The hierarchy of equivalence relations on the natural numbers under computable reducibility

Let $i_{0},\ldots,i_{n-1}$ be a system of pairwise $E$-inequivalent natural numbers. Then $=_{n}$ is easily seen to be reducible to $E$ by the map $f(k)=i_{k}$ for $k<n$, and $f(k)=i_{n-1}$ for $k\geq n$.

Now suppose that $E$ is $\Pi^{0}_{1}$ and that $E$ has infinitely many classes. Then $=_{\NN}$ is reducible to $E$ by the map which, on input $n$, begins enumerating $E$-incomparable elements. When a system of $n$ pairwise $E$-incomparable elements is found, we map $n$ to the largest one. ∎

Begin by letting $i_{1},\ldots,i_{n}$ be a maximal system of pairwise $E$-inequivalent natural numbers. If $E$ is $\Sigma^{0}_{1}$, then given natural numbers $a,b$ we can decide whether $a\mathrel{E}b$ as follows. Begin enumerating $E$-equivalent pairs until it is discovered that $a\mathrel{E}i_{j_{1}}$ and $b\mathrel{E}i_{j_{2}}$. Then $a\mathrel{E}b$ if and only if $j_{1}=j_{2}$.

On the other hand, if $E$ is $\Pi^{0}_{1}$ then we can enumerate $E$-inequivalent pairs until we find that $a\not\mathrel{E}i_{j}$ for all $j$ other than some $j_{1}$, and $b\not\mathrel{E}i_{j}$ for all $j$ other than some $j_{2}$. Then again, $a\mathrel{E}b$ if and only if $j_{1}=j_{2}$.

Finally, if $E$ is computable then given $a$, a program can order the equivalence classes which occur below $a$ by their least elements, and map $a\mapsto j$ if $a$ is in the $j^{\text{th}}$ class. Note that in the case that $E$ has just $n$ classes, then this is also a reduction to $=_{n}$. ∎

If $f\colon\NN\rightarrow\NN$ is a 1-reduction from $A$ to $B$, then it is easy to see that $f$ is a also a computable reduction from $E_{A}$ to $E_{B}$. On the other hand, suppose that $f$ is a reduction from $E_{A}$ to $E_{B}$. Then $f(A)$ is either contained in $B$, or else it is a singleton. If $f(A)$ were a singleton, say $\{n\}$, then $A=f^{-1}(\{n\})$ would be computable, contradicting our hypothesis. Hence $f(A)\subseteq B$ and $f(\NN\smallsetminus A)\subseteq\NN\smallsetminus B$, and so $f$ is a many-one reduction from $A$ to $B$. Moreover, $f$ is already injective on $\NN\setminus A$.

Now, we will adjust $f$ on $A$ to obtain a $1$-reduction $g$ from $A$ to $B$ as follows. Let $g(0)=f(0)$, and inductively let $g(n+1)=f(n+1)$ so long as $f(n+1)$ is distinct from $g(0),\ldots,g(n)$. On the other hand, if $f(n+1)=g(k)$ for some $k\in 0,\ldots,n$, then we must have that $f(n+1)\in B$. Hence we simply enumerate $B$ in search of a new element $a\in B$ which is distinct from $g(0),\ldots,g(n)$, and let $f(n+1)=a$. Then $g$ is as desired. ∎

Consider the relation $E_{A}$ where $A$ is a simple set (see [Soa87, Theorem V.1.3]). That is, $A$ is a c.e. co-infinite set whose complement contains no infinite c.e. sets. Since $A$ is not computable, neither is $E_{A}$, and it follows that $E_{A}$ is not reducible to $=_{\NN}$. On the other hand, if $f$ is a computable reduction from $=_{\NN}$ to $E_{A}$ then there exists $n\in\NN$ such that $f(\NN\smallsetminus\{n\})\subseteq A^{c}$. But this is a contradiction, since $f(\NN\smallsetminus\{n\})$ is c.e. and infinite. ∎

For any program $e$, let $E_{e}$ be the equivalence relation obtained by taking the transitive closure of whatever relation is enumerated by $e$, together with the diagonal for reflexivity. That is, we interpret the set $W_{e}$ as pairs, and we set $E_{e}$ to be the smallest equivalence relation containing these pairs. Thus, $E_{e}$ is the $e^{\text{th}}$ c.e. equivalence relation, and every c.e. equivalence relation arises this way.

Now define the universal relation by $(e,a)\mathrel{U}_{ce}(e^{\prime},a^{\prime})$ if and only if $e=e^{\prime}$ and $a\mathrel{E_{e}}a^{\prime}$. Thus, we have divided $\NN$ into slices, and put $E_{e}$ on the $e^{\text{th}}$ slice. This relation is c.e., since we may computably enumerate approximations to $U_{ce}$ by running all programs and taking the transitive closure of what has been produced so far. It is universal for c.e. relations since $E_{e}$ reduces to $U_{ce}$ by mapping $a\mapsto(e,a)$.

This construction generalizes to oracles as follows. If $z$ is any oracle, we have the notion $E_{e}^{z}$ and we can define the relation $U_{ce}^{z}$ defined by performing the above construction relative to $z$. The relation $U_{ce}^{z}$ is $z$-c.e., and every $E_{e}^{z}$ computably reduces to $U_{ce}^{z}$ (without need for $z$) by the map $a\mapsto(e,a)$, as before. ∎

Clearly if $E$ is the orbit equivalence relation induced by a computable action, then it is c.e.  Conversely, suppose that $E$ is a c.e. relation, enumerated by a program $e$. Let $\Gamma$ be a fixed computable copy of the free group $F_{\omega}$ with generators $x_{1},x_{2},\ldots$. If $e$ enumerates a pair $(n,n^{\prime})$ into $E$ at stage $s$, then we let $i$ be a code for the triple $(s,n,n^{\prime})$ and let $x_{i}$ act by swapping $n$ and $n^{\prime}$ and leaving the other generators fixed. Clearly the orbit relation arising from this action is precisely $E$. Moreover the action is computable, since each generator codes a bound which tells how long to run $e$ to find out which elements it swaps. ∎

To see that $=^{ce}$ is in $\Pi^{0}_{2}$, note that $W_{e}=W_{e^{\prime}}$ if and only if for all numbers $n$ enumerated into $W_{e}$, there is a stage by which $n$ is also enumerated into the other $W_{e^{\prime}}$ (and vice versa). To show that it is $\Pi^{0}_{2}$ complete, we use the fact that the set $\mathrm{TOT}$ of programs which halt on all inputs is $\Pi^{0}_{2}$ complete (see [Soa87, Theorem IV.3.2]). The set $\mathrm{TOT}$ is $m$-reducible to the $=^{ce}$-equivalence class of $W_{e_{0}}=\NN$ by the following function: $f(e)$ is the program which outputs $n$ just in case $e$ halts on input $n$. ∎

Let $E$ be an arbitrary c.e. relation. Since each equivalence class $[n]_{E}$ of $E$ is c.e., we can define a reduction from $E$ to $=^{ce}$ by mapping each $n$ to a program $f(n)$ which enumerates $[n]_{E}$. On the other hand, there can’t be a reduction from $=^{ce}$ to $E$ since $=^{ce}$ is $\Pi^{0}_{2}$ complete and $E$ is c.e. ∎

It is not difficult to build a reduction from $=^{ce}$ to $E_{0}^{ce}$. For instance, given a program $e$, we can define the program $f(e)$ as follows. Whenever $n$ is enumerated into $W_{e}$, the program $f(e)$ enumerates codes for the pairs $(n,0),(n,1),(n,2),\ldots$ (or more accurately, it arranges to periodically add more and more of these). Then clearly we have that $W_{e}$ and $W_{e^{\prime}}$ differ if and only if $W_{f(e)}$ and $W_{f(e^{\prime})}$ differ infinitely often.

On the other hand, there cannot be a computable reduction from $E_{0}^{ce}$ to $=^{ce}$, since $E_{0}^{ce}$ is $\Sigma^{0}_{3}$ complete while $=^{ce}$ is just $\Pi^{0}_{2}$. To see that $E_{0}$ is $\Sigma^{0}_{3}$ complete, note that by [Soa87, Corollary IV.3.5], even its equivalence class $\mathrm{COF}=\left\{\,e\mid W_{e}\textrm{ is cofinite}\,\right\}$ is $\Sigma^{0}_{3}$ complete. ∎

The reductions are just the same as the classical ones from the Borel theory. We will quickly give the reductions so that the reader may verify they are computable in our sense. To begin, notice that $E_{0}$ is reducible to $E_{3}$ by the map: $A\mapsto\NN\times A$, where each column of the image of $A$ looks exactly like $A$ itself. To reduce $E_{0}$ to $E_{1}$, map $A$ to $\left\{\,\langle x,y\rangle\mid y\geq x~\&~y\in A\,\right\}$, so that the column $x$ equals $A-\{0,\ldots,x-1\}$. For $E_{0}\leq E_{2}$, we let $I_{n}$ be any fixed c.e. partition of $\NN$ into sets such that $\sum_{i\in I_{n}}1/i\geq 1$. Then it is easy to see that $E_{0}\leq E_{2}$ via the map $A\mapsto\bigcup_{n\in A}I_{n}$.

To show that $E_{3}\leq Z_{0}$, we first observe that $E_{0}$ reduces to $Z_{0}$ via the map $f(A)=\bigcup_{n\in A}[2^{n},2^{n+1})$. (Indeed, if $A\mathrel{E_{0}}B$ then $f(A)\mathrel{\triangle}f(B)$ is finite and hence has density zero. Conversely, if $A\mathrel{E_{0}}B$ is infinite, then $(f(A)\mathrel{\triangle}f(B))\cap N$ will be $\geq 1/2$ infinitely often, and hence $f(A)\mathrel{\triangle}f(B)$ does not have density zero.) Now, fix a partition $I_{n}$ of $\NN$ into sets such that the density of $I_{n}$ is exactly $1/2^{n}$, and let $\pi_{n}$ denote the unique isomorphism $\NN\cong I_{n}$. Then we can reduce $E_{3}$ to $Z_{0}$ using the map $(A_{n})\mapsto\bigcup\pi_{n}(f(A_{n}))$.

Finally, to reduce $E_{3}$ to $E_{\sf set}$, suppose we are given a sequence $(A_{n})$. For each column $A_{n}$ and each $s\in 2^{<\NN}$, we will place the set $1^{n}\raise 1.0pt\hbox{${}^{\frown}$}0\raise 1.0pt\hbox{${}^{\frown}$}s\raise 1.0pt\hbox{${}^{\frown}$}(A\smallsetminus\left|s\right|)$ as a column of $f((A_{n}))$. It is not difficult to check that this mapping is as desired. ∎

We use a direct arithmetic complexity argument. Observe that $E_{0}^{ce}$, $E_{1}^{ce}$, and $E_{2}^{ce}$ are all easily seen to be $\Sigma^{0}_{3}$. On the other hand, we shall show that $E_{3}^{ce}$ is not $\Sigma^{0}_{3}$. In fact, we shall show that $E_{3}^{ce}$ has one equivalence class which is $\Pi^{0}_{3}$ complete, namely, the set $F$ consisting of indices $e$ for c.e. subsets of $\NN\times\NN$ such that every column of $W_{e}$ is finite. (As a matter of fact, $E_{3}^{ce}$ is $\Pi^{0}_{4}$-complete, but we will not prove this here.)

To begin, consider any set of the form $A=\left\{\,x\mid(\forall n)(\exists N)(\forall k)\phi(x,n,N,k)\,\right\}$, where $\phi$ is computable. For each $x$, we shall write down a program $e_{x}$ which enumerates a c.e. subset of $\NN\times\NN$ such that $x\in A$ if and only if $e_{x}\in F$. For each $n$, the program $e_{x}$ considers each $N$ in turn, and checks whether for all $k$, $\phi(x,n,N,k)$ holds. Whenever we find a counterexample $k$ for the current $N$, we enumerate one more element into the $n^{\text{th}}$ column of $W_{e_{x}}$. Observe that if $x\in A$, then for every $n$, we will eventually find the $N$ that works, and so the $n^{\text{th}}$ column of $W_{e_{x}}$ will be finite. Hence in this case $e_{x}\in F$. If $x\notin A$, then for some $n$, we will have to change our mind about $N$ infinitely often, and so the $n^{\text{th}}$ column of $W_{e_{x}}$ will be all of $\NN$. Hence in this case $e_{x}\notin F$. ∎

On inputs $i$ and $j$, let the output $f(i,j)$ be (a Gödel code for) the program which enumerates the set $W_{f(i,j)}$ we now describe. For each element $\langle c,k\rangle$ and each stage $s+1$, enumerate $\langle c,k\rangle$ into $W_{f(i,j),s+1}$ if the following hold:

• $\circ$

  $\langle c,k\rangle\in W_{i,s}\smallsetminus W_{j,s}$; and

• $\circ$

  $(\forall n<k)~[\langle c,n\rangle\in W_{i,s}\iff\langle c,n\rangle\in W_{j,s}]$.

Additionally, if $\langle c,k\rangle\in W_{f(j,i),s}\cap W_{i,s}\cap W_{j,s}$, then enumerate $\langle c,k\rangle$ into $W_{f(i,j),s+1}$. This is the entire construction.

Now fix $c$, and suppose that $W_{i}$ and $W_{j}$ agree on their $c$${}^{\text{th}}$ columns: $(\forall k)[\langle c,k\rangle\in W_{i}\iff\langle c,k\rangle\in W_{j}]$. Then, if $\langle c,k\rangle$ ever entered $W_{f(i,j)}$, it did so either because it had already entered $W_{f(j,i)}$, or else because $\langle c,k\rangle\in W_{i,s}\smallsetminus W_{j,s}$, in which case it must later have entered $W_{j}$ as well (by their agreement on the $c$${}^{\text{th}}$ column), and therefore was then enumerated into $W_{f(j,i)}$ as well. Thus $W_{f(i,j)}$ and $W_{f(j,i)}$ agree with each other on their $c$${}^{\text{th}}$ columns.

On the other hand, suppose $W_{i}$ and $W_{j}$ disagree on their $c$${}^{\text{th}}$ columns, and let $\langle c,k\rangle$ be the least point lying in just one of them. If $\langle c,k\rangle\in W_{i}\smallsetminus W_{j}$, then by its minimality, it will eventually be enumerated into $W_{f(i,j),s+1}$ at some stage $s+1$. It will never appear in $W_{f(j,i)}$. Moreover, the only numbers $k^{\prime}$ such that $\langle c,k^{\prime}\rangle$ can be enumerated into either $W_{f(i,j)}$ or $W_{f(j,i)}$ at stages $>s+1$ are those which either sat in $W_{f(j,i),s+1}$ or $W_{f(i,j),s+1}$ already, or else have $k^{\prime}<k$, because once $\langle c,k\rangle$ has appeared in $W_{i,s+1}$, no $\langle c,k^{\prime}\rangle$ with $k^{\prime}>k$ could ever again be the least difference between the $c$${}^{\text{th}}$ columns of $W_{i}$ and $W_{j}$. Therefore, $W_{f(i,j)}$ and $W_{f(j,i)}$ do differ on at least one element of their $c$${}^{\text{th}}$ columns, but do not differ by more than finitely many elements. (In fact, each of these columns is finite.)

A symmetric argument holds when the least difference lies in $(W_{j}-W_{i})$. Therefore, if $W_{i}~E_{1}~W_{j}$, then each of the finitely many columns on which they differ contains only finitely many differences between $W_{f(i,j)}$ and $W_{f(j,i)}$, and none of the other columns contain any differences at all between them. Thus $W_{f(i,j)}~E_{0}~W_{f(j,i)}$. Conversely, if $W_{i}$ and $W_{j}$ differ on infinitely many columns, then every one of those columns contains at least one difference between $W_{f(i,j)}$ and $W_{f(j,i)}$ as well, and so $W_{f(i,j)}$ and $W_{f(j,i)}$ are not $E_{0}$-related. Thus we have proven Proposition 3.9. ∎

We define a computable function $g$ implicitly by building the c.e. sets $W_{g(i)}$ uniformly for $i\in\omega$. Here we give some intuition for the construction, before presenting it in full. The slice $C^{cj}$ is dedicated to making sure that, if the set $W_{j}$ differs on column $c$ from all the sets $W_{i}$ with $i<j$, then there should be at least one difference between $W_{g(j)}$ and each $W_{g(i)}$. At each stage $s+1$ in the construction, we will consider all $c,j\in\omega$ and all $i<j$, and define two auxiliary elements: $m^{cij}_{s}$, which represents the least element of the $c$${}^{\text{th}}$ column $\omega^{c}$ which lies in the symmetric difference of $W_{i,s}$ and $W_{j,s}$, and $x^{cj}_{s}\in C^{cj}$, the element we are currently using to distinguish $W_{g(i)}$ from $W_{g(j)}$ on the slice $C^{cj}$. The former will converge to the least element of $(W_{i}\mathbin{\triangle}W_{j})\cap\omega^{c}$, or diverge if this set is empty. Each time it changes, the current $x^{cj}_{s}$ enters $W_{g(i)}\cap W_{g(j)}$ and a new $x^{cj}_{s+1}$ is chosen and added to $W_{g(j)}$. The same will happen occasionally on behalf of some $k>j$, as described below, but in the end, $W_{g(i)}$ and $W_{g(j)}$ will agree on the slice $C^{cj}$ if and only if $W_{i}$ and $W_{j}$ agree on the column $\omega^{c}$.

We also must ensure that $(W_{g(i)}\mathbin{\triangle}W_{g(j)})\cap C^{cj}$ is finite, since $W_{j}$ and some $W_{i}$ might be equal on all other columns (hence might be $E_{1}$-related). The $C^{cj}$ slice is not intended to differentiate any $W_{i}$ and $W_{i^{\prime}}$ with $i^{\prime}<i<j$; indeed, we want to have $W_{g(i)}\cap C^{cj}=W_{g(i^{\prime})}\cap C^{cj}$ for all such $i$ and $i^{\prime}$, leaving it to $C^{ci}$ to differentiate between them if necessary. Moreover, if any $i<j$ has $W_{i}^{c}=W_{j}^{c}$, then we will let $C^{ci}$ do the job, and $W_{g(j)}\cap C^{cj}$ will be equal to $W_{g(k)}\cap C^{cj}$ for every $k$, so that we do not repeat the process of adding differences between the same sets.

A set $W_{k}$ with $k>j$ might equal $W_{j}$ on the $c$${}^{\text{th}}$ column, or might equal some $W_{i}$ with $i<j$ there. In the former case we would want to build $W_{g(k)}$ to be equal on $C^{cj}$ to $W_{g(j)}$, but in the latter case we would want it equal to $W_{g(i)}$ instead – which could pose difficulties if $W_{g(j)}\neq W_{g(i)}$ on $C^{cj}$. Our solution is to use the least difference (if any) between $W_{i}^{c}$ and $W_{j}^{c}$ to guess which is the case. If this minimum element $m^{cij}=\lim_{s}m^{cij}_{s}$ lies in both $W_{j}$ and $W_{k}$, for instance, then we can be sure that $W_{k}^{c}\neq W_{i}^{c}$, and if similar outcomes hold for each minimum (for every $i<j$), then we will make $W_{g(k)}$ look like $W_{g(j)}$ on $C^{cj}$. On the other hand, if there is some minimum element which shows that $W_{k}^{c}\neq W_{j}^{c}$, then on this slice we will make $W_{g(k)}$ look like the sets $W_{g(i)}$ (which are the same on this slice for all $i<j$). We do this for only finitely many $k>j$, namely those $\leq c$, to avoid having to add infinitely many elements to the sets $W_{g(j)}$ and $W_{g(i)}$, for that could wipe out the difference we would like to build between those sets. For each single $k$, there are only finitely many values $j$ and columns $c$ for which this restriction will stop $W_{k}$ from being considered in the construction of $W_{g(k)}$ on $C^{cj}$; whenever either $j\geq k$ or $c>k$, $W_{k}$ will be used in that construction. On the finitely many slices $C^{cj}$ in which it was not considered, $W_{g(k)}$ will have only a finite difference from any other set $W_{g(n)}$ anyway, which will have no effect on the question of whether $W_{g(k)}~E_{0}~W_{g(n)}$.

For the slice $C^{cj}$, there are two basic outcomes possible. First, suppose some $i<j$ satisfies $W_{i}^{c}=W_{j}^{c}$. Then there will be infinitely many stages $s$ at which $m^{cij}_{s+1}$ is either $\neq m^{cij}_{s}$ or undefined (if the symmetric difference is empty at that stage). At every one of these stages, the current $x^{cj}_{s}$, which already sat in $W_{g(j)}$, will be added to every set $W_{g(k)}$ with $k\neq j$, and the newly chosen $x^{cj}_{s+1}$ will be added to $W_{g(j)}$. Since this happens infinitely often, we wind up with $C^{cj}\subseteq W_{g(k)}$ for every $k\in\omega$.

On the other hand, suppose every $i<j$ has $W_{i}^{c}\neq W_{j}^{c}$. Then every sequence $\langle m^{cij}_{s}\rangle_{s\in\omega}$ will converge to a limit $m^{cij}$, the minimum of $W_{i}^{c}\mathbin{\triangle}W_{j}^{c}$. Once we reach a stage $s_{0}$ at which all $W_{i,s_{0}}\mathord{\upharpoonright}_{m^{cij}+1}=W_{i}\mathord{\upharpoonright}_{m^{cij}+1}$ and also $W_{j,s_{0}}\mathord{\upharpoonright}_{m^{cij}+1}=W_{j}\mathord{\upharpoonright}_{m^{cij}+1}$ for all $i$, the values $m^{cij}_{s}$ will never change again, and therefore will never cause $x^{cj}_{s_{0}}$ to enter any $W_{g(i)}$. The indices $k$ with $j<k\leq m$ (if there are any) may yet cause this element to enter the sets $W_{g(i)}$. There are only finitely many such $k$, however, and each one causes this to happen at no more than one stage after $s_{0}$ (namely, the unique stage $s+1$, if one exists, such that $W_{j,s}^{c}$ and $W_{k,s}^{c}$ agree on the set of minima $\left\{\,m^{cij}\mid i<j\,\right\}$, but $W_{j,s+1}^{c}$ and $W_{k,s+1}^{c}$ fail to agree on that set). Therefore, in this second outcome, there will exist a limit $x^{cj}=\lim_{s}x^{cj}_{s}$, which will lie in $W_{g(j)}$, will not lie in any $W_{g(i)}$ with $i<j$, and will lie in $W_{g(k)}$ (for $j<k\leq m$) if and only if $W_{k}$ and $W_{j}$ contain exactly the same elements from the set of minima. Moreover, this $x^{cj}$ will be the only element of $C^{cj}$ on which $W_{g(j)}$ differs from any set $W_{g(k)}$. So in this case we have accomplished the goal of establishing a single difference on the slice $C^{cj}$ between $W_{g(i)}$ and $W_{g(j)}$ for each $i<j$, while not differentiating $W_{g(i)}$ from $W_{g(i^{\prime})}$ on this slice for any $i^{\prime}<i<j$. The point of our treatment of the sets $W_{g(k)}$ with $j<k\leq m$ was discussed above, and will appear below in Lemma 3.13. (The sets $W_{g(k)}$ with $k>m$ agree on $C^{cj}$ with all $W_{g(i)}$ for $i<j$, and hence differ by at most the element $x^{cj}$ from $W_{g(j)}$.)

Now we give the formal construction. At stage $0$, every element $m^{cij}_{0}$ is undefined, and $x^{cj}_{0}$ is the least element of the slice $C^{cj}$. We also enumerate $x^{cj}_{0}$ into the set $W_{g(j),0}$ (which as yet contains nothing other than this element).

At stage $s+1$, for each fixed $j$ and $c$ and for every $i<j$, we find the least element $m^{cij}_{s+1}$ (if any) of the symmetric difference $(W_{i,s}^{c}\mathbin{\triangle}W_{j,s}^{c})$. Suppose first that there exists either an $i<j$ such that this symmetric difference is empty, or an $i<j$ such that $m^{cij}_{s+1}\neq m^{cij}_{s}$. Then we enumerate the $x^{cj}_{s}$ into every slice $W_{g(k),s+1}$, (we shall see that it was already in $W_{g(j),s}$), define $x^{cj}_{s+1}$ to be the next-smallest element of $C^{cj}$, and enumerate this $x^{cj}_{s+1}$ into $W_{g(j),s+1}$.