z)))$, which we know how to handle, and similarly for the other two conjuncts. Finally,
$$w=\p{\p{x,z},x+z}$$
can be written as an identity using no further variables, as $\p{\p{x,z},x+z}$ is a polynomial with rational coefficients in $x$ and $z$ (and $x+z$). QED

**Corollary:** Every $\Delta_0$ formula in $n$ free variables is equivalent to a $\Delta_0$ formula using $\max\{3,n+1\}$ distinct variables.

Comments:

We only used elementary properties of the pairing function, hence the equivalence is not just true in $\N$, it is provable in a very weak theory: IOpen is certainly enough (probably $\mathrm{PA}^-$ will suffice with a little care).

The result still holds for bounded formulas in a richer language with extra binary functions or relations. Even more generally, if we expand the language with $k$-ary functions and relations with $k>2$, the result holds with $k+1$ in place of $3$.

The transformation more-or-less preserves quantifier complexity: for any $i\ge1$, an $E_i$ formula in $n$ free variables is equivalent to an $E_i$ formula using $\max\{3,n+1\}$ distinct variables, and similarly for $U_i$. However, we need to make sure the $E_i$ and $U_i$ classes are defined in such a way as to explicitly allow conjunctions and disjunctions; of course, we cannot reduce the number of distinct variables in prenex formulas.

doescollapse. $\endgroup$ – Emil Jeřábek Mar 4 '17 at 8:01